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HIGHER NATIONAL DIPLOMA

IN

INFORMATION TECHNOLOGY

STUDY PACK

OPERATIONS RESEARCH(Version: January 2004)

COPYRIGHTS RESERVED

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HIGHER NATIONAL DIPLOMA IN INFORMATION TECHNOLOGY

SUBJECT: OPERATIONS RESEARCH

CODE: 710 / 04 /S02

AIM OF THE SUBJECT:

1. To provide various mathematical tools for analysis of systems in the Business environment.

2. To provide techniques which would be needed in analysis of Business Systems to managerial decision-making.

3. To provide tools to solve complex business problems.

4. To represent the concepts of “efficiency” and “scarcity in well defined mathematical model of a given situation.

5. To provide Science Techniques of the derivation of computational methods for solving such models.

DESIGN LENGTH:

THEORY 190: HOURSLABORATORY 10: HOURSTOTAL 200: HOURS

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CODE: 710 / 04 /S02

UNIT 1 APPROXIMATIONS

HOURS: 20

OBJECTIVE

At the end of the unit the student should be able to:

Apply given Technologies in estimating solutions in Business environment.

1.1 Newton-Raphson iteration method for solving polynomial equations.

1.2 Trapezium Rule for approximating a definite integral.

1.3 Simpson Rule for approximating a definite integral

1.4 Maclaurin Series expansion.

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CODE: 710 / 04 /S02

UNIT 2 LINEAR PROGRAMMING

HOURS: 20

OBJECTIVE

To formulate and solve linear programming models using the Graphical and other techniques.

2.1 Model formulation.

2.2 Solution of L.P model by graphical and simplex method.

2.3 Duality and the Dual Simplex method.

2.4 Sensitivity Analysis using the Simplex method.

2.5 Assignment problem.

2.6 Transportation problem.

UNIT 3: NON-LINEAR FUNCTIONS:

HOURS: 20.

OBJECTIVE

To develop an optimization theory using differential calculus to determine maximum and minimum points for Non Linear Functions.

3.1 Overview of differentiation and integration.

3.2 Continuous functions and partial derivatives.

3.3 Necessary and sufficient conditions for extreme points.

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3.3.1 The gradient vector

3.3.2 The Hessian matrix



CODE: 710 / 04 /S02

UNIT 4 PROJECT MANAGEMENT WITH PERT/CPM

HOURS: 20

OBJECTIVETo select a method which is appropriate in a given situation to find the shortest route between two given nodes in a network and determine the route and its length.

4.1 Introduction.

4.2 The project network diagram.

4.3 Critical Path Analysis.

4.4 Determination of the critical path.

4.5 Project activity crushing.

4.6 Probabilistic time duration of activities.

UNIT 5: RANDOM VARIABLES AND THEIR PROBABILITY:

HOURS: 20.

OBJECTIVETo appreciate the properties of a probability function and be able to apply it in Business environment.

5.1 Introduction 5.2 Probability density functions (pdf)

5.3 Discrete random variables.

5.4 Continuous random variables.

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5.5 Cumulative density functions (CDF)

5.6 Relations among probability distributions

5.7 Joint probability distributions with two variables



CODE: 710 / 04 /S02

UNIT 6 DECISION THEORY

HOURS: 20

OBJECTIVE

Apply the concept of optimistic approach, conservative approach, and minimax regret approach to decision-making problems.

6.1 Decisions under risk

6.1.1 Expected value criterion

6.1.2 Expected profit with perfect information

6.1.3 Minimizing expected loss.

6.1.4 Expected value of perfect information

6.1.5 Decision trees in multistage decision-making

6.2 Decisions under uncertainty

6.2.1 Laplace criterion

6.2.2 Minimax/Maximin criterion

6.2.3 Savage Minimax Regret criterion

6.2.4 Hurwicz criterion

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UNIT 7: THEORY OF GAMES:

HOURS: 20

OBJECTIVES:

Use analytical criteria for decisions under uncertainty with the assumptions that “nature” is the opponent.

7.1 Pure strategy games

7.1.1 Two-Person Zero Sum games

7.1.2 Saddle point solutions

7.2 Mixed strategy games

7.2.1 Solution of games by Linear Programming.

7.2.2 Graphical solution of (2*n) and (m*2) games.

7.2.3 Solution of (m*n) games by the Simplex methods.

UNIT 8: INVENTORY MODELLING:

HOURS: 20

OBJECTIVES:

To determine the reorder point and cost of inventory for a given model.

8.1 Generalized inventory model

8.2 Deterministic Models.

8.3 The Economic Order Quantity (EOQ) model

8.4 Inventory with backordering

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8.5 Economic Production Quantity model

8.6 Probabilistic models

8.7 Continuous Review model.

UNIT 8: REGRESSION ANALYSIS AND CORRELATION:

HOURS: 20

OBJECTIVES:

To identify the relationships between variables and forecast future trends.

9.1 Simple Linear Regression Analysis.

9.1.1 Scatter plots and Line of best fit.

9.1.2 Least Squares equation regression model.

9.1.3 Forecasting using Regression techniques.

9.1.4 Making inferences about parameters a and b.

9.2 Correlation Coefficient.

9.2.1 Coefficient of determination.

9.2.2 Interpret the value of the correlation coefficient.

UNIT 10 INTRODUCTION TO QUEUEING THEORY

HOURS: 20

OBJECTIVE:

To identity a Queuing problem and calculate parameters related to the queue.

10.1 Queuing processes of Single Server models.

10.2 Definitions and Notations.

10.3 Relationships between

10.3.1 Expected Waiting Time per Customer in the system.

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10.3.2 Expected Waiting Time per Customer in the queue.

10.3.3 Expected Number of Customer in the system.

10.3.4 Expected Number of Customer in the Queue.

UNIT 2 LINEAR PROGRAMMING

HOURS: 20

LINEAR PROGRAMMING

Is a technique used to determine how best to allocate personnel, equipment, materials, finance, land, transport e.t.c. , So that profit are maximized or cost are minimized or other optimization criterion is achieved.Linear programming is so called because all equations involved are linear. The variables in the problem are Constraints. It is these constraints, which gives rise to linear equations or Inequalities.The expression to the optimized is called the Objective function usually represented by an equation.

Question 1:A furniture factory makes two products: Chairs and tables. The products pass through 3 manufacturing stages; Woodworking, Assembly and Finishing.The Woodworking shop can make 12 chairs an hour or 6 tables an hour.The Assembly shops can assembly 8 chairs an hour or 10 tables an hour.The Finishing shop can finish 9 chairs or 7 tables an hour.The workshop operates for 8 hours per day. If the contribution to profit from each Chair is $4 and from each table is $5, determine by Graphical method the number of tables and chairs that should be produced per day to maximize profits.

Solution:Let number of chairs be X.Let number of tables be Y.

Objective Function is:P = 4X + 5Y.

Constraints:WW: X/12 + Y/6 <= 8

X + 2Y <= 96 when X=0 Y= 48 when Y=0 X= 96

AW: X/8 + Y/10 <= 85X + 4Y <= 320 when X=0 Y=80 when Y=0 X=64

FNW: X/9 + Y/7 <= 87X + 9Y <= 504 when X=0 Y=56 when Y=0 X=72

X>=0 Y>=0

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100

90

80

70

60

50

40

30

20 P------ this gives the maximum point

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010 20 30 40 50 60 70 80 90 100

Question 2:Mr. Chabata is a manager of an office in Guruwe; he decides to buy some new desk and chairs for his staff.He decides that he need at least 5 desk and at least 10 chairs and does not wish to have more than 25 items of furniture altogether. Each desk will cost him $120 and each chair will cost him $80. He has a maximum of $2400 to spend altogether.Using the graphical method, obtain the maximum number of chairs and desk Mr. Chabata can buy.

Solution:Let X represents number of Desk.Let Y represents number of Chairs.


Constraints:X >= 5Y>= 10

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X + Y <= 25

30

25

20PPoint P (10,15) gives the maximum point

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10

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05 10 15 20 25 30 35 40

The Optimum Solution = 120 * 10 + 80 * 15 = 1200 + 1200 = 2400

Question 3:A manufacturer produces two products Salt and Sugar. Salt has a contribution of $30 per unit and Sugar has $40 per unit. The manufacturer wishes to establish the weekly production, which maximize the contribution. The production data are shown below:

Production UnitMachine Hours Labour Hours Materials in Kg

Salt 4 4 1Sugar 2 6 1Total available per unit 100 180 40

Because of the trade agreement sales of Salt are limited to a weekly maximum of 20 units and to honor an agreement with an old established customer, at least 10 units of Sugar must be sold per week.

Solution:Let X represents Salt.Let Y represents Sugar.


Constraints:4X + 2Y <= 100 {machine hours}

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4X + 6Y <= 180 {Labour hours}X + Y <= 40 {material}Y>= 10X<= 20X>=0; Y>=0;

SOLVING LINEAR PROGRAMMING USING SIMPLEX METHOD:

The graphical outlined above can only be applied to problems containing 2 variables. When 3 or more variables are involved we use the Simplex method. Simplex comprises of series of algebraic procedures performed to determine the optimum solution.In Simplex method we first convert inequalities to equations by introducing a Slack variable.A Slack variable represents a spare capacity in the limitation.

STEPS TO FOLLOW IN SIMPLEX METHOD:

i. Obtain the pivot column as the column with the most positive indicator row.ii. Obtain pivot row by dividing elements in the solution column by their corresponding

pivot column entries to get the smallest ratio. Element at the intersection of the pivot column and pivot row is known as pivot elements.

iii. Calculate the new pivot row entries by dividing pivot row by pivot element. This new row is entered in new tableau and labeled with variables of new pivot column.

iv. Transfer other row into the new tableau by adding suitable multiplies of the pivot row (as it appears in the new tableau) to the rows so that the remaining entries in the pivot column becomes zeroes.

v. Determine whether or not this solution is optimum by checking the indicator row entries of the newly completed tableau to see whether or not they are any positive entries. If they are positive numbers in the indicator row, repeat the procedure as from step 1.

vi. If they are no positive numbers in the indicator row, this tableau represents an optimum solution asked for, the values of the variables together with the objective function could then be stated.

Question 1:Maximize Z = 40X + 32YSubject to:

40X +20Y <= 6004X + 10Y <= 1002X + 3Y <= 38 Using the Simplex method

Solution:To obtain the initial tableau, we rewrite the Objective Function as:

Z = 40X + 32Y

Introducing Slack variables S1, S2, S3 in the 3 inequalities above we get:

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40X + 20Y + S1 = 6004X + 10Y + S2 = 1002X + 3Y + S3 = 38X >= 0Y >=0.

TABLEAU 1:Pivot Element

X Y S1 S2 S3 Solution

S1 20 1 0 0 600

S2 4 10 0 1 0 100

S3 2 3 0 0 1 38

Z 40 32 0 0 0 0 Indicator Row

Pivot Column

TABLEAU 2:


X 0.5 0.025 0 0 15

S2 0 8 -0.1 1 0 40

S3 0 2 -0.05 0 1 8

Z 0 12 -1 0 0 -600

TABLEAU 3:


X 1 0 0.0375 0 -0.25 13

S2 0 0 0.1 1 -4 8

Y 0 0.025 0 0.5 4

Z 0 0 -0.7 0 -6 -648 Indicator Row

Pivot Column13

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Conclusion:Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X = 13 and Y = 4 giving Z = 648.

NB:a) If when selecting a pivot column we have ties in the indicator row, we then

select the pivot column arbitrary.b) If all the entries in the selected pivot column are negative then the objective

function is unbound and the maximum problem has no solution.c) A minimization problem can be worked as a maximization problem after

multiply the objective function and the inequalities by –1.d) Inequalities change their signs when multiplied by negative number.

Question 2:Maximize Z = 5X1 + 4X2

Subject to:2X1 +3X2 <= 17X1 + X2 <= 73X1 + 2X2 <= 18 Using the Simplex method

Solution:Max Z = 5X1 + 4X2

Subject to:

2X1 + 3X2 + S1 = 17X1 + X2 + S2 = 73X1 + 2X2 + S3 = 18X1 >= 0X2>=0.

TABLEAU 1:

X1 X2 S1 S2 S3 Solution

S1 2 3 1 0 0 17

S2 1 1 0 1 0 7

S3 2 0 0 1 18

Z 5 4 0 0 0 0 Indicator Row

Pivot Column

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TABLEAU 2:


S1 0 5/3 1 0 -2/3 5

S2 0 1/3 0 1 -1/3 1

X1 2/3 0 0 1/3 6

Z 0 2/3 0 0 -5/3 -30

TABLEAU 3:


X2 0 3/5 0 -2/5 3

S2 0 0 -1/5 1 -1/5 0

X1 1 0 -2/5 0 9/15 4

Z 0 0 -2/5 0 -7/5 -32

Pivot Column

Conclusion:Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X1= 4 and X2 = 3 giving Z = 32.

Question 3:A company can produce 3 products A, B, C. The products yield a contribution of $8, $5 and $10 respectively. The products use a machine, which has 400 hours capacity in the next period. Each unit of the products uses 2, 3 and 1 hour respectively of the machine’s capacity.There are only 150 units available in the period of a special component, which is used singly in products A and C.200 kgs only of a special Alloy is available in the period. Product A uses 2 kgs per unit and Product C uses 4kgs per units. There is an agreement with a trade association to produce no more than 50 units of product in the period.The Company wishes to find out the production plan which maximized contribution.

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Solution:Maximize Z = 8X1 + 5X2+ 10X3

Subject to:

2X1 + 3X2 + X3 <= 400 {machine hour}X1 + X3 <= 150 {component}2X1 + 4X3 <= 200 {Alloy}

X2 <=50 {Sales}X1 >= 0X2 >= 0X3 >= 0.

Introducing slack variables:Maximize Z = 8X1 + 5X2+ 10X3

Subject to:

2X1 + 3X2 + X3 + S1= 400X1 + X3 + S2 = 1502X1 + 4X3 + S3 = 200

X2 + S4 =50X1 >= 0X2 >= 0X3 >= 0.

TABLEAU 1:

X1 X2 X3 S1 S2 S3 S4 Solution

S1 2 3 1 1 0 0 0 400

S2 1 0 1 0 1 0 0 150

S3 2 0 0 0 1 0 200

S4 0 1 0 0 0 0 1 50

Z 8 5 10 0 0 0 0 0 Indicator Row

Pivot ColumnNB: Ignore S4 in finding pivot row.

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TABLEAU 2:


S1 3/2 3 0 1 0 -1/4 0 350

S2 1/2 0 0 0 1 -1/4 0 100

X3 1/2 0 0 0 1/4 0 50

S4 0 1 0 0 0 0 1 50

Z 3 5 0 0 0 -5/2 0 -500

Pivot Column

TABLEAU 3:


S1 3/2 0 0 1 0 -1/4 -3 200

S2 1/2 0 0 0 1 -1/4 0 100

X3 1/2 0 4 0 0 1/4 0 50

X2 0 0 0 0 0 1 50

Z 3 0 0 0 0 -5/2 -5 -750

Pivot ColumnTABLEAU 4:


S1 0 0 -3 1 0 -1 -3 50

S2 0 0 -1 0 1 -1/2 0 50

X1 0 2 0 0 1/2 0 100

X2 0 1 0 0 0 0 1 50

Z 0 0 -6 0 0 -4 -5 -1050

Pivot Column

Conclusion:

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Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X1= 100 and X2 = 50 giving Z = 1050. Two slack variable S1 =0 and S2 =0. This means that there is no value to be gained by altering the machine hours and component constraints.

GENERAL RULE :

Constraints only have a valuation when they are fully utilized.These valuations are known as the SHADOW Prices or Shadow Costs or Dual Prices or Simplex Multipliers.A constraint only has a Shadow price when it is binding i.e. fully utilized and the Objective function would be increased if the constraint were increased by 1 unit.When solving Linear Programming problems by Graphical means the Shadow price have to be calculated separately. When using Simplex method they are an automatic by product.

MIXED CONSTRAINTS:

This involves constraints containing a mixture of <= and >= varieties. Using Maximization problem we use “Less than or equal to” type. (<=).Faced with a problem which involves a mixture of <= and >= variety. The alternative solution to deal with “Greater than or equal to” (>=) type is to multiply both sides by –1 and change the inequality sign.

Question 4:Maximize Z = 5X1 + 3X2+ 4X3

Subject to:

3X1 + 12X2 + 6X3 <= 6606X1 + 6X2 + 3X3 <= 12306X1 + 9X2 + 9X3 <= 900

X3 >=10Solution:The only constraint that need to be changed is X3 >=10 by multiply by –1 both sides and we get:

-X3 <= -10

Maximize Z = 5X1 + 3X2+ 4X3

Subject to:

3X1 + 12X2 + 6X3 + S1 = 6606X1 + 6X2 + 3X3 + S2 = 12306X1 + 9X2 + 9X3 <+ S3 = 900

-X3+ S4 = -10

TABLEAU 1:

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X1 X2 X3 S1 S2 S3 Solution

S1 3 12 6 1 0 0 600

S2 6 6 3 0 1 0 1200

S3 9 9 0 0 1 900

Z 5 3 4 0 0 0 0 Indicator Row

Pivot Column

FINAL TABLEAU:

X1 X2 X3 S1 S2 S3 Solution

S1 0 15/2 3/2 1 0 -1/2 150

S2 0 -3 -6 0 1 -1 300

X1 3/2 3/2 0 0 1/6 150

Z 0 -9/2 -7/2 0 0 -5/6 -750

Pivot Column

Conclusion:Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X1= 150 producing Z = $750. Plus production to satisfy constrain (d) 20 units of X3

producing $ 40 contribution.Therefore Total solution is 150 units of X1 and 10 units of X3 giving $790. NB: Maximize Z = 5(150) + 3(0)+ 4(10)

= $790.

Question 5:Maximize Z = 3X1 + 4X2

Subject to:

4X1 + 2X2 <= 1004X1 + 6X2 <= 180X1 + X2 <= 40

X1 <= 20X2 >=10

Solution:The only constraint that need to be changed is X2 >=10 by multiply by –1 both sides and we get: -X2 <= -10Maximize Z = 3X1 + 4X2

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Subject to:4X1 + 2X2 + S1 = 100 {1}4X1 + 6X2 + S2 = 180 {2}X1 + X2 + S3 = 40 {3}

X1 + S4 = 20 {4}-X2 + S5 =-10 {5}

TABLEAU 1:X1 X2 S1 S2 S3 S4 S5 Solution

S1 4 1 0 0 0 0 100

S2 4 6 0 1 0 0 0 180

S3 1 1 0 0 1 0 0 40

S4 1 0 0 0 0 1 0 20

S5 0 -1 0 0 0 0 1 -10

Z 3 4 0 0 0 0 0 0 Indicator Row

Pivot ColumnThe problem is then solved by the usual Simplex iterations. Each iteration improves on the one before and the process continues until optimum is reached.


X2 0 0 0 0 0 -1 10

S2 4 0 1 0 0 0 2 80

S3 4 0 0 1 0 0 6 120

S4 1 0 0 0 1 0 1 30

S5 1 0 0 0 0 1 0 20

Z 4 0 0 0 0 0 4 -40

This shows 10X2 being produced and $40 contribution. The first four constraints have surpluses of 80, 120, 30 and 20 respectively. Not optimums as there are still positive values in Z row.


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X2 0.667 1 0 0.167 0 0 0 30

S2 2.667 0 1 -0.333 0 0 0 40

S3 -0.333 0 0 -0.167 0 0 0 10

S4 1 0 0 0 1 1 0 20

S5 0.333 0 0 0.167 0 0 20

Z 0.333 0 0 -0.667 0 0 0 -120

This shows 30X2 being produced and $120 contribution. All constraints have surpluses except Labour hours. Not optimum as there is a positive value in Z row.


X1 0 0.375 0.125 0 0 0 15

X2 0 1 -0.25 0.25 0 0 0 20

S3 0 0 -0.125 -0.125 1 0 0 5

S4 0 0 -0.375 0.125 0 1 0 5

S5 0 0 -0.25 0.25 0 0 1 10

Z 0 0 -0.125 -0.625 0 0 0 -125

Conclusion:Since the indicator row is negative the solution is optimum with 15X1 and 20X2 giving $125 contribution.Shadow prices are X1 = $0.125 and X2 = $0.625.Non-binding constraints are {3}, {4}, {5} with 5, 5 and 10 spare respectively.

DUALITY:

There is a dual or inverse for every Linear Programming problem. Because solving Simplex problem in Maximization is quite simple and straightforward, it is usually to convert a Minimization problem into Maximization problem using dual.The dual or inverse of Linear Programming problem is obtained by making the constraints in the inequalities coefficient of the new objective function.

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The cofficiences of the original inequalities are combined with the cofficiences of the original objective function as the constraints.

Question 6:Minimize Z = 40X1 + 50X2

Subject to:3X1 + 5X2 >= 1505X1 + 5X2 >= 2003X1 + X2 >= 60

X1, X2 >=0

Solution:The Dual Linear Programming problem is as follows:

Maximize P = 150Y1 + 200Y2 +60 Y3

Subject to:3Y1 + 5Y2 + 3Y3 <= 405Y1 + 5Y2 + Y3 <= 50 Y1>=0, Y2>=0, Y3>=0.

Y1 Y2 Y3 S1 S2 Solution

S1 3 3 1 0 40S2 5 5 1 0 1 50P 150 200 60 0 0 0 Indicator Row

Pivot Column


Y2 3/5 1 3/5 1/5 0 8S2 0 -2 -1 1 10P 30 0 -60 -40 0 -1600


Y2 0 1 1.2 0.5 -0.3 5Y1 0 -1 -0.5 0.5 5P 0 0 -30 -25 -15 -1750

Conclusion:Since the indicator row is negative the solution is optimum with 5Y1 and 5Y2 giving $1750 contribution.

Question 7:22

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Minimize Z = 16X1 + 11X2 Subject to:2X1 + 3X2 >= 35X1 + X2 >= 8X1, X2 >=0 Using the Dual problem.

Solution:The Dual Linear Programming problem is as follows:

Maximize P = 3Y1 + 8Y2

Subject to:2Y1 + 5Y2 <= 163Y1 + Y2 <= 11 Y1>=0, Y2>=0.

Y1 Y2 S1 S2 Solution

S1 2 1 0 16S2 3 1 0 1 11Z 3 8 0 0 0 Indicator Row

Pivot Column

Y1 Y2 S1 S2 Solution

Y1 0.4 0.2 0 3.2S2 2.6 0 -0.2 1 7.8Z -0.2 0 -1.6 0 -25.6

Conclusion:Since the indicator row is negative the solution is optimum. Hence P = 3Y1 + 8Y2 is maximum when Y1 = 3.2 and Y2= 0 and P = 25.6In the primary problem, the solution correspond the slack variable values in the final tableau.i.e.

X1= S1 = 1.6X2= S2 = 0.

Hence Z = 16*1.6 + 11*0 => 25.6

TERMS USED WITH LINEAR PROGRAMMING :

FEASIBLE REGION:

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Represents all combinations of values of the decision variables that satisfy every restriction simultaneous.The corner point of the feasible region gives what is known as BASIC FEASIBLE SOLUTION i.e. the solution that is given by the coordinates at the intersection of any two binding constraints.

BINDING CONSTRAINTS:

Is an inequality whose graph forms the bounder of the feasible region.

NON BINDING CONSTRAINTS:

Is an inequality, which does not conform to the feasible region.

DUAL PRICE / SHADOW PRICES:

It is important that management information to value the scarce resources. These are known as Dual price / Shadow price. Derived from the amount of increase (or decrease) in contribution that would arise if one more (or one less) unit of scare resource was available.

ASSIGNMENT PROBLEM:

This is the problem of assigning any worker to any job in such a way that only one worker is assigned to each job, every job has one worker assigned to it and the cost of completing all jobs is minimized.

STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:

a. Layout a two way table containing the cost for assigning a worker to a job.b. In each row subtract the smallest cost in the row from every cost in the row. Make

a new table.c. In each column of the new table, subtract the smallest cost from every cost in the

column. Make a new table.d. Draw horizontal and vertical lines only through zeroes in the table in such a way

that the minimum number of lines is used.e. If the minimum number of lines that covers zeroes is equal to the number of rows

in the table the problem is finished.

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f. If the minimum number of lines that covers zeroes is less than the number of rows in the table the problem is not finished go to step g.

g. Find the smallest number in the table not covered by a line.i. Subtract that number from every number that is not covered by a line.

ii. Add that number to every number that is covered by two lines.iii. Bring other numbers unchanged. Make a new table.

h. Repeat step d through step g until the problem is finished.

Question 1:Use the assignment method to find the minimum distance assignment of Sales representative to Customer given the table below: What is the round trip distance of the assignment.

Sales Representative Customer Distance (km)A 1 200A 2 400A 3 100A 4 500B 1 1000B 2 800B 3 300B 4 400C 1 100C 2 50C 3 600C 4 200D 1 700D 2 300D 3 100D 4 250

1 2 3 4A 200 400 100 500B 1000 800 300 400C 100 50 600 200D 700 300 100 250

TABLEAU 2:

1 2 3 4A 100 300 0 400B 700 500 0 100C 50 0 550 150D 600 200 0 150

TABLEAU 3:

1 2 3 4

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A 50 300 0 300B 650 500 0 0C 0 0 550 50D 550 200 0 50

TABLEAU 4:

1 2 3 4A 0 250 0 300B 600 450 0 0C 0 0 600 100D 500 150 0 50

Conclusion:Since the number of lines is now equal to number of rows, the problem is finished with the following assignment:SALES REP CUSTOMER DISTANCE

A 1 200B 4 400C 2 50D 3 100

750 km

Therefore total round Trip distance = 750 km * 2 => 1500 km

Question 2:A foreman has 4 fitters and has been asked to deal with 5 jobs. The times for each job are estimated as follows.

A B C D

1 6 12 20 122 22 18 15 203 12 16 18 154 16 8 12 205 18 14 10 17

Allocate the men to the jobs so as to minimize the total time taken.

Solution:Insert a Dummy fitter so that number of rows will be equal to number of column.

TABLEAU 1:

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A B C D DUMMY1 6 12 20 12 02 22 18 15 20 03 12 16 18 15 04 16 8 12 20 05 18 14 10 17 0

TABLEAU 2:

A B C D DUMMY1 0 4 10 0 02 16 10 5 8 03 6 8 8 3 04 10 0 2 8 05 12 6 0 5 0

TABLEAU 3:

A B C D DUMMY1 0 4 10 0 3

2 13 7 2 5 03 3 5 5 0 04 10 0 2 8 35 12 6 0 5 3

Conclusion:Since the number of lines is now equal to number of rows, the problem is finished with the following assignment: FITTERS JOBS TOTALS

A 1 6B 4 8C 5 10D 2 15Dummy 2 0

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THE ASSIGNMENT TECHNIQUE FOR MAXIMIZING PROBLEMS:

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Maximizing assignment problem typically involves making assignments so as to maximize contributions.

STEPS INVOLVED :

a) Reduce each row by largest figure in that row and ignore the resulting minus signs.

b) The other procedures are the same as applied to minimization problems.

Question 3:A foreman has 4 fitters and has been asked to deal with 4 jobs. The times for each job are estimated as follows.

W X Y Z

A 25 18 23 14B 38 15 53 23C 15 17 41 30D 26 28 36 29

Allocate the men to the jobs so as to maximize the total time taken.

Solution:

TABLEAU 1: W X Y Z

A 0 7 2 7B 15 38 0 30C 26 24 0 11D 10 8 0 7

TABLEAU 2:

W X Y ZA 0 0 2 0B 15 31 0 23C 26 17 0 4D 10 1 0 0

TABLEAU 3:

W X Y ZA 0 0 3 1

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B 14 30 0 23C 25 16 0 4D 9 0 0 0

TABLEAU 4:

W X Y ZA 0 0 7 1B 10 26 0 19C 21 12 0 0D 9 0 4 0

Conclusion:Since the number of lines is now equal to number of rows, the problem is finished with the following assignment:

A W 25B Y 53C Z 30D X 28

$136 39

Question 4:A Company has four salesmen who have to visit four clients. The profit records from previous visits are shown in the table and it is required to Maximize profits by the best assignment.

A B C D

1 6 12 20 122 22 18 15 203 12 16 18 154 16 8 12 20

Solution:

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TABLEAU 1: W X Y Z

1 6 12 20 122 22 18 15 203 12 16 18 154 16 8 12 20

TABLEAU 2: W X Y Z

1 14 8 0 82 0 4 7 23 6 2 0 34 4 10 8 0

TABLEAU 3: W X Y Z

1 14 6 0 82 0 2 7 23 6 0 0 34 4 10 8 0

Conclusion:Since the number of lines is now equal to number of rows, the problem is finished with the following assignment:

4 D 202 A 221 C 203 B 16

$78 39

TRANSPORTATION PROBLEM:

This is the problem of determining routes to minimize the cost of shipping commodities from one point to another. The unit cost of transporting the products from any origin to any destination is given. Further more, the quantity available at each origin and quantity required at each destination is known.

STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:

Arrange the problem in a table with row requirements on the right and column requirements at the bottom. Each cell should contain the unit cost approximates to the shipment.

30

Obtain an initial solution by using the North West Corner rule. By this method one begins at the up left corner cell and works up to the lower right corner. Place the quantity of goods in the first cell equal to the smallest of the rows or column totals in the table. Balance the row and column respectively until you reach the lower right hand cell.

Find cell values for every empty cell by adding and subtract around the closing loop.

If all empty cell have + values the problem is finished. If not pick the cell with most – (negative) value. Allocate a quantity of goods to that cell by adding and subtract the small value of the column or row entries in the closed loop. The closed loop techniques involves the following steps:

Pick an empty cell, which has no quantity of goods in it. Place a + sign in the empty cell. Use only occupied cells for the rest of the closed loop. Find an occupied cell that has occupied values in the same row or

same column and place a – (negative) sign in this cell. Go to the next occupied cell and place + sign in it. Continue in this manner until you return to the unoccupied cell in

which you started. A closed loop exists for every empty cell as long as they are occupied

cell equal to number of rows + number of column – 1.

Question 1:A firm has 3 factory (A, B, C) and 4 warehouses (1, 2, 3, 4). The capacities of the factories and the requirements of the warehouse are in the table below.

FACTORY CAPACITY WAREHOUSE REQUIREMENTSA 220 1 160B 300 2 260C 380 3 300

4 180

The cost of shipping one unit from each factory to each warehouse is given below.

FACTORY WAREHOUSE COST $ A 1 3 A 2 5

A 3 6A 4 5B 1 7B 2 4B 3 9B 4 6C 1 5C 2 12

31

C 3 10C 4 8

Using the Transportation method, find the least cost shipping schedule and state what it is ?.Solution:

TABLEAU 1:1 2 3 4 Capacity

A 160 60 220

B 200 100 300

C 200 180 380

Req 160 260 300 180 900

This is the initial solution, which costs(160 * 3) + (60 * 5) + (200 * 4) + (100 * 9) + (200 * 10) + (180 * 8)= $5920.00


A 160 60 220

B 260 40 300

C 200 180 380

Req 160 260 300 180 900

The costs = (160 * 3) + (60 * 6) + (260 * 4) + (40 * 9) + (200 * 10) + (180 * 8)= $5680.00


A 220 220

B 260 40 300

C 160 40 180 380

Req 160 260 300 180 900

The costs = (160 * 5) + (40 * 10) + (260 * 4) + (40 * 9) + (220 * 6) + (180 * 8)= $5360.00

32

3 5

105 812

9

6

67 4

5

- 1

5

72

- 3- 4

3 5

105 812

9

6

67 4

5

- 1

5

7-2

14

3 5

105 812

9

6

67 4

5

- 1

3

7

2 14


A 220 220

B 260 40 300

C 160 80 140 380

Req 160 260 300 180 900

The costs = (160 * 5) + (80 * 10) + (260 * 4) + (40 * 6) + (220 * 6) + (140 * 8)= $5320.00

Conclusion:Since all the cell values are positive the solution is optimum with the following allocations:

A Supplies 220 to 3B Supplies 260 to 2C Supplies 160 to 1

C Supplies 80 to 3C Supplies 140 to 4

With a minimum cost of $ 5320.00

HEXCO NOV’93:

Below is a transportation problem where costs are in thousand of dollars.

SOURCES DESTINATIONS

A B C CAPACITIES

X 14 13 15 500Y 16 15 12 400Z 20 15 16 600

REQUIREMENTS 700 300 500

i. Solve this problem fully indicating the optimum delivery allocations and the corresponding total delivery cost. [6 marks]

ii. There are two optimum solutions. Find the second one [4 marks].33

3 5

105 812

9

6

67 4

5

14

6

2 13

iii. Solve the same problem considering XA is an infeasible (prohibited / impossible) route and find the new total transportation cost [7 marks].

iv. If under consideration is a road network in a war zone, what is the simple economic effect of bombing a bridge between X and A? [3 marks].

Solution: PART (i)TABLEAU 1:

A B C Capacity

X 500 500

Y 200 200 400

Z 100 500 600

Req 700 300 500 1500

The initial solution = (500 * 14) + (200 * 16) + (15 * 200) + (100 * 15) + (500 * 16) = $22700 0000

TABLEAU 2:A B C Capacity

X 500 500

Y 200 200 400 400

Z 300 300 600

Req 700 300 500 1500

Hence delivery allocations are:

X Supplies 500 to AY Supplies 200 to AY Supplies 200 to C

Z Supplies 300 to BZ Supplies 300 to C

34

3 5

105 12

9

6

7 4

0

-4

4

1

3 5

105 12

9

6

7 4

4

4

0

5

With a minimum cost of (500 * 14) + (200 * 16) + (15 * 300) + (200 * 12) + (300 * 16) = $21900 0000

PART (ii)The existence of an alternative least cost solution is indicated by a value of zero in an unoccupied cell in the final table. We add and subtract the smallest quantity in the column or row of the zero to get the alternative.


X 500 500

Y 400 400 400

Z 200 300 100 600

Req 700 300 500 1500


X Supplies 500 to AY Supplies 400 to CZ Supplies 200 to A

Z Supplies 300 to BZ Supplies 100 to C


PART (iii)


X --- 300 200 500

Y 400 400 400

35

3 5

105 12

9

6

7 4

4

40

5

3 5

9

6

7 45 0

Z 300 300 600

Req 700 300 500 1500


X Supplies 300 to BX Supplies 200 to CY Supplies 400 to A

Z Supplies 300 to AZ Supplies 300 to C


PART (iv)

The simple economic effect of bombing the bridge between X and A= 24100 0000 – 21900 0000= 2200 000

DUMMIES:

This is an extra row or column in a transportation table with zero cost in each cell and with a total equal to the difference between total capacity and total demand.In an unbalance transportation problem a dummy source or destination is introduced.

HEXCO NOV’98:

The transport manager of a company has 3 factories A, B and C and four warehouse I, II, III and IV is faced with a problem of determining the way in which factories should supply warehouses so as to minimize the total transportation costs.In a given month the supply requirements of each warehouse, the production capacities of the factories and the cost of shipping one unit of product from each factory to each warehouse in $ are shown below.

FACTORY WAREHOUSES

I II III IV PRO AVAIL

A 12 23 43 3 6

36

105 121

B 63 23 33 53 53C 33 1 63 13 17

REQUIREMENTS 4 7 6 14 31

You are required to determine the minimum cost transportation plan [20 marks].

Solution:TABLEAU 1:

I II III IV Dummy Capacity

A 4 2 6

B 5 6 14 28 53

C 17 17

Req 4 7 6 14 45 76

This is the initial solution, which costs(4 * 12) + (2 * 23) + (5 * 23) + (6 * 33) + (14 * 53) + (28 * 0) + (17 * 0)= $1149.00

TABLEAU 2:I II III IV Dummy Capacity

A 4 2 6

B 7 6 12 28 53

C 17 17

Req 4 7 6 14 45 76

The costs=(4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (12 * 53) + (28 * 0) + (17 * 0)= $1049.00

TABLEAU 3:I II III IV Dummy Capacity

A 4 2 6

B 7 6 40 53

C 12 5 1737

12 23

6333 131

33

43

5363

30

51

-22

21

- 51

10

0

023

030 -40

12 23

6333 131

33

43

5363

350

1

-22

-29

50

60

0

023

030 -40

12 23

6333 131

33

43

5363

310

41

-22

11

16

20

0

023

030

40

Req 4 7 6 14 45 76

The costs=(4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (40 * 0) + (12 * 13) + (5 * 0)= $569.00

TABLEAU4:I II III IV Dummy Capacity

A 4 2 6

B 2 6 45 53

C 5 12 17

Req 4 7 6 14 45 76

The costs=(4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1)

= $459.00


Factory A Supplies Warehouse IFactory A Supplies Warehouse IVFactory B Supplies Warehouse II

Factory B Supplies Warehouse IIIFactory B Supplies Warehouse DummyFactory C Supplies Warehouse IIFactory C Supplies Warehouse IV

With a minimum cost of (4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1) = $459.00

HEXCO MARCH’2000:

A well-known organization has 3 warehouse and 4 Shops. It requires transporting its goods from the warehouse to the shops. The cost of transporting a unit item from a warehouse to a shop and the quantity to be supplied are shown below.

TO DESTINATION

I II III IV TOTAL SUPPLY

SOURCE A 10 0 20 11 15SOURCE B 12 7 9 20 25

38

12 23

6333 131

33

43

5363

332

19

2211

32

42

0

023

052

18

SOURCE C 0 14 16 18 5

TOTAL DEMAND 5 15 15 10

Use any method to find the optimum transportation schedule and indicate the cost [14marks].

DEGENERATE SOLUTION :

It involves working a transportation problem if the number of used routes is equal to:Number of rows + Number of column – 1.

However if the number of used routes can be less than the required figure we pretend that an empty route is really used by allocating a zero quantity to that route.

MAXIMIZATION PROBLEMS :

Transportation algorithm assumes that the objective is to minimize cost. However it is possible to use the method to solve maximization problem by either:

Multiply all the units’ contribution by – 1. Or by subtracting each unit contribution from the maximum contribution in

the table.

UNIT 3: NON-LINEAR FUNCTIONS:

HOURS: 20.

NON -LINEAR FUNCTIONS:

MARGINAL DISTRIBUTION:

PARTIAL INTEGRATION:

Partial integration is a function with more than one variable or finding the probability of a function with more than one variables i.e. f(X1, X2, X3, ….Xn) and is just the rate at which the values of a function change as one of the independent variables change and all others are held constant.

Question 1:

If f(x, y) = 2(x + y –2xy) given the intervals 0<= x<=1, 0<=y<=1.Find the marginal distribution of x = f(x).Find the marginal distribution of y = f(x).

Solution:

39

Pr {0<=x<=1} = 01 2(x + y –2xy)x= 2 01 (x + y –2xy)x

= 2 [x2/2 + xy + x2y]01

= 2 [½ + y – y]= 2[½]= 1

Pr {0<=y<=1} = 01 2(x + y –2xy)y= 2 01 (x + y –2xy)y

= 2 [xy +y2/2 + xy2]01

= 2 [x + ½ – x]= 2[½]= 1

Question 1:

If f(X1, X2) =(X21X2 + X3

1X22 + X1) given the intervals 0<= X1<=2, 1<=X2 <=3.

Find the marginal distribution of x = f(x).Find the marginal distribution of y = f(x).Find the Expected value of X1 (E(X1)).Find the variance of X1 (Var (X1)).

Solution:Pr {0<=X1<=2} = 02 (X2

1X2 + X31X2

2 + X1)x= [X3

1X2 /3+ X41X2

2 /4+ X21/2]0

2

= [8X2 /3+ 4X22 + 2] – [0]

= 8X2 /3+ 4X22 + 2

Pr {1<=X2<=3} = 13 (X21X2 + X3

1X22 + X1)y

= [X21X2

2 /2+ X31X3

2 /3+ X1X2]13

= [9X21 /2+ 27X3

1 /3+ 3X2]13 –[X2

1 /2+ X32 /3+ X1]

= 9X21 /2+ 27X3

1 /3+ 3X2 – X21 /2- X3

2 /3 - X1

= 8X21 /2+ 26X3

1 /3+ 2X2

Expected value of E(X1) =02 X. f(X1)x = 02 X(X2

1X2 + X31X2

2 + X1)x = 02 (X3

1X2 + X41X2

2 + X21)x

= [X41X2 /4+ X5

1X22 /5+ X3

1/3]02

40

= [4X2 + 32X22 /5 + 8 /3] – [0]

= 4X2 + 32X22 /5 + 8 /3

Variance of X1 = Var (X1) = 02 ([X1 - E(X1)]2 . f(X1)x= 02 ([X1 - 4X2 + 32X2

2 /5 + 8 /3]2 * (X21X2 + X3

1X22 + X1)x.

Question 2:A manufacturing company produces two products bicycles and roller skates. Its fixed costs production is: $1200 per week. Its variables costs of production are: $40 for each bicycle produced and $15 for each pair of roller skates. Its total weekly costs in producing x bicycles and y pairs of roller skates are therefore c= cost.C(x, y) = 1200 + 40x + 15y for example; in producing x = 20 bicycles and y = 30 pairs of roller skates/ week.The manufacture experiences total cost of:

C(20, 30) = 1200 + 40(20) + 15(30)= 1200 + 800 + 450= 2450.

Question 3:A manufacturing of Automobile tyres produces 3 different types: regular, green and blue tyres. If the regular tyres sell for $60 each, the green tyres for $50 each and the blue tyres for $100 each. Find a function giving the manufacture’s total receipts or revenue from the of x regular tyres and y green tyres and z blue tyres.R(x, y, z) = 60x + 50y +100z.

Solution:Since the receipts of the sale of any tyre type is the price per tyre times the number of tyres sold:The total receipts are:R(x, y, z) = 60x + 50y + 100zFor example receipts from the sell of 10 tyres of each type would be:R(10,10,10) = 60(10) + 50(10) + 100(10)

= 600 + 500 + 1000= $2100

PARTIAL DIFFERENTIATION :

For a function “f” of a single variable, the derivative f measures the rate at which the values of f(x) change as the independent variable x change.A partial derivative of a function i.e. f(X1, X2, X3.. Xn) of several variables is just the rate at which the values of the function change as one of the independent variable changes and all others are held constant.

Question 3:For the function f(x, y) = X3 + 4X2Y3

+ Y2

41

Find f / x

f / y

f(-2; 3)

Solution:

f / x = 3X2 + 8XY3

f / y = 12X2Y2 + 2Y

f(-2; 3) = X3 + 4X2Y3 + Y2

= (-2)3 + 4(-2)2(3)3 + (3)2

= -8 + 16(27) +9= 433

Question 4:A company produces electronic typewriters and word processors, it sells the electronic typewriters for $100 each and word processors for $300 each. The company has determined that its weekly sales in producing x electronic writers and y word processors are given by the following joint cost function.C(x, y) = 200 + 50x +8y + X2 + 2Y2

Find the numbers of x and y of machines that the company should manufacture and sell weekly in order to maximize profits.

Solution:Revenue function is given by:

R(x, y) = 100x + 300y

Profit = Revenue – Cost.Then Profit function is given by:P(x, y) = R(x, y) – C(x, y)

= (100x + 300y) – (200 + 50x +8y + X2 + 2Y2)= 50x + 292y – 200 - X2 - 2Y2

To find the critical points of turning points of x and y. We set the partial derivative = 0.Thus

p / x => 50 – 2x = 0 50 = 2x x = 25

p / y => 292 – 4y =0 292 = 4y

42

y = 73

The production schedule for maximum profit is therefore x = 25 type writers and y = 73 word processors which yields a profit of

P = 50(25) + 292(73) – 200 – 625 – 2(73)2

= 1250 + 21316 – 200 – 625 – 1065 = 22566 – 11493 = $11083

NECESSARY AND SUFFICIENT CONDITIONS FOR EXTREMA :

The necessary condition or the GRADIENT VECTOR of the extrema determines the turning points or critical points of a function.Let X0 be a variable representing the turning point and represented mathematically as:X0 = (A0, B0, … N0).

In general form; a necessary condition or gradient vector for X0 to be an extrema point of f(x) is that the gradient () f (X0) = 0.

Question 1:Given f(X1, X2, X3) =(X1 + 2X3 + X2X3 – X2

1 - X22 - X2

3)Find the gradient vector for X0 i.e. f (X0) = 0. Solution:The necessary condition (gradient vector) f (X0) = 0 is given by:

f / x1 => 1 - 2X1 = 0. 1 - 2X1 = 0. [1]

f / x2 => X3 - 2X2 = 0. X3 - 2X2 = 0. [2]

f / x3 => 2 + X2 - 2X3 = 0. 2 + X2 - 2X3 = 0. [3]

(a) Finding X1 is given by 1 = 2X1 X1 = ½

(b) Equation 2 is given by X3 - 2X2 = 0.43

X3 = 2X2.

(c) On equation 3 where therefore substitute X3 with 2X2.Thus 2 + X2 - 2X3 = 0.

2 + X2 – 2(2X2) = 0. 2 + X2 – 4X2 = 0. 2 – 3X2 = 0. X2 = 2/3.

Therefore X3 = 2X2. X3 = 2(2 / 3) X3 = 4/3.

Therefore X0 = (½, 2/3, 4/3)

A sufficient condition for X0 a point to be extremism is that the HESHIAN matrix (denoted by H) evaluate at X0 is:

i. Positive definite when X0 is a Minimum point.ii. Negative definite when X0 is a Maximum point.

The Hessian matrix is achieved by finding the 2nd Partial derivation of the first Partial derivative of each equation with respect to all variables defined.

Thus the Hessian matrix is evaluated at the point X0H/X0 = 2f/ X

21,

2f/ X1 X22,

2f/ X1 X23

2f/ X2 X

21,

2f/ X22,

2f/ X2 X23

2f/ X3 X

21,

2f/ X3 X22,

2f/ X23

f / x1 => 1 - 2X1 = 0.

f / x2 => X3 - 2X2 = 0.

f / x3 => 2 + X2 - 2X3 = 0.

To establish the sufficiency the function has to have:

H/X0 = -2 0 0 0 -2 1 0 1 -2

Since the Hessian matrix is 3 by 3 matrix then: Find the 1st Principal Minor determinant of 1 by 1 matrix in the Hessian matrix. Find the 2nd Principal Minor determinant of 2 by 2 matrix in the Hessian matrix. Find the 3rd Principal Minor determinant of 3 by 3 matrix in the Hessian matrix.

44

The Positive definite when X0 is a Minimum point is evaluated as: When 1st PMD = + ve. When 2nd PMD = +ve. When 3rd PMD = +ve.

Or When 1st PMD = - ve. When 2nd PMD = +ve. When 3rd PMD = +ve.

Thus 3 by 3 Hessian matrix the number of positive number should be greater than one. (Should be two or more).

The Negative definite when X0 is a Maximum point is evaluated as: When 1st PMD = - ve. When 2nd PMD = - ve. When 3rd PMD = - ve.

Or When 1st PMD = + ve. When 2nd PMD = - ve. When 3rd PMD = - ve.

Thus 3 by 3 Hessian matrix the number of negative number should be greater than two. (Should be two or more).

H/X0 = -2 0 0 0 -2 1 0 1 -2

Thus the 1st PMD of (-2) = -2

Thus the 2nd PMD of –2 00 -2

= (-2 * -2) – (0 * 0)= 4

The 3rd PMD = -2 0 0 0 -2 1 0 1 -2

= -2 –2 1 - 0 0 1 + 0 0 -21 -2 0 -2 0 1

= -2 {(-2 * -2) – (1 * 1)} – 0 (0 – 0) + 0 (0 – 0)= - 2 (3)= - 6

45

Thus the PMD is equal to –2, 4 and –6 and H/X0 is negative definite and X0 = (½, 2/3, 4/3) represents a Maximum point.

Question 2:Given f(X1, X2, X3) =(-X1 + 2X3 - X2X3 + X2

1+ X22 - X2

3)i. Find the gradient vector for X0 i.e. f (X0) = 0.

ii. Determine the nature of the turning points using Hessian Matrix.

Solution:The necessary condition (gradient vector) f (X0) = 0 is given by:

f / x1 => -1 + 2X1 = 0. -1 + 2X1 = 0. [1]

f / x2 => -X3 + 2X2 = 0. -X3 + 2X2 = 0. [2]

f / x3 => 2 - X2 - 2X3 = 0. 2 - X2 - 2X3 = 0. [3]

(b) Finding X1 is given by -1 = 2X1 X1 = ½

(b) Equation 2 is given by -X3 + 2X2 = 0.X3 = 2X2.

(c) On equation 3 where therefore substitute X3 with 2X2.Thus 2 - X2 - 2X3 = 0.

2 - X2 – 2(2X2) = 0. 2 - X2 – 4X2 = 0. 2 – 5X2 = 0. X2 = 2/5.

Therefore X3 = 2X2. X3 = 2(2 / 5) X3 = 4/5.

46

Therefore X0 = (½, 2/5, 4/5)

H/X0 = 2f/ X21,

2f/ X1 X22,

2f/ X1 X23

2f/ X2 X

21,

2f/ X22,

2f/ X2 X23

2f/ X3 X

21,

2f/ X3 X22,

2f/ X23

H/X0 = 2 0 0 0 2 -1 0 -1 -2

Thus the 1st PMD of (2) = 2

Thus the 2nd PMD of 2 00 2

= (2 * 2) – (0 * 0)= 4

The 3rd PMD = 2 0 0 0 2 -1 0 -1 -2

= -2 2 -1 - 0 0 -1 + 0 0 2-1 -2 0 -2 0 -1

= 2 {(2 * -2) – (-1 * -1)} – 0 (0 – 0) + 0 (0 – 0)= 2 (-4) - (1)= 2 (-5)= - 10

Thus the PMD is equal to 2, 4 and –10 and H/X0 is positive definite and X0 = (½, 2/5, 4/5) represents a Minimum point.

NON LINEAR ALGORITHMS : (COMPUTATIONS)

THE GRADIENT METHOD S:

The general idea is to generate successive iterative points, starting from a given initial point, in the direction of the fast and increase (maximization of the function).The method is based on solving the simultaneous equations representing the necessary conditions for optimality namely f (X0) = 0.

47

Termination of the gradient method occurs at the point where the gradient vector becomes null. This is only a necessary condition for optimality suppose that f(x) is maximized.Let X0 be the initial point from which the procedure starts and define f (Xk) as the gradient of f at Kth point Xk.This result is achieved if successive point Xk and Xk+1 are selected such that Xk+1 = Xk + rk f (Xk) where rk is a parameter called Optimal Step Size.

The parameter rk is determined such that Xk+1 results in the largest improvement in f. In other words, if a function h(r) is defined such that h(r) = f(Xk )+ rk f (Xk). This function is then differentiated and equate zero to the differentiatable function to obtain the value of rk.

Question 3:Consider maximizing f(X1, X2) =(4X1 + 6X2 - 2X2

1- 2X2X1 - 2X22)

And let the initial point be given by X0(1, 1).

Hint in X0(1, 1) X1 =1 and X2 = 1

Solution:Find f (X0) = (f/x1, f/x2)

= (4 – 4X1 – 2X2; 6 – 2X1 - 4X2)

1st iteration

Step 1: Find f (X0) = (4 – 4 – 2; 6 – 2 - 4)= (-2; 0)

Step 2: Find Xk+1 = Xk + rk f (Xk) X0+1 = X0 + rk f (X0) X1 = (1, 1) + r(-2; 0) (1, 1) + (-2r, 0) (1 + -2r, 1) (1 – 2r, 1)

Thus h(r) = f(Xk )+ rk f (Xk) = f(X0 )+ r f (X0) = f(1 – 2r; 1) = 4(1 – 2r) + 6(1) – 2(1 – 2r)2 – 2(1)(1 – 2r) - 2(1)2. = 4(1 – 2r) + 6 – 2(1 – 2r) 2 – 2(1 – 2r) - 2. = 4(1 – 2r) – 2(1 – 2r) + 6 – 2 – 2(1 – 2r)2. = (4 – 2)(1 – 2r) + 4 – 2(1 – 2r)2. = 2(1 – 2r) – 2(1 – 2r)2 + 4. = – 2(1 – 2r)2 +2(1 – 2r) + 4. = – 2(1 – 2r)2 + 2 – 4r + 4.

h1(r) = 0 – 2(1 – 2r)2 + 2 – 4r + 4 = 0. - 4 * -2(1 – 2r) – 4 = 0.

48

8(1 – 2r) + - 4 = 0. 8 – 16r – 4 = 0 4 –16r = 0. r = ¼

The optimum step size yielding the maximum value of h(r) is h1 = ¼. This gives X1= (1 –2(¼); 1)

= (1 - ½; 1)= (½; 1)

UNIT 4 PROJECT MANAGEMENT WITH PERT/CPM

HOURS: 20

PROJECT MANAGEMENT :

TERMS USED IN PROJECT MANAGEMENT:

PROJECT :Is a combination of interrelated activities that must be executed in a certain order before the entire task can be completed.

ACTIVITY :Is a job requiring time and resource for its completion.

ARROW :Represents a point in time signifying the completion of some activities and the beginning of others.

NETWORK :Is a graphic representation of a project’s operation and is composed of activities and nodes.

RULES FOR CONSTRUCTING NETWORK DIAGRAM:

Each activity is represented by one and only one arrow in the network. No two activities can be identified by the same head and tail events. If activities

A and B can be executed simultaneously, then a dummy activity is introduced either between A and one end event or between B and one end event. Dummy activities do not consume time or resources. Another use of the dummy activity: suppose activities A and B must precede C while activity E is preceded by B only.

To ensure the correct precedence relationships in the network diagram, the following questions must be answered as every activity is added to the network:

What activities must be completed immediately before this activity can start.

49

What activities must follow this activity? What activities must occur concurrently with this activity?

Question 1:

ACTIVITY PRECEDED BY DURATION (Weeks) A Initial activity 10B A 9C A 7D B 6E B 12F C 6G C 8H F 8I D 4J G, H 11K E 5L I 7

Find the critical path and the time for completing the project.

Solution:

D I 6 4

B 9 E 12 L 7

K A 5 10

dummy J 11 C 7

G 8

F H 6 8

50

00 0

110 10

219 25

317 17

425 31

531 37

725 31

623 23

831 31

929 35

1042 42

EARLISET START TIME :

Represents all the activities emanating from i. Thus ESi represent the earliest occurrence time of event i. Earliest finish time is given by:

EF = Max {ESi + D}

LATESET COMPLETION TIME :

It initiates the backward pass. Where calculations from the “end” node and moves to the “start” node.Latest start time is given by:

LSi = Min {LF – D}

DETERMINATION OF THE CRITICAL PATH :

A Critical path defines a chain of critical that connects the start and end of the arrow diagram. An activity is said to be critical if the delay in its start will cause a delay in the completion date of the entire project. Or it is the longest route, which the project should follow until its completion date of the entire project.

The critical path calculations include two phases:

FORWARD PASS:Is where calculations begin from the “start” node and move to the “end” node. At each node a number is computed representing the earliest occurrence time of the corresponding event.

BACKWARD PASS:Begins calculations from the “end” node and moves to the “start” node. The number computed at each node represents the latest occurrence time of the corresponding event.

DETERMINITION OF THE FLOATS :

A Float or Spare time can only be associated with activities which are non critical. By definition activities on the critical path cannot have floats.There are 3 types of floats.

TOTAL FLOAT :

This is the amount of time a path of activities could be delayed without affecting the overall project duration.

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Total Float = Latest Head Time – Earliest Tail time – duration.= LS – ES.= LF – ES – D= LF – EF or EC.

FREE FLOAT :This is the amount of time an activity can be delayed without affecting the commencement of a subsequent activity at its earliest start time.Free Float = Earliest Head Time – Earliest Tail Time – Duration.

= LF – ES – D= ESj – ESi – D.

INDEPENDENT FLOAT :This is the amount of time an activity can be delayed when all preceding activities are completed as late as possible and all succeeding activities completed as early as possible.Independent Float = EF – LS – D.

NORMAL EARLIEST TIME LATEST TIME TOTAL FLOAT

ACTIVITY TIME ES EF = ES + D LS = LF – D LF =LS – ES

A 10 0 10 0 10 0B 9 10 19 16 25 6C 7 10 17 10 17 0D 6 19 25 25 31 6E 12 19 31 25 37 6F 6 17 23 17 23 0G 8 17 25 23 31 6H 8 23 31 23 31 0I 4 25 29 31 35 6J 11 31 42 31 42 0K 5 31 36 37 42 6L 7 29 36 35 42 6

Question 2:Draw the network for the data given below then find the critical path as well total float and free float.

ACTIVITY (I, J) DURATION

52

00 0

(0, 1) 2(0, 2) 3(1, 3) 2(2, 3) 3(2, 4) 2(3, 4) 0(3,5) 3(3, 6) 2(4, 5) 7(4, 6) 5(5, 6) 6

Solution:

2

2 2 3

3 6

Dummy 3

7 5

2

ACTIVITY D ES EF=ES+D LS=LF-D LF TOTAL FREE Float Float

(0, 1) 2 0 2 2 4 2 2

(0, 2) 3 0 3 0 3 0 0(1, 3) 2 2 4 4 6 2 2(2, 3) 3 3 6 3 6 0 0(2, 4) 2 3 5 4 6 1 1(3, 4) 0 6 6 6 6 0 0(3,5) 3 6 9 10 13 4 4(3, 6) 2 6 8 17 19 11 11(4, 5) 7 6 13 6 13 0 0

53

12 4

23 3

36 6

619 19

46 6

513 13

(4, 6) 5 6 11 14 19 8 8(5, 6) 6 13 19 13 19 0 0

PERT ALGORITHM :

PROBABILISTIC TIME DURATION OF ACTIVITIES .

The following are steps involved in the development of probabilistic time duration of activities.

Make a list of activities that make up the project including immediate predecessors.

Make use of step 1 sketch the required network. Denote the Most Likely Time by Tm, the Optimistic Time by To and

Pessimistic time by Tp. Using beta distribution for the activity duration the Expected Time Te

for each activity is computed by using the formula: Te = (To + 4Tm + Tp) / 6.

Tabulate various times i.e. Expected activity times, Earliest and Latest times and the EST and LFT on the arrow diagram.

Determine the total float for each activity by taking the difference between EST and LFT.

Identify the critical activities and the expected date of completion of the project.

Using the values of Tp and To compute the variance (2) of each activity’s time estimates by using the formula: 2 = {{Tp – To} / 6}2.

Compute the standard normal deviate by: Zo = (Due date – Expected date of Completion) / Project variance.

Use Standard normal tables to find the probability P (Z <= Zo) of completing the project within the scheduled time, where Z ~ N(0,1).

Question 3:A project schedule has the following characteristics:

Activity Most Likely Time Optimistic Time Pessimistic Time

1 – 2 2 1 32 – 3 2 1 32 – 4 3 1 53 – 5 4 3 5

54

4 – 5 3 2 44 – 6 5 3 75 – 7 5 4 66 – 7 7 6 87 – 8 4 2 67 – 9 6 4 88 – 10 2 1 39 – 10 5 3 7

I. Construct the project network.II. Find expected duration and variance for each activity.

III. Find the critical path and expected project length.IV. What is the probability of completing the project in 30 days.

Solution:

4 5 6

2 3 7 4 5

2

3 2

5

Expected job Time Te = (To + 4Tm + Tp) / 6.

Variance 2 = {{Tp – To} / 6}2.

Activity Tm To Tp Te 2

1 – 2 2 1 3 2 0.1112 – 3 2 1 3 2 0.1112 – 4 3 1 5 3 0.4453 – 5 4 3 5 4 0.1114 – 5 3 2 4 3 0.1114 – 6 5 3 7 5 0.4455 – 7 5 4 6 5 0.1116 – 7 7 6 8 7 0.111

55

10 0

22 2

34 8

58 12

717 17

923 23

1028 28

45 5

610 10

821 26

7 – 8 4 2 6 4 0.4457 – 9 6 4 8 6 0.4458 – 10 2 1 3 2 0.1119 – 10 5 3 7 5 0.445

Critical path (*) comprises of activities (1 –2), (2 - 4), (4 –6), (6 –7), (7 –9) and (9 –10)

Expected project length is = 28 days.

Variance 2 = 0.111 + 0.445 + 0.445 + 0.111 + 0.445 + 0.445= 2.00 (on critical path only)

(iv) Probability of completing the project in 30 days is obtained by:Zo = (Due date – Expected date of Completion) / Project variance.= (30 – 28) / 2.= 1.414 (Look this from Normal tables)

Now from Standard Normal tables Z= 0.4207.P (t <= 30) = P (Z <= 1.414)

= 0.5 + 0.4207= 0.9207

0 1.414

This shows that the probability of meeting the scheduled time will be 0.9207

COST CONSIDERATIONS IN PERT / CPM:

The cost of a project includes direct costs and indirect costs. The direct costs are associated with the individual activities and the indirect costs are associated with the overhead costs such as administration or supervision cost. The direct cost increase if the job duration is to be reduced whereas the indirect costs increase if the job duration is to be increased.

TIME COST OPTIMIZATION PROCEDURE :

The process of shortening a project is called Crashing and is usually achieved by adding extra resources to an activity. Project crashing involves the following steps:

56

Critical Path: Find the normal critical path and identify the critical activities. Cost Slope: Calculate the cost slope for the different activities by using the Formula: COST SLOPE = Crash cost – Normal cost.

Normal Time – Crash Time. Ranking: Rank the activities in the ascending order of cost slope. Crashing: Crash the activities in the critical path as per the ranking i.e.

activities having lower cost slope would be crashed first to the maximum extent possible. Calculate the new direct cost by cumulatively adding the cost of crashing to the normal cost.

Parallel Crashing: As the critical path duration is reduced by the crash in step 3 other paths become critical i.e. we get parallel critical paths. This means that project duration can be reduced by simultaneous crashing of activities in the parallel critical paths.

Optimal Duration. Crashing as per Step 3 and step 4 an optimal project is determined. It would be the time duration corresponding to which the total cost (i.e. Direct cost plus Indirect cost) is a minimum.

Question 4:

For the network given below find the optimum cost schedule for the completion of the project:

JOB NORMAL CRASH

TIME COST $ TIME COST $

1 – 2 10 60 8 1202 – 3 9 75 6 1502 – 4 7 90 4 1503 – 4 6 100 5 1403 – 5 9 50 7 803 – 6 10 40 8 704 – 5 6 50 4 705 – 6 7 70 5 110

Solution:JOB COST SLOPE

*1 – 2 30 --->(4) =(120 –60) / (10 – 8) *2 – 3 25 ---> (3) 2 – 4 20

3 – 4 40 ---> (5)3 – 5 153 – 6 15

57

4 – 5 10 ---> (1) 5 – 6 20 ---> (2)

10

9

6 9 710

7

6

The critical path = 1 –2, 2 –3, 3 –4, 4 –5 and 5 –7.Expected project Length = 38 days. Associated with 38 days the minimum direct project cost

= 60 + 75 + 90 + 100 + 50 + 40 + 50 + 70= $535

In order to reduce the project duration we have to crash at least one of the jobs on the critical path. This is being done because crashing of the job not on the critical path does not reduce the project length.

1 st Crashing :

On critical path the minimum cost slope is job 4 –5 and is to be crashed at extra cost of $10 per day.

10

9

6 9 710

58

10 0

319 19

531 31

638 38

425 25

210 10

10 0

210 10

319 19

636 36

7

4

Duration of project = 36 days and Total cost = $535 + $10 * 2 = $555.

2nd Crashing:

Now crash job 5 –6 and is to be crashed at extra cost of $20 per day.

10

9

6 9 510

7

4


3rd Crashing:

Now crash job 2 –3 for 3 days and is to be crashed at extra cost of $25.

10

6

6 9 510

7

59

529 29

425 25

10 0

529 29

425 25

210 10

319 19

634 34

10 0

526 26

422 22

210 10

316 16

631 31

4


4th Crashing:

Now crash job 1 –2 for 2 days and is to be crashed at extra cost of $30.

10

6

6 9 58

7

4


5th Crashing:

Final crash job 3 –4 for 1 day and it is to be crashed at extra cost of $40 and two critical paths occurs.

10

6

5 9 58

7

4

60

10 0

524 24

420 20

28 8

314 14

629 29

10 0

523 23

419 19

28 8

314 14

628 28

Duration of project = 28 days and Total cost = $730+ $40 * 1 = $770.

Optimum Duration of project = 28 days and Total Cost = $770.

UNIT 5: RANDOM VARIABLES AND THEIR PROBABILITY:

HOURS: 20.

RANDOM VARIABLES AND PROBABILITY FUNCTIONS (DISCRETE ):

Given a Sample space S = {1,2,3,4,5,6} we may therefore use the variable such as X to represent an outcome in the sample space such a variable is called Random variable.When the outcome in a sample space are represented by values in a random variable the assignment of probabilities to the outcome can be thought of as a function for which the domain is the sample space, we refer to this as the probability function written as Pr.We use the following notation with probability function Pr {X = a} which means the probability associated with the outcome a while Pr {X in E} means the probability associated with event E.Given S = {1,2,3,4,5,6}

a. Find the probability of S = 3.b. Find the probability of S = 5.c. Find the probability of X in E when E = 1,2.3.

Solution:i. P (S = 3) = 1/6.

ii. P (S = 5) = 1/6.iii. P (X in E) = ½.

PROBABILITY DENSITY FUNCTIONS :

PROPERTIES OF DISCRETE RANDOM VARIABLE :

a. It is a discrete variable.b. It can only assume values x1, x2. …xn.c. The probabilities associated with these values are p1, p2. …pn.Where P(X = x1) = p1.

P(X = x2) = p2...P(X = xn) = pn.

Then X is a discrete random variable if p1 + p2. …pn = 1.

This can be written as ∑ P(X = x) =1.all x

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Question 1:The P.d.f. of a discrete random variable Y is given by P (Y=y) = cy2, for y = 0,1,2,3,4.Given that c is a constant, find the value of c.

Solution:

Y 0 1 2 3 4P(Y =y) 0 c 4c 9c 16c

= ∑ P(X = x) =1. all x

1 = c + 4c + 9c + 16c 1 = 30c c = 1/30.

Question 2:The Pdf. of a discrete random variable X is given by P (X=x) = a(¾)x , for x = 0,1,2,3...Find the value of the constant a.

Solution:

= ∑P(X = x) =1. all x

P(X = 0) = a(¾)0.P(X = 1) = a(¾)1.P(X = 2) = a(¾)2.P(X = 3) = a(¾)3 and so on.

So ∑P(X = x) =a + a(¾) + a(¾)2 + a(¾)3 + … all x

= a( 1 + ¾ + (¾)2 + (¾)3 + …)

= a ( 1/1- ¾) -> (sum of an infinite G.P with first term 1 and common ratio ¾)

= a(4)

4a = 1

a = ¼

62

EXPECTED VALUE/ MEAN / AVERAGE :

For a random variable X associated with a sample space {x1, x2. …xn} the concept of expected value is the generalization of the average of numbers {x1, x2. …xn}.

EXPECTED VALUE WITH SAME PROBABILITIES:

The expected value of X with same probabilities is given by:

E(x) = X1 + X2 + …Xn/n.

Question 3:Given that an die is thrown 6 times and the recordings are as follows then calculate the expected mean or mean score

Score x 1 2 3 4 5 6P(X = x) 1/6

1/61/6

1/61/6

1/6

Solution:E(x) = X1 + X2 + …Xn/n.

= 1 + 2 + 3 + 4 + 5 + 6/6= 21/6= 7/2= 3.5

EXPECTED VALUE WITH DIFFERENT PROBABILITIES:

The expected value of X with different probabilities is given by:

E(x) = P1 * X1 + P2 * X2 + …Pn * Xn.

Question 4:Given a random variable X which has a Pdf shown below. Calculate the expected mean.

X -2 -1 0 1 2P(X = x) 0.3 0.1 0.15 0.4 0.05

Solution:E(x) = P1 * X1 + P2 * X2 + …Pn * Xn

= -2 * 0.3 + -1 * 0.1 + 0 * 0.15 + 1 * 0.4 + 2 * 0.05= -0.2

Question 5:

63

A venture capital firm is determined based on the past experience that for each $100 invested in a high technology startup company; a return of $400 is experienced 20% of time. A return of $100 is experienced 40% of the time and zero (0) total loss is experienced 40% of the time.What is the firm’s expected return based on this data?.

Solution:S = {400, 100, 0}E(x) = P1 * X1 + P2 * X2 + …Pn * Xn

= 400 * 0.2 + 100 * 0.4 + 0 * 0.4. = 80 + 40 + 0= $120

Question 6:Given that an unbiased die was thrown 120 times and the recordings are as follows then calculate the expected mean or mean score.

Score x 1 2 3 4 5 6Frequency f 15 22 23 19 23 18 Total = 120

Solution:

E(x) = ∑fx/∑f.

= (15 + 44 + 69 + 76 + 115 +108)/120.= 3.558

Question 7:The random variable X has Pdf P(X=x) for x = 1,2,3.

X 1 2 3P(X = x) 0.1 0.6 0.3

Calculate:a. E(3).b. E(x).c. E(5x).d. E(5x + 3).e. 5E(x) + 3.f. E(x2).g. E(4x2 - 3).h. 4E(x2) – 3.

Solution:

X 1 2 3

64

5x 5 10 155x + 3 8 13 18x2 1 4 94x2 – 3 1 13 33P(X = x) 0.1 0.6 0.3

a. E(3) = ∑P(X = x). all x

= ∑3P(X = x). all x

= 3(0.1) + 3(0.6) + 3(0.3). = 3.

b. E(x) =∑xP(X = x). all x

=1(0.1) +2(0.6) +3(0.3)= 2.2

c. E(5x) = ∑5xP(X = x). all x

= 5(0.1) + 10(0.6) +15(0.3)= 11

d. E(5x + 3) = ∑(5x + 3)P(X = x). all x

= 8(0.1) + 13(0.6) + 18(0.3)= 14

e. 5E(x) + 3 = 5(2.2) + 3 = 14

f. E(x2) = ∑ x2P(X = x). all x

= 1(0.1) + 4(0.6) + 9(0.3)= 5.2

65

g. E(4x2 - 3) = ∑(4x2 – 3)P(X = x). all x

= 1(0.3) + 13(0.6) + 33(0.3) = 17.8

h. 4E(x2) – 3 = 4(5.2) – 3=20.8 – 3= 17.8

VARIANCE :

The expected value of a random variable is a measure of central tendency i.e. what values are mostly likely to occur while Variance is a measure of how far apart the possible values are spread again weighted by their respective probabilities.The formula for variance is given by:

Var (x) = E(x – μ)2 this can be reduced to

Var (x) = E(x2) - μ2

Question 8:The random variable X has probability distribution shown below.

x 1 2 3 4 5P(X =x) 0.1 0.3 0.2 0.3 0.1

Find:i. μ = E(x).

ii. Var(x) using the formula E(x – μ)2

iii. E(x2)iv. Var(x) using the formula E(x2) - μ2

Solution:

i. E(x) =μ = ∑xP(X = x). all x

=1(0.1) + 2(0.3) + 3(0.2) + 4(0.3) + 5(0.1)= 3

66

ii. Var (x) = E(x – μ)2

=∑(x – 3)2P(X = x). all x

X 1 2 3 4 5(x – 3) -2 -1 0 1 2(x – 3)2 4 1 0 1 4P(X = x) 0.1 0.3 0.2 0.3 0.1

= 4(0.1) + 1(0.3) + 0(0.2) + 1(0.3) + 4(0.1)= 1.4

iii. E(x2) = ∑x2P(X = x). all x

= 1(0.1) + 4(0.3) + 9(0.2) + 16(0.3) + 25(0.1)= 10.4

iv. Var(x) = E(x2) - μ2

= 10.4 – 9= 1.4

STANDARD DEVIATION :

Is the square root of its variance given by the following formula:

δ = √Var (x).

Question 9:From the question given above find the standard deviation for part (iv).

δ = √Var (x).δ = √1.4= 1.183215957= 1.18

CUMULATIVE DISTRIBUTION FUNCTION :

When we had a frequency distribution, the corresponding Cumulative frequencies were obtained by summing all the frequencies up to a particular value.In the same way if X is a discrete random variable, the corresponding Cumulative Probabilities are obtained by summing all the probabilities up to a particular value.

If X is a discrete random variable with Pdf P(X = x) for x = x1, x2. …xn then the Cumulative distribution function is given by:

67

F(t) = P(X <= t) = ∑t P(X = x).

x = x1

The Cumulative Distribution is sometimes called Distribution function.Question 10:The probability distribution for the random variable X is given below then constructs the Cumulative distribution table.

X 0 1 2 3 4 5 6P(X =x) 0.03 0.04 0.06 0.12 0.4 0.15 0.2

Solution: F(t) = ∑t P(X = x).

x = x1

SoF(0) = P(X <= 0) = 0.03F(1) = P(X <= 1) = 0.03 + 0.04 = 0.07F(2) = P(X <= 2) = 0.03 + 0.04 + 0.06 = 0.13 and so on.

The Cumulative Distribution table will be as follows:

X 0 1 2 3 4 5 6F(x) 0.03 0.07 0.13 0.25 0.65 0.8 1

Question 11:For a discrete random variable X the Cumulative distribution function F(x) is given below:

X 1 2 3 4 5F(x) 0.2 0.32 0.67 0.9 1

Find:a) P(x = 3).b) P(x > 2).

Solution:a. F(3) = P(x = 3) => P(x = 1) + P(x = 2) + P(x = 3)

= 0.67F(2) = P(x <= 2) => P(x = 1) + P(x = 2)

= 0.32

Therefore P(x = 3) = 0.67 – 0.32

= 0.35

b. P(x > 2) = 1 – P(x <= 2)= 1 – F(2)

68

= 1 – 0.32= 0.68

PROBABILITY DISTRIBUTION (CONTINUOUS RANDOM VARIABLES ):

A random variable X that can be equal to any number in an interval, which can be either finite or infinite length, is called a Continuous Random Variable.

PROBABILITY DENSITY FUNCTIONS :

There are 2 essential properties of Pdf: Because probabilities cannot be negative. The integral of a function must be non-

negative for all choices of interval [a, b] i.e. f(x) >= 0 for all values in the sample space for the random variable X.

Since the probability associated with the entire sample space is always 1. The integral of f(x) of the entire sample space = 1.

Question 1:

A continuous random variable has Pdf f(x) where f(x) = kx, 0<= x <= 4.i. Find the value of constant k.

ii. Sketch y = f(x).iii. Find P(1 <= X <= 2½).

Solution:

i. ∫ f(x) ∂x = 1. all x

∫04 kx ∂x = 1.

[kx2/2]04 = 1.

8k = 1

k = ⅛

ii. Sketch of y = f(x).

69

½ y = ⅛x

0 4

P(1<= x <= 2½) = ∫12½[⅛x]∂x.

= [x2/16]12½

= 0.328

Question 2:

A continuous random variable has Pdf f(x) where

Kx 0<= x <= 4.f(x)= k(4 – x) 2<= x <= 4

0 otherwise

a) Find the value of constant k.b) Sketch y = f(x).

Solution:

D =>∫ab

P ∂x + ∫ab

Q ∂x = 1.

=>∫02 kx ∂x + ∫2

4 k(4 – x) ∂x = 1.

=> [kx2/2]02 + [4xk - kx2/2]2

4 = 1.

=> [4k/2] – [0] + {[16k - 16k/2] – [8k - 4k/2]} = 1.

[2k] + {[8k] – [6k]} = 1. 4k = 1 k = ¼

c. Sketch y = f(x).

X 0 1 2 3 4Y 0 ¼ ½ ¾ 1

F(x) = kx.70

X 2 3 4Y ½ ¼ 0

F(x) = k(4 – x)

1

¾

½

¼

0 1 2 3 4

EXPECTED VALUE OR MEAN (CONTINUOUS RANDOM VARIABLE) :

For a continuous random variable X defined within finite interval [a, b] with continuous Pdf f(x) then the expected value or mean is given by:

E(x) = ∫ab x. f(x) ∂x

Question 3:

A continuous random variable has Pdf f(x) where

6/7x 0<= x <= 1.f(x)= 6/7x(2 – x) 1<= x <= 2

0 otherwise

i. Find E(x).ii. Find E(x2).

Solution:

D =>∫ab P ∂x + ∫a

b Q ∂x = 1.

E(x) = ∫ab

x. f(x) ∂x

E(x) = ∫01 6/7 x2 ∂x + ∫1

2 6/7 x2(2 – x)∂x

= 6/7[x3/3]01 + 6/7[2/3x3 – x4/4]1

2

= 6/7[⅓] + 6/7{16/3 – 4 – (⅔ - ¼)}

71

= 6/7[5/4]

= 15/14

E(x2) = ∫ab

x2. f(x) ∂x

E(x2) = ∫01 6/7 x3 ∂x + ∫1

2 6/7 x3(2 – x)∂x

= 6/7[x4/4]01 + 6/7[x4/2 – x5/5]1

2

= 6/7[¼] + 6/7{8 - 32/5 – (½ - 1/5)}

= 6/7[31/20]

= 93/70

VARIANCE AND STANDARD DEVIATION :

The variance and Standard Deviation associated with a continuous random variable X on the sample space [a, b] is given by:

Var (x) = ∫ab [x - E(x)]2 . f(x) ∂x or Var(x) = E(x2) - μ2 ∂x

and Standard deviation = √Var (x).E(x) = μ.

Question 4:A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4.Find:

a) E(x).b) E(x2).c) Var (x).d) The standard deviation of x.e) Var(3x +2).

Solution:

i. E(x) = ∫ab

x. f(x) ∂x

E(x) = ∫04 ⅛x2 ∂x

= ⅛[x3/3]04

72

= 8/3

ii. E(x2) = ∫ab

x2. f(x) ∂x

=∫04 ⅛x3 ∂x

= ⅛[x4/4]04

= ⅛(64)

= 8

iii. Var(x) = E(x2) - μ2 ∂x

= E(x2) - E2(x) ∂x

= 8 – (8/3)2

= 8/9

iv. Standard Deviation => = √Var (x).= √8/9

= 2√2/3

v. Var(3x + 2) = 9 Var(x) this has been obtained form the concept Var (ax)= Var a2(x).

= 9 (8/9)= 8

MODE :

The Mode is the value of X for which f(x) is greatest in the given range of X. It is usually to draw a sketch of y = f(x) and this will give an idea of the location of the Mode.For some Probability Density functions it is possible to determine the mode by finding the maximum point of the curve y = f(x) from the relationship f1(x) = 0.

f1(x) = d/∂x * f(x).

Question 5:A continuous random variable has Pdf f(x) where f(x) = 3/80(2 + x)(4 – x), 0<= x<= 4.

a) Sketch y = f(x).b) Find the mode.

73

Solution:X 0 1 2 3 4Y 24/80

27/8024/80

15/80 0a)

27/80 Mode

24/80 f(x) = 3/80(2 + x)(4 – x)

15/80

0 1 2 3 4

b) The mode => f(x) = 3/80(2 + x)(4 – x)

= 3/80(8 + 2x – x2) f1(x) = (2+ 2x) f1(x) = 0. 0 = 2+ 2x 2x = 2 x = 1

MEDIAN :

The median splits the area under the curve y = f(x) into 2 halves so if the value of the Median is m. Therefore the formula for the median is given by:

∫am

f(x) ∂x = 0.5.

F(m) = 0.5

Question 6:A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4.Find:

a. The median m.

Solution:

m => ∫am f(x) ∂x = 0.5.

F(m) = 0.5

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f(x) = ⅛x ∂x

0.5 = m2/16

m2 = 8

m = 2.83

CUMULATIVE DISTRIBUTION FUNCTION: F(x)

When considering a frequency distribution the corresponding cumulative frequencies were obtained by summing all the frequencies up to a particular value.In the same way if X is a continuous random variable with Pdf f(x) defined for a<=x<=b then the Cumulative Distribution Function is given by F(t):

F(t) = P(X <= t) = ∫at f(x) ∂x

PROPERTIES OF CDF:

F(b) = ∫ab f(x) ∂x = 1.

If f(x) is valid for - <= x <= then F(t) = ∫-t f(x) ∂x where the interval is

taken over all values of x <= t. The Cumulative distribution function is sometimes known as just as the

distribution function.

Question 6:A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0<= x<= 4.

Find:i. The Cumulative distribution function F(x).

ii. Sketch y = F(x).iii. Find P(0.3 <=x<= 1.8).

Solution:

i. F(t) = ∫at f(x) ∂x

F(t) = ∫0t ⅛x ∂x

= ⅛[x2/2]0t

= t2/16

75

F(t) = t2/16 0<=t<=4

NB: (1) F(4) = 42/16 = 1

0 x <= 0. F(x) = x2/16 0<= x <= 4

1 x >= 4

ii. Sketch y = F(x).

X 0 1 2 3 4Y 0 1/16

1/29/16 1

1F(x) = 1

9/16

1/2 F(x)= x2/16

1/16

0 1 2 3 4

iii. P(0.3 <= x <= 1.8) = F(1.8) – F(0.3)

F(1.8) = (1.8)2/16

= 0.2025

F(0.3) = (0.3)2/16

= 0.005625

Therefore P(0.3 <= x <= 1.8) = F(1.8) – F(0.3)= 0.2025 – 0.005625= 0.196875= 0.197

Question 7:A continuous random variable has Pdf f(x) where

x/3 0<= x <= 2.f(x)= -2x/3 +2 2<= x <= 3

0 otherwise

a. Sketch y = f(x).76

b. Find the Cumulative distribution function F(x).c. Sketch y = F(x).d. Find P(1 <= X <= 2.5)e. Find the median m.

Solution:i. Sketch y = f(x).

X 0 1 2Y 0 ⅓ ⅔

X 2 3Y ⅔ 0

⅔

y = x/3

⅓ y =-2x/3 +2

0 1 2 3

ii. CDF = F(t) = ∫0t x/3∂x

= [x2/6]0t

= t2/6

F(t) = x2/6 0<=x<=2

NB: F(2) = 22/6 = ⅔

F(t) = F(2) + (Area under the curve y = -2x/3 +2 between 2 and t)So

F(t) = F(2) + ∫2t

(-2x/3 +2) ∂x

= F(2) + [-x2/3 + 2x]2t

= ⅔ + {-t2/3 +2t – ( -4/3 + 4)}

= -t2/3 +2t – 22<= t <= 3

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NB: F(2) = -9/3 + 6 – 2 = 1

Therefore CDF = x2/6 0<= x <= 2.

f(x)= - x2/3 +2x -2 2<= x <= 3.1 x >= 3.

iii. Sketch of y = F(x).

y = 1 1

y = - x2/3 +2x -2 2/3

y = x2/6

1/3

0 1 2 3

iv. P(1 <= X <= 2.5) = F(2.5) – F(1) as 2.5 is in the range 2<= x <=3.

F(2.5) = - x2/3 +2x –2 F(2.5) = - (2.5)2/3 +2(2.5) –2 = 11/12

F(1) = x2/6 as 1 is in the range 0 <= x <= 2.

F(1) = x2/6

F(1) = 12/6

= 1/6

Therefore P(1 <= X <= 2.5) = F(2.5) – F(1)= 11/12 - 1/6

= 0.75

v. m => ∫am

f(x) ∂x = 0.5 where m is the median.

F(2) = ⅔ so the median must lie in the range 0 <= x <= 2. F(m) = m2/6

78

m2/6 = 0.5m2 = 3.m = 1.73

OBTAINING THE PDF FROM THE CDF :

The Probability Density Function can be obtained from the Cumulative Distribution function as follows:

Now F(t) = ∫at f(x) ∂x a<= t <= b.

Sof(x) = d/∂x * F(x). = F1(x).

NB: The gradient of the F(x) curve gives the value of f(x).

Question 8:A continuous random variable has Pdf f(x) where

0 x <= 0.F(x)= x3/27 0<= x <= 3

1 x >= 3.

Find the Pdf of X, f(x) and sketch y = f(x).

Solution:a. f(x) = d/∂x * F(x).

= d/∂x(x3/27).= 3x2/27

= x2/9

Therefore the Pdf is equal to:

x2/9 0<=x<=3f(x) =

0 otherwise.

b. Sketch of y = f(x).

1 y = x2/9

0 1 2 3

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Question 9:A continuous random variable X takes values in the interval 0 to 3.It is given that P(X > x) = a + bx3, 0 <= x <= 3.

i. Find the values of the constants a and b.ii. Find the Cumulative distribution function F(x).

iii. Find the Probability density function f(x).iv. Show that E(x) = 2.25.v. Find the Standard deviation.

Solution:a. P(X > x) = a + bx3, 0 <= x <= 3.

So P(X > 0) = 1 and P(X > 3) = 0.i.e. a + b(0) = 1 and a + b(27) = 0Therefore a = 1 and 1 + 27b = 0.

B = -1/27.

So P(X > x) = 1 - x3/27, 0 <= x <= 3.

b. Now P(X <= x) = x3/27 (CDF)

X3/27 0<=x<=3F(x) =

1 x > 3.

c. f(x) = d/∂x * F(x).= d/∂x(x3/27).= 3x2/27

= x2/9

d. E(x) = ∫ab

x. f(x) ∂x

=∫03 x. x2/9∂x

=∫03 x3/27∂x

= [x4/36]03

= 2.25

e. Var(x) = ∫ab [x - E(x)]2 . f(x) ∂x = ∫a

b x2.f(x)∂x - E2(X)

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=∫03 x4/9∂x – 2.252.

=[x5/45]03 - 5.0625.

= 0.3375

f. δ = √ Var (x).= √ 0.3375

= 0.581

RELATIONSHIPS AMONG PROBABILITY DISTRIBUTIONS :

JOINT PROBABILITY DISTRIBUTION :

Question 1:

2(X + Y - 2XY) 0<= X<=1, 0<= Y<=1Given f(X, Y) = 0 Otherwise

i. Show that this is a PDF.ii. Find P(0 <= X <=½), (0 <= Y <=¼).

iii. Find CDF.

Solution:

a) =∫ab∂X∫a

b∂Y

=∫01∂X∫0

1∂Y [2(X + Y - 2XY)]

= 2∫01∂X [(XY + Y2/2 - XY2)]0

1

= 2∫01∂X [(X + ½ - X)]

= 2[(X2/2 + ½X - X2/2)]01

= 2[½ + ½ - ½]= 2 * ½= 1

b) =∫ab

∂X∫ab

∂Y

=∫0½

∂X∫0¼

∂Y [2(X + Y - 2XY)]

81

= 2∫0½∂X[(XY + Y2/2 - XY2)]0

¼

= 2∫0½∂X[(¼X + 1/32 - 1/16X)]

= 2∫0½∂X[(3/16X + 1/32)]

= 2[(3/32X2 + 1/32X)] 0½

= 2[(3/128 + 1/64)]

= 2 * 5/128

= 5/64

c) F(u, v) =∫ab

∂u∫ab

∂v

=∫0x∂U∫0

y∂V[2(U + V – 2UV)]

= 2∫0x∂U[(UV + V2/2 - UV2)]0

y

= 2∫0x∂U[(UY + Y2/2 - UY2)]

= 2[(YU2/2 + UY2/2 - U2X2/2)]0X

= 2[(X2Y/2 + XY2/2 - X2X2/2)]

= X2Y + XY2 - X2X2

NB: Given CDF to get PDF we differentiate the function. Given PDF to get CDF we integrate the function.Given a PDF to prove that it is a PDF integrate until the answer is 1.Given PDF the marginal distribution is found by integrating.Given CDF the marginal distribution is found by differentiating.

Question 1:

Given the CDF = x2 + 3xyz + z.Find the corresponding PDF.

Solution:To get the corresponding PDF we differentiate with respect to x, y, z.

f(x, y, z) =x2 + 3xyz + z= d/∂x(2x + 3yz).= d/∂y(3z).= d/∂z (3).

= 3

UNIT 6 DECISION THEORY

HOURS: 20

DECISION THEORY :82

DECISION UNDER RISK :

A. THE EXPECTED VALUE CRITERION :

Muchadura buys and sell tomatoes at $3 and $8 a case respectively. Because tomatoes are perishable, they must be bought a day and sold after that, they become valuable.A 90 days observation of a business reveals the following information in cases.

Daily Sales Number Of Days Sold Probability Of Demand

10 cases 18 (18 / 90) = 0.211 cases 36 (36 / 90) = 0.412 cases 27 (27 / 90) = 0.313 cases 9 (9 / 90) = 0.1

Total number of days 90 = 1

The Probability of demand = Number of days sold / Total number of days.

There are only 4 sales volumes but their sequence in UNKNOWN. The problem is how many cases must Muchadura stock for selling the following day. If he stocks more or less than the demand on any day he will suffer some losses.Example:

If he stocks 13 cases on a particular day and the customer demand on that particular day is 10 cases, then he gets a profit of 3 cases.

Let buying price = $3.Let Selling price = $8.

NB: This method if getting Maximum profit = Selling Price – Buying Price per case = $5. Then 10 * 5 = $50 and perishable ones = 3 * 3 = $9 and then the Maximum profit = $50 - $9 = $41.For the 3 unsold cases he suffer a loss for each equivalent to the Buying price. All such losses are calculated in a conditional profit table. I.e. a table of profit on a given supply and demand levels as illustrated below.

STOCK ACTION EXPECTED PROFIT

Possible ProbabilityDemand of Demand Profits

10 11 12 13 10 11 12 13

83

10 50 47 44 41 0.2 10 9.4 8.8 8.211 50 55 52 49 0.4 20 22 20.8 19.612 50 55 60 57 0.3 15 16.5 18 17.113 50 55 60 65 0.1 5 5.5 6 6.5

Total 1 50 53.4 53.6 51.4

Muchadura must stock 12 cases which give the highest Expected profit of $53.60 on a daily bases to get the maximum profit over a long time.

B. EXPECTED PROFIT WITH PERFECT INFORMATION:

Suppose Muchadura had perfect information i.e. complete and accurate information about the future. Sales demand would still vary from 10 to 13 cases in their respective probabilities. But Muchadura would not know in advance how many cases would be needed each day.The table below shows the conditional profits value in such situation.



10 11 12 13 10 11 12 13

10 50 0.2 1011 55 0.4 2212 60 0.3 1813 65 0.1 6.5

Total 1 10 22 18 6.5

Maximum possible profit = 10 + 22 + 18 + 6.5.= $56.50

C. MINIMISING EXPECTED LOSS:

There are two types of Losses: Overstocking Under stocking.

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OVERSTOCKING:Means you will loss the buying amount.

UNDER STOCKING:It means you will loss the profit.

STOCK ACTION EXPECTED LOSS

Possible ProbabilityDemand of Demand Loss

10 11 12 13 10 11 12 13

10 0 3 6 9 0.2 0 0.6 1.2 1.811 5 0 3 6 0.4 2 0 1.2 2.412 10 5 0 3 0.3 3 1.5 0 0.913 15 10 5 0 0.1 1.5 1 0.5 0

Total 1 6.5 3.1 2.9 5.1

The Expected Minimum Loss is $2.9. The values above the zero diagonal are due to overstocking and those below are due to under stocking.

D. THE EXPECTED VALUE OF PERFECT INFORMATION:

The expected value of perfect information is calculated as:The expected Profit with perfect information – The expected value Criterion.= 56.50 – 53.60= 2.9In general the expected value of perfect information = Minimum Expected Loss.

E. ITEMS WITH A SALVAGE VALUE:

Main items do not become completely worthless after their prime deaths. They will have reduced values (Salvage values). The Salvage values must be considered when computing conditional profits or losses.

85

Consider the previous case when the cost price was $3 and selling price was $8. Any unsold cases will be disposed / salvage of at $4 the profit will be $4 - $3 = $1.



10 11 12 13 10 11 12 13

10 50 51 52 53 0.2 10 10.2 10.4 10.611 50 55 56 57 0.4 20 22 22.4 22.812 50 55 60 61 0.3 15 16.5 18 18.313 50 55 60 65 0.1 5 5.5 6 6.8

Total 1 50 54.2 56.8 58.2

Muchadura must stock 13 cases, which give the maximum profit of $58.20.

DECISION MAKING UNDER UNCERTAINITY :

Under conditions of uncertainty, only payoffs are known and nothing is known about the likelihood of each state of nature.Different persons have suggested several decisions rules for making decisions under such situations:

A. MAXIMIN (PESSIMISTIC) :

It is based upon the consecutive approach to assume that the worst possible is going to happen. The decision maker considers each alternative and locates the minimum payoff for each and then selects that alternative which maximizes the minimum payoffs.

STEPS INVOLVED:

i. Determine the minimum assured payoffs for each alternative.ii. Choose that alternative which corresponds to the Maximum of the above

minimum payoff.

B. MAXIMAX CRITERIA (OPTIMISTIC) :

Is based upon extreme optimistic, the decision maker selects that particular strategy which corresponds to the maximum of the maximum payoffs of each strategy.

86

STEPS INVOLVED:

i. Determine the maximum possible payoff for each alternative.ii. Select that alternative which corresponds to the maximum of the above

maximum payoff.

C. HURWICZ :

In order to overcome the disadvantage of extreme pessimistic (Maximin) and extreme Optimism (Maximax) criteria. Hurwicz introduced the concept of co-efficient of optimism or pessimism as .Therefore the Hurwicz concept is given by:

Maximin j {V (ai, j)} + (1 - ) Maximax j {V(ai, j)}

D. LAPLACE CRITERION :

This criterion is based on what is known as the principal if insufficient reason. Since the probabilities associated with the occurrences of 1, 2….n are unknown, we do not have enough information to conclude that these probabilities will be different.

For if this is not the case, we should be able to determine these probabilities and the situation will no longer be a decision under uncertainty. Thus because of insufficient reason to believe otherwise; the states of 1, 2….n are equally likely to occur.The Laplace principal assumes that of 1, 2….n are likely to occur therefore the probability = 1/n where n = number of occurrences.

E. MAXIMIN REGRET CRITERION (SALVAGE) :

This is a less conservative criterion. In this case a new loss matrix is created such that V(Ai, Qj) is replaced by R(Ai, Qj) which is defined by:

Max {V (Ai, Qj) – V(Ai, Qj)} if V is profitR(Ai, Qj)=

V(Ai, Qj) – Min {V(Ai, Qj)} if V is loss

R(Ai, Qj) is the difference between the best choice in column Qj and the values of V(Ai, Qj) in the same column.R(Ai, Qj) is a regret matrix.

Question 1:Given the following data:

Customers CategoryQ1 Q2 Q3 Q4

87

Supplier A1 5 10 18 25Level A2 8 7 8 23

A3 21 18 12 21A4 30 22 9 25

Using the above data:Find

i. Maximin Criterion.ii. Maximax Criterion.

iii. Hurwicz Criterion.iv. Laplace Criterion.v. (Salvage) Maximin Regret Criterion.

Solution:

a) Maximin Criterion. Customers Category

Q1 Q2 Q3 Q4 Maximin

Supplier A1 5 10 18 25 5 Level A2 8 7 8 23 7

A3 21 18 12 21 12A4 30 22 9 25 9

Maximin = 12The optimal decision is to take supply level A3

b) Maximax Criterion. Customers Category

Q1 Q2 Q3 Q4 Maximax


A3 21 18 12 21 21A4 30 22 9 25 30

Maximax = 30The optimal decision is to take supply level A4

c) Minimax Criterion. Customers Category

Q1 Q2 Q3 Q4 Minimax


88

A3 21 18 12 21 21A4 30 22 19 25 30

Minimax = 21The optimal decision is to take supply level A3

d) Hurwicz Criterion. Customers Category

Q1 Q2 Q3 Q4

Supplier A1 5 10 18 25 Level A2 8 7 8 23

A3 21 18 12 21A4 30 22 19 25

Hurwicz: = Maximin j {V (ai, j)} + (1 - ) Maximax j {V(ai, j)}

Let assume that α = ½.

Maximin Maximax Maximin + (1 - ) Maximax5 25 157 23 1512 21 16.59 30 19.5

The optimal decision is to take supply level A4

e) Laplace Criterion. Customers Category

Q1 Q2 Q3 Q4


A3 21 18 12 21A4 30 22 19 25

Laplace: = 1/n

Expected costs for different actions A1, A2, A3 and A4 are:

E(A1) = ¼(5 + 10 + 18 + 25) = 14.5E(A2) = ¼(8 + 7 + 8 + 23) = 11.5E(A3) = ¼(21 + 18 + 12 + 21) = 18

89

E(A1) = ¼(30 + 22 + 19 + 25) = 24

The optimal decision is to take supply level A2

f) Savage Minimax Regret Criterion. Customers Category

Q1 Q2 Q3 Q4


A3 21 18 12 21A4 30 22 19 25

Q1 Q2 Q3 Q4 Max{ R(Ai, Qj)}

A1 0 3 10 4 10 A2 3 0 0 2 3 Minimax

A3 16 11 4 0 16A4 25 15 11 4 25

The optimal decision is to take supply level A2 which gives Minimax.

Question 2:A firm has to decide on its advertising campaign. It has a choice between Tv, Newspaper, Poster and Radio advertising. The return on the advertising medium is measured by the number of potential customers who have the opportunity to see each medium. This will depend on the type of the weather. The figures in the table are the numbers of potential customers in thousands. The firm has funds for using only to one medium.

Poor Moderate Weather Good Excellent

Tv 200 190 170 130Newspaper 180 160 150 130Poster 110 140 140 190Radio 210 190 160 110

Decide which medium the firm should use based on the following information:i. Maximin Criterion.

ii. Maximax Criterion.iii. Hurwicz Criterion.iv. Laplace Criterion.v. (Salvage) Maximin Regret Criterion.

90

Solution:

a) Maximin Criterion.

Poor Mod Good Excellent Maximin

Tv 200 190 170 130 130 Newspaper 180 160 150 130 130

Poster 110 140 140 190 110Radio 210 190 160 110 110

Maximin = 130

The best medium to use for advertising is Tv or Newspaper since they have the highest number of (possible) potential customers of 130

b) Maximax Criterion.

Poor Mod Good Excellent Maximax

Tv 200 190 170 130 200 Newspaper 180 160 150 130 180

Poster 110 140 140 190 190Radio 210 190 160 110 210

Maximax = 210

The best medium to use for advertising is the Radio since it has the highest number of (possible) potential customers of 210.

c) Minimax Criterion.

Poor Mod Good Excellent Minimax

Tv 200 190 170 130 200 Newspaper 180 160 150 130 180

Poster 110 140 140 190 190Radio 210 190 160 110 210

91

Minimax = 180

The best medium to use for advertising is the Newspaper with 180 potential customers.

d) Hurwicz Criterion.

Poor Mod Good Excellent

Tv 200 190 170 130 Newspaper 180 160 150 130

Poster 110 140 140 190Radio 210 190 160 110

Hurwicz: = Maximin j {V (ai, j)} + (1 - ) Maximax j {V(ai, j)}

Let assume that α = 0.7

Maximin Maximax Maximin + (1 - ) Maximax130 200 151130 180 145110 190 109.7110 210 140

The best medium to use for advertising is the Tv since it has the highest number of (possible) potential customers of 151

g) Laplace Criterion.


Tv 200 190 170 130 Newspaper 180 160 150 130

Poster 110 140 140 190Radio 210 190 160 110

92

Laplace: = 1/n

Expected costs for different actions A1, A2, A3 and A4 are:

E(Tv) = ¼(200 + 190 + 170 + 130) = 172 500E(Newspaper) = ¼(180 +160 + 150 + 130) = 155 000E(Poster) = ¼(110 + 140 + 140 + 190) = 145 000E(Radio) = ¼(210 + 190 + 160 + 110) = 167 500

The best medium to use for advertising is Tv since it has the highest number of (possible) potential customers who amount to 172 500

h) Savage Minimax Regret Criterion.


Tv 200 190 170 130 Newspaper 180 160 150 130

Poster 110 140 140 190Radio 210 190 160 110

Poor Mod Good Excellent Max {R(Ai, Qj)}

Tv 90 50 30 20 90 Newspaper 70 20 10 20 70 Minimax

Poster 0 0 0 80 80Radio 100 50 20 0 100

The best medium to use for advertising is Newspaper, which minimize maximum view ship.

UNIT 7: THEORY OF GAMES:

HOURS: 20

GAME THEORY :

It deals with Decision under Uncertainty which involves 2 or more intelligent opponents in which each opponent aim to optimize his / her own decision at the expense of the other

93

opponent. Typical examples include launching an advertisement campaign. (2) Competing for products or planning war techniques for opposing enemy.In game theory an opponent is referred to as a Player. Each player has the number of choices, limited or unlimited called Strategies. The outcomes payoffs of a game are summarized as functions of different strategies for each player.A game with 2 players were a gain of one player is equal to the loss of another player is called a Two Person Zero Sum game.To illustrate two person zero sum game considers a coin-matching situation in which each of the players A and B select a Head H or Tail T.If the outcome match Head or Tail Player A wins $1 from Player B otherwise Player A loses $1 to Player B.In this game each player has 2 strategies Head or Tail which yields the following 2*2 game matrix expressed in terms of payoff to Player A.

Player B H T

Player A HT

OPTIMUM SOLUTION OF TWO PERSON ZERO SUM GAME :

Is obtained by using minimax Maximin criterion according to which Player A (whose strategies represents rows) select a strategy (mixed or Pure) which maximize his minimum gains, the minimum being taken over all the strategies of Player B.In similar way Player B selects his strategy that minimize his maximum loses.

VALUE OF THE GAME :

Is the maximum guaranteed gain to Player A or the minimum possible loss to Player B denoted by V.When Maximin possible value is equal to Minimax value of the corresponding pure strategies, the game is said to have the Optimum Strategies and has a Saddle point.

Player B1 2 3 4

MaximinPlayer A 1 2

2 53 -4

94

1 -1-1 1

8 2 9 56 7 187 3 -4 10

5

Minimax 8 5 9 18

The game has a Saddle point which exists at row 2 column 2 and the value of the game = 5.

Player B1 2 3

MaximinPlayer A 1 1

2 13 1

Minimax 6 3 6

NB: The optimum solution which shows that Maximin <> Minimax means it is a mixed strategy i.e. there is no agreement (poor strategy). Below and above show that there is no Saddle point which means Maximin <> Minimax.

Player B1 2 3 4

MaximinPlayer A 1 -10

2 13 24 -1

Minimax 8 7 15 4

Consider the following game G.

Player B B1 B2 Maximin

Player A A1 2

A2 -1

Minimax 2 6

Maximin = Minimax which is a condition strictly determinable game hence the game is strictly determinable whatever maybe. The value of the game = 2 with the best strategy for Player A from A1 and the best strategy for Player B from B1.

Question 1:Find the range of values of P and Q, which will render the entry (2,2) a Saddle point of the game.

Player B B1 B2 B3

MaximinPlayer A A1 2

95

1 3 62 1 36 2 1

5 -10 9 06 7 8 18 7 15 23 4 -1 4

2 6-2 -1

2 4 510 7 q4 p 8

A2 7A3 4

Minimax 10 7 8

Ignoring values of P and Q determine the Maximin and Minimax value of the payoff matrix. The Maximin =f and Minimax = f thus there exists a Saddle point at position (2,2). This impose a condition on P as P<= f and on Q as Q >= f.Hence the required range for P and Q is = 7<= Q, P<= 7.

Question 2:For what values of is the game with the following payoff. Payoff is strictly determinable.

Player B B1 B2 B3

MaximinPlayer A A1 2

A2 -7A3 -2

Minimax -1 6 2

This shows that the value of the game (V) lies between –1 and 2.That is –1 <= V <= 2. For a strictly determinable we have –1 <= <= 2.

GRAPHICAL SOLUTION OF 2 BY N & M BY 2 GAMES :

The graphical solution of 2 by N and M by 2 is only applicable to games in which at least on of the Players have 2 strategies only.

Consider the following 2 by N games.

Player B

96

6 2-1 -7-2 4

Y1 Y2 Y3

Player A X1

X2

The expected payoff of the corresponding to the pure strategies of B are given below:

This shows that A’s average payoff varies linearly with X1. According to the Minimax criterion for mixed strategies games Player A should select the value V that minimize his maximum expected playoffs. This may be done by plotting a straight line as function of X1.

Question 3:Consider the following 2 by 4 games.

Player B1 2 3 4

Player A 1

2

97

A11 A12 A1n

A21 A22 A2n

B’s Pure Strategies A’s Expected Payoff

(A11 –A12) X1 + A21

(A21 – A22) X1 + A22

N (A2n –A2n) X1 + A2n

2 2 3 -14 3 2 6

Solution:

6 65 54 43 32 2

1 1 X1 0 1/2 0 X1

- 1 -1 -2 -2

This is a point of integration of any 2 of the lines or more i.e. lines 2, 3 and 4.A’s optimum strategy is X1 = ½ and X2 = 5/2 and the value of the game is obtained by substituting X1 in the equations of any of the lines passing the intersection of the point.The value of the game = 5/2.

Question 4:Consider the following 2 by 3 game, find B’s pure strategies and A’s expected payoff using the graphically method.

Player B 1 2 3

Player A 1

2

98


(2 – 4) X1 + 4 (2 – 3) X1 + 33 (3 – 2) X1 + 24 (-1 –7) X1 + 6


- 2X1 + 4 -X1 + 3

3 X1 + 24 - 7X1 + 6

1 3 118 5 2

Solution:

11 1110 109 98 87 76 65 54 H 43 32 2

1 1 X1 0 0 X1

- 1 Maximin -1 -2 -2

Since Player A wishes to Maximize his Minimum expected payoff we consider the highest point of intersection on the lower elevation of A’s expected payoff equations.This point it represents the maximum value of the game for A.Lines 2 and 3 passes through it defines the relative moves B2 and B3 it along B needs to play.

The solution of the original 2 * 3 game reduces to a 2 by 2 payoff matrix.

Player B B1 B2

Player A A1 X1+ X2 = 1

A2 Y1 + Y2 = 1

99


- 7X1 + 8 - 2X1 + 5

3 9X1 + 2

3 11

5 2

Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained by: X1 = 3 – 11 / (3 + 2) – (5 + 11) = 8 / 11

X2 = 1 - 8 / 11 = 3 / 11

The value of the game V = (2 * 3) – (5 * 11) / (3 + 2) – (5 + 11)

= 49 / 11

Question 4:Consider the following 2 by 4 game, find B’s pure strategies and A’s expected payoff using the graphically method.

Player B 1 2 3 4

Player A 1

2100

1 3 -3 72 5 4 -6

Solution:

11 1110 10

9 98 87 76 65 54 43 32 2

1 1 X1 0 0 X1

- 1 H -1 -2 -2 -3 Maximin -3 -4 -4 -5 -5 -6 -6


Player B B1 B2


A2 Y1 + Y2 = 1

101


- X1 + 2 - 2X1 + 5

3 - 7X1 + 44 13X1 - 6

-3 7

4 -6

Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained by: X1 = -3 – 7 / (-3 + -6) – (4 + 7) = 11 / 20

X2 = 1 - 11 / 20 = 9 / 20

Y1 = -6 – 7 / (-3 + -6) – (4 + 7) = 13 / 20

Y2 = 1 - 13 / 20 = 7 / 20

The value of the game V = (-2 * -6) – (7 * 4) / (3 + 2) – (5 + 11)

= 40 / 20 = 2

MINIMAX PROBLEMS :

Obtain the optimum strategy for both persons and the value of the game for zero person sum game whose payoff matrix is as follows:

Player B 1 2

Player A 123456

Solution:

102

1 -33 5

-1 6 4 1 2 2 -5 0

A’s Pure Strategies B’s Expected Payoff

- 4Y1 - 3 - 2Y1 + 5

3 - 6Y1 + 64 5Y1 + 15 2Y1 + 26 -5Y1 + 0

11 1110 10

9 Minimax 98 87 H 76 65 54 43 32 2

1 1 Y1 0 0 Y1

- 1 -1 -2 -2 -3 -3 -4 -4 -5 -5 -6 -6 -7 -7


Player B B1 B2


A2 Y1 + Y2 = 1

Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained by: X2 = 3 – 5 / (3 + 1) – (4 + 5) = 2 / 5

X1 = 1 - 2 / 5 = 3 / 5

Y1 = 1 – 5 / (3 + 1) – (4 + 5) = 4 / 5

Y2 = 1 - 4 / 5 = 1 / 5

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3 5

4 1

The value of the game V = (1 * 3) – (4 * 5) / (3 + 1) – (4 + 5)

= 17 / 5

GENERAL SOLUTION OF M * N RECTANGULAR GAMES :

In a rectangular game with M * N payoff matrix if there does exists any saddle point and is not possible to reduce the size of the game to 2 by 2 payoff matrix.The following methods are generally used to solve the game:

Linear programming method. Iterative Method.

Question 5:Solve the following game; consider the game problem in which B’s linear programming problem.

Player B 1 2 3

Maximin Player A 1 2

2 23 1

Minimax 8 8 8

Player B Y1 Y2 Y3

Player A X1

X2

X3

B’s linear programming problem is as follows:

Maximize Z = Y1 + Y2 + Y3

Subject to:8Y1 + 4Y2 + 2Y3 <= 1 {1}2Y1 + 8Y2 + 4Y3 <= 1 {2}Y1 + 2Y2 + 8Y3 < = 1 {3}Y1 , Y2 , Y3 >= 0

TABLEAU 1:

104

8 4 22 8 41 2 8

8 4 22 8 42 2 8

Y1 Y2 Y3 S1 S2 S3 Solution

S1 4 2 1 0 0 1

S2 2 8 4 0 1 0 1

S3 1 2 8 0 0 1 1

Z 1 1 1 0 0 0 0

PC

TABLEAU 2:Y1 Y2 Y3 S1 S2 S3 Solution

Y1 0.5 0.25 0.125 0 0 0.125

S2 0 7 3.5 -0.25 1 0 0.75

S3 0 1.5 7.75 -0.125 0 1 0.875

Z 0 0.5 0.75 -0.125 0 0 -0.125

PC

TABLEAU 3:


Y1 1 0 0 0.107 -0.072 0 0.072

Y2 0 0.5 0.036 0.143 0 0.107

S3 0 0 7 -0.179 -0.215 1 0.715

Z 0 0 0.5 -0.143 -0.072 0 -0.1785

TABLEAU 3:


Y1 1 0 0 0.107 -0.072 0 0.072

Y2 0 1 0 0.049 0.159 -0.072 0.057

Y3 0 0 -0.026 -0.031 0.143 0.102

105

8

1

1

1

Z 0 0 0 -0.156 -0.087 -0.072 -0.2295

Conclusion:Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, Y1= 0.072, Y2= 0.057 and Y1= 0.102 producing Z = $0.2295.

UNIT 8: INVENTORY MODELLING:

HOURS: 20

INVENTORY SYSTEM MODEL :

Inventory deals with manufacturing sufficient stock of goods (parts and raw materials) to ensure a smooth operation of a business activity.

GENERALIZED INVENTORY MODEL :

The ultimate objective of inventory model is to answer the following two questions: How much to order. When to order.

The answer to the 1st question is expressed in terms of economic order quantity (EOQ), which is the optimum amount that should be ordered every time an order is placed and may vary with time depending on the situation under consideration.

The 2nd part depends on the type of the inventory type. If the inventory system requires periodic review i.e. equal time intervals (e.g. every week or month). The time for acquiring a new order usually coincides with the beginning of each time interval.If a system is of continuous review type a reorder point is usually specified by an inventory level at which a new order must be placed.We can express the solution of the general inventory problems as follows:

PERIODIC REVIEW CASE:

Receive a new order of the amount specified by the order quantity at equal intervals of order time, which can be weekly or monthly.

CONTINUOUS REVIEW CASE:

When the inventory level reaches the re-order point places an order whose size is equal to the quantity.The order quantity and re-order point are normally determined by minimizing the total inventory cost, the point which can be expressed as a function of its principle components in the following manner.

106

Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.

PURCHASING COST :

Purchasing cost is the price of the quantity of goods to be supplied. The purchasing cost becomes an important factor when the commodity unit prices become dependant on the size of the order. This situation is normally expressed in terms of quantity discount or price break where the unit price of an item decrease with the increase of order quantity.

SETUP COST / ORDERING COST :

Represents a fixed charge incurred when an order is placed. Thus to satisfy the demand for a given time period, ordering of smaller quantities will result in a higher set up cost during the period than if the demand is satisfied by placing large orders.

HOLDING COST / CARRYING COST :

Represents the cost of carrying inventory or represents the cost of maintaining inventory e.g. warehouse rental, security, handling, depreciation, stock maintenance, insurance and loses of interest on capital.

SHORTAGE COST :

Is a penalty incurred when we run out of stock of the needed commodity. It generally includes costs due to the loss of customer’s goodwill as well as potential loss in income. We need to minimize the total cost involved in stocking operations by determining the optimum quantity for the replenishment order Q.

INVENTORY NOTATION :

Tc – Total inventory cost per unit time.

D – Total demand per unit time.

Ci – Cost of production or purchase per unit expressed as $ per unit.

Q – Quantity per order.

Q* - Optimum quantity = EOQ.

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H – Holding cost or Carrying cost per unit of inventory per time.

T – Time between orders or length of inventory cycle Q/D.

K – Cost of processing an order or Ordering cost / Setup cost.

L – Leading time.

Cs – Shortage cost.

Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.

Tc = CiD + QH/2 + DK/Q + Cs.

We need to minimize Tc depending on the quantity Q hence by differentiation we get

(Tc)/Q = (QH/2)/ Q + (DK/Q)/Q + (CiD)/Q + (Cs)/Q

0 = H/2 – DKQ2

0 = H/2 – DK/Q2

Multiply both sides by 2Q2 we get 0 = HQ2 – 2DK

0 = HQ2 – 2DKHQ2 = 2DKQ2 = 2DK/H

Q* = 2DK/H (This is the formula for EOQ)

Economic Order Quantity is a deterministic model referred to as Wilson Economic Lot Size.

EOQ ASSUMPTIONS :

Tc = CiD + QH/2 + DK/Q + Cs.

The above model will hold for the following assumptions:

1) Demand should be constant.2) Replenishment should be instantaneous.

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3) Shortages are not allowed.4) Fixed quantity order per inventory order level should be maintained.

Question 1:The demand for an item is 18000 units per month and the holding cost per unit is $14.40 per year and the cost of ordering is $400. No shortages are allowed and the replenishment rate is instantaneous.

a) Determine the optimum order quantity.b) The total cost per year of the inventory if the cost of one unit is $1.c) The number of orders per year.d) The time between orders.

Solution:Since all the information asked for are years. Convert all information in months to years.D = 18000 * 12 = 216000 per year.H = $14.40 per year. K = $400.

EOQ => Q* = 2DK/H

2* 21600 * 400/14.40

172800000/14.40

12000000

3464.101615

3464

Tc = CiD + QH/2 + DK/Q.

= (1 * 216000) + (3464 * 14.40)/2 + (216000 * 400)/3464

= 216000 + 24940.80 + 24942.26

= 265883.06

The number of orders per year = D/Q

=216000/3464= 62.36 orders

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The time between orders = Q/D

= 3464/216000= 0.016= 0.02

Question 2:The demand for a particular item is 18000 units per year. Holding cost per unit is $1.20 per year and the cost of procurement is $400. No shortages are allowed and the replenishment rate is instantaneous.

a.Determine the optimum order quantity.b.The number of orders per year.c.The time between orders.d.The total cost per year of the inventory if the cost of one unit is $1.

Solution:D = 18000 per year.H = $1.20 per year. K = $400.

EOQ => Q* = 2DK/H

2* 18000 * 400/1.20

14400000/1.20

12000000

3464.101615

3464

The number of orders per year = D/Q

=18000/3464= 5.196 orders

The time between orders = Q/D or 365/5.196 = 70.24 days

= 3464/18000= 0.1924= 0.19 days


= (1 * 18000) + (3464 * 1.20)/2 + (18000 * 400)/3461

110

= 18000 + 206.40 + 2078.52

= 22156.92

Question 3:Suppose a company has soft drinks products. It has a constant annual demand rate of 3600 cases. A case of soft drinks cost the Harare drinks company $3. Ordering cost is $20 per order and holding cost is charged at 25% of the cost per unit. There are 250 working days per year and the lead-time is 5 days. Identify the following aspects of inventory policy:

i. Economic order quantity.ii. Re-order point.

iii. Cycle time (Length) = Q/D.iv. The total annual cost.

Solution:D = 3600.H = 25% of $3. K = $20 per order.Ci = $3 per unit.

i. EOQ => Q* = 2DK/H

2* 3600 * 20/0.75

144000/0.75

192000

438.178

438

ii. Re-order point = Daily Demand * Lead-time.

= 3600/250 * 5= 72 crates

iii. Cycle time = Q/Daily demand.

Daily Demand is given by = 3600/250 = 14.40

Cycle Time = 438/14.40

= 30.42 days

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iv. Tc = CiD + QH/2 + DK/Q.

= (3 * 3600) + (438 * 0.75)/2 + (20 * 3600)/438= 10800 + 164.25 + 164.38356= 11128.633= 11128.63

` RE-ORDER POINT :

Is a point in which a new order should be placed.

GENERAL RULES TO DETERMINE THE RE-ORDER POINT :

i. Determine the number of days the order should be placed (Cycle length) = Q/D.ii. Determine the number of days which is the minimum level to re-order new entities

and is given by Lead-time – Cycle Length (Q/D) where Lead-time is the time between the placement of an order and its receipt.

iii. Determine the level of inventory (stock) to re-order new items and is given by: Number of days which gives minimum level to order new items * Demand.

Question 4:The daily demand for a commodity is approximately 100 units. Every time an order is placed a fixed cost of $100 is incurred and daily holding cost per unit inventory $0.02.If the lead-time is 12 days.Determine:

i. Economic order quantity.ii. Re-order point.

iii. Cycle time.iv. Minimum inventory level.v. Level of inventory stock.

Solution:D = 100.H = $0.02. K = $100.

i. EOQ => Q* = 2DK/H

2* 100 * 100/0.02112

20000/0.02

1000000

1000

ii. Re-order point = Demand * Lead-time= 100 * 12= 1200

iii. Cycle time = Q/D = 1000/100

= 10.

vi. Minimum inventory level = Lead-time – Cycle Length.= 12 – 10= 2 days

vii. Level of inventory stock = Minimum inventory level * Demand.= 2 * 100= 200 orders.

EOQ WITH GRADUAL REPLENISHMENT / ECONOMIC PRODUCTION LOT SIZE:

Suppose a demand of an item in a company is D units per unit time and the company can produce them at a rate of R units per unit time where R>D. The cost of set-up is $K and the holding cost per unit time is $H. We wish to determine the Optimum manufacturing quantity and the total cost per year assuming the cost of one unit is $Ci. It is assumed that shortages are not allowed.

EOQ => Q* = 2DK/H(1 – D

/R)

Question 5:A company uses 100000 units per year that cost $3 each. The carrying costs are 1% per month and the ordering cost are $2.50 per year.What is the economic order quantity made the items themselves on a machine with a potential capacity of 600000 units per year.

Solution:D = 100000.K = $2.50.

113

H = 1% = 0.01 * 12 * 3 = 0.36

EOQ => Q* = 2DK/H(1 – D

/R)

= 2* 100000 * 2.50/0.36(1 – 100000

/600000)

= 500000/0.36(1 – 1/6)

=1666666.667

= 1290.994449

=1291.

Question 6:A company manufactures an item, which is also used in the company. The demand of this item is 18000 units per year and the production rate is 3000 per month. The cost of set up is $500. The holding cost of 1 unit per month is $0.15.Determine the optimum manufacturing quantity and the total cost per year assuming the cost of 1 unit is $2 and shortages are not allowed.

Solution:D = 18000 per year.H = $0.15 * 12 = $1.8 per year. K = 500Ci = $2R = 3000 * 12 = 36000 per year.

EOQ => Q* = 2DK/H(1 – D

/R)

= 2* 18000 * 500/1.8 *(1 – 18000

/36000)

114

= 18000000/0.36(1 – 0.5)

== 18000000/0.18

=3162.27766

= 3162


= (2 * 18000) + (3162 * 1.8)/2 + (18000 * 500)/3162= 36000 + 2845.8 + 2846.29981= 41692.09981= 41692

INVENTORY CONTROL (TYPES OF CONTROL SYSTEMS) :

There are 2 two basic inventory control systems: Reorder Level Periodic Review system.

PERIODIC REVIEW SYSTEM :

This is the reviewing of all stocks at fixed intervals periodically. Therefore absolute stocks can be eliminated. Variable quantities can be ordered.

THE REORDER LEVEL SYSTEM :

The reorder level system results in fixed quantities being ordered at variable intervals dependent upon demand. The reorder level system usually has three (3) control levels namely:

The reorder level.

115

The minimum level. The maximum level.

NB: The Reorder Level is calculated so that if the worst

anticipated position occurs, Stock would be replenished in time.

The Minimum Level is calculated so that management will be warned when demand is above average. There may be no danger but the situation needs watching.

The Maximum Level is calculated so that management will be warned when demand is the minimum anticipated and quencequently the stock level is likely to rise.

Question 7:The following is an illustration of a company with a simple manual reorder level system.

Normal Usage = 110 per day.Minimum Usage = 50 per day.Maximum Usage = 140 per day.Lead time = 25 – 30 days.EOQ = 5000

Using the above data calculate the various control levels i.e. Reorder level. Minimum Usage. Maximum Usage. [20 marks].

Solution:a) Reorder Level = Maximum Usage * Maximum Lead time.

= 140 * 30= 4200 units

b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage)

= 4200 – (27.5 * 110)= 4200 – 3025= 1175 units

c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage)= 4200 + 5000 – (25 * 50)= 7950

Question 8:

116

The following is an illustration of a company with a simple manual reorder level system.Normal Usage = 560 per day.Minimum Usage = 250 per day.Maximum Usage = 750 per day.Lead time = 15 – 20 days.EOQ = 10000

Using the above data calculate the various control levels i.e. Reorder level. Minimum Usage. Maximum Usage. [20 marks].

Solution:a) Reorder Level = Maximum Usage * Maximum Lead time.

= 750 * 20= 15000 units

b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage)

= 15000 – (17.5 * 560)= 15000 – 9800= 5200 units

c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage)= 15000 + 10000 – (15 * 250)= 21250

UNIT 8: REGRESSION ANALYSIS AND CORRELATION:

HOURS: 20

CORRELATION :

Correlation is a measure of strength of a linear relationship between 2 sets of numbers. There are 2 measures of correlation namely:

The Pearson Product Moment Correlation Coefficient represented by r. The Spearman’s Rank Correlation Coefficient represented by R.

A coefficient of +1 indicates a perfect Positive relationship whilst a coefficient of – 1 shows perfect negative correlation.Hence the correlation coefficient will have values ranging between – 1 and + 1 inclusive.A coefficient of zero implies no correlation.

SCATTER DIAGRAM:

117

It is a way of representing a set of data by a scatter of plots or dots. One variable is plotted on the X-axis and another on the Y-axis. Normally variable X is the one controlled over (independent variable) and Y variable is the variable that you are interested in (dependent variable).

POSITIVE CORRELATION: NEGATIVE CORRELATION:* * * * * * *

Delivery * * * *Time * * *

* * * ** * * ** *

Number of deliveries Temperature

NO CORRELATION: NON-LINEAR CORRELATION:* * * * * *

* * * * * * **Salary * * * * * * **

* * * * * * **

* * * * * * * *** * * * * * * **

Age * *

Quantity produced

The first diagram is a positive because number of deliveries increase so apparently does the delivery time.The second is a negative correlation because as air temperature increases the heating cost falls.For the third correlation shows that there is no correlation existing between salary and age of an employee.For non-linear correlation it suggest that quantity produced and efficiency are correlated but not linearly.

Given two sets of data represented by the variables X and Y the Product Moment Correlation Coefficient is given by:

r = (X - ) (Y – Ý)

(X - )2 * (Y – Ý)2

OR

n ΣXY - ΣX · ΣY r =

2 2 2 2 [n Σ X - (ΣX)] [n Σ Y - (ΣY)]

118

Question 1:

Calculate r for the data given below:

X 15 24 25 30 35 40 45 65 70 75Y 60 45 50 35 42 46 28 20 22 15

Solution:

X Y XY X2 Y2

15 60 900 225 360024 45 1080 576 202525 50 1250 625 250030 35 1050 900 122535 42 1470 1225 176440 46 1840 1600 211645 28 1260 2025 78465 20 1300 4225 40070 22 1540 4900 48475 15 1125 5625 225424 363 12835 21926 15123


2 2 2 2 [n Σ X - (ΣX)] [n Σ Y - (ΣY)]

10 * 12835 - 424 * 363 r =

10* 21926 - 179776 * 10 * 15123 - 131769

128350 - 153912 r =

119

39484 * 19461

- 25562 r =

27719.9959

r = - 0.922150238 r = - 0.92

Question 2:

Calculate r for the data given below:

X 1 2 3 6 5 6Y 5 10.5 15.5 25 16 22.5

Solution:

X Y XY X2 Y2

1 5 5 1 252 10.5 21 4 110.253 15.5 46.5 9 240.254 25 100 16 6255 16 80 25 2566 22.5 135 36 506.2521 94.5 387.5 91 1762.75


2 2 2 2 [n Σ X - (ΣX)] [n Σ Y - (ΣY)]

6 * 387.5 - 21 * 94.5 r =

6* 91 - 441 * 6 * 1762.75 – 8930.25

120

2325 – 1984.5 r =

546 - 441 * 10576 – 8930.25

340.5 r =

415.7598467

r = 0.819

(B) THE SPEARMAN’S RANK CORRELATION COEFFICIENT:

When two sets of data are given positions or ranked, we use Spearman’s Rank Correlation Coefficient represented by R. It is given by: R = 1 - 6 Σd2

n(n2 -1)

Where d = difference between pairs of ranked values.n = number of pairs of ranking.1 = is an independent number.

Question 2:A group of eight students were tested in Maths and English and the ranking in the two subjects are given below:

Student A B C D E F G HMaths 2 7 6 1 4 3 5 8English 3 6 4 2 5 1 8 7

Solution:

121

Students Maths English D D2

A 2 3 - 1 1B 7 6 1 1C 6 4 2 4D 1 2 -1 1E 4 5 -1 1F 3 1 2 4G 5 8 -3 9H 8 7 1 1

TOTAL = 22

R = 1 - 6 Σd2

n(n2 -1)

R = 1 – 6 * 22 8(82 -1)

R = 1 - 132 8* 63

R = 1 – 0.2619 = 0.738

REGRESSION :

This is a statistical technique that can be used for short to medium term forecasting. It seeks to establish the line of “best fit” for some observed data.These are 2 ways of obtaining the trend of a set of data:

The graphical method. The method of least squares.

THE GRAPHICAL METHOD :

In this method a scatter diagram is plotted and a straight line is produced through the middle of the points representing the trend line.This line can then be used (when extended) to predict a future value.

y

* Trend line or line of best fit. * * *

122

* * * * * *

* x

METHOD OF LEAST SQUARES :

This method gives an equation of the trend line or regression line. The equation is given by:

Y = A + Bx Where B = n ΣXY - ΣX · ΣY

n ΣX2 - (ΣX)2

and A = Y - BX

x is an independent variable.n is the number of items in the list.

Question 3:Consider the following data:

X 1 2 3 4 5 6Y 6 4 3 5 4 2

i. Find the equation of the regression line.ii. Find Y when X = 4 using the equation.

Solution:

X Y XY X2

1 6 6 12 4 8 43 3 9 94 5 20 164 4 20 256 2 21 3621 24 75 91

123

B = n ΣXY - ΣX · ΣY n ΣX2 - (ΣX)2

B = 6(75) – 21 * 24 6 (91) – (21)2

= 450 – 504 546 – 441

= -18/35

X = X/n 21/6 = 3.5

Y = Y/n 24/6 = 4

A = Y - BX

= 4 – (7/2 * - 54/105)= 4 + 9/5

=29/5

Y = A + Bx = 29/5 – 18/35X

(ii) When X = 4 Y = A + Bx

= 29/5 – 18/35* 4= 29/5 – 72/35

= 131/35 or 326/35

Question 4:For data below calculate the equation of the regression Y upon X.

X 98 78 74 80 80 83 95 100 97 75Y 14 9 10 11 10 11 12 13 11 9

Solution:X Y XY X2

98 14 1372 960478 9 720 608474 10 740 547680 11 880 6400

124

80 10 800 640083 11 913 688995 12 1140 9025100 13 1300 1000097 11 1067 940975 9 675 5625860 110 9589 74912

B = n ΣXY - ΣX · ΣY n ΣX2 - (ΣX)2

B = 10(9589) – 860 * 110 10 (74912) – (860)2

= 95890 – 94600 749120 – 739600= 0.1355

X = X/n 860/10 = 86

Y = Y/n 110/10 = 11

A = Y - BX= 11 – (0.1355 * 86)= 11 – 11.653= - 0.6533

Y = A + Bx = - 0.6533 + 0.1355X

UNIT 10 INTRODUCTION TO QUEUEING THEORY

HOURS: 20

QUEUEING MODELS :

It involves a customer’s standpoint where a waiting line is created. Customers arrive at the facility and they join a waiting line (or Queue). The server chooses a customer from the waiting line to begin service. Upon the completion of a service, the process of choosing a new (waiting) customer is repeated.It is assumed that no time is lost between the completion of a service and the admission of a new customer into the facility.

125

The major actors in queuing situation are Customer and the Server. The interaction between the Customer and the Serve are of interest only in as far as it relates to the period of time the customer needs to complete a service. From the standpoint of customer arrivals we will be interested in the time intervals that separate successive arrivals.

The customer arrivals and service times are summarized in terms of probability distributions normally referred to as: arrivals and service time distribution.These distributions may represent situations where customers arrive and are served individually {banks or supermarkets}.Customers may arrive and or be served in groups are referred to as Bulk queues.In queue model the manner of choosing customer from the waiting line to start service is referred to as Service discipline. The Service discipline occurs as FCFS rule (first come first service), LCFS rule (last come first served) and SIRO (service in random order).Customers arriving at a facility may be put in Priority queues such that those with higher priority will receive preference to start service first.

The facility may include more than one server thus allowing as many customers as the number of servers to be serviced simultaneously this is referred as Parallel service.Facilities which comprise a number of series stations through which the customer may pass before service is completed (e.g. processing of a product on a sequence of machines) is referred to as Queues in series or Tandem queues.

A combination of Parallel and Series is called Network queues.Calling source is the source from which calls for service (arrivals of customer) are generated.Queue size the number of limited customers that may be allowed at a facility may be because of space limitation (e.g. car spaces in a drive-in bank).

PURE BIRTH MODEL / PROCESS :

Pure birth model is process situation where customer arrive and never leave. E.g. issuing of birth certificates for newborn babies. These certificates are normally kept as permanent records in a Central office administered by the state health department.The birth of new babies and issuing of birth certificates is a completely random process that can be described ay a Poisson distribution.

Assuming that is the rate at which birth certificates are issued, the pure birth process of having n arrivals (birth certificates) during the time period t is given as:

Pn(t) = (t)n e-t

n!

126

Where n = 0, 1, 2, 3… and is the rate of arrival per unit time with the expected number of arrivals during t being equal to t.

Question 1:Suppose that the birth in a state are spaced over time according to a an exponential distribution with one birth occurring every 7 minutes on the average.

Solution:Since the average inter arrival (inter birth) time = 7 minutes the birth rate in the state is computed as:

= 24{hours} * 60{minutes} = 205.7 births per day. 7

The number of birth in the state per year =:

t = 205.7 * 365 = 75.080 births per year

The probability of no births in any one-day is computed as:

Pn(t) = (t)n e-t

n!

P0(1) = (205.7 * 1)0 e-205.7 * 1 0. 0!

Suppose that we are interest in the probability of issuing 45 birth certificates by the end of a period of 3 hours given that 35 certificates were issued in the first 2 hours.We observe that since birth occur according to a Poisson process the required probability reduces to having 45 –35 = 10 births in one (3 – 2) hours.

Given = 60/7 = 8.57 births per hour.

P10(1) = (8.57 * 1)10 e-8.57 * 1 10!

P10(1) = (2137018192) * (0.000189712)

3628800

0.11172

Question 1:

127

Suppose that the clerk who enters the information from the birth certificate into a computer normally wait at least 5 certificates to be accumulated. What it the probability that the clerk will be waiting a new batch every hour.

Solution: = / = births per hour.

P5(1) = (* 1)5 e- * 1 5!

0.92868

PURE DEATH MODEL / PROCESS :

Pure death model is process situation where customers are withdrawals. Consider the situation of stocking units of an item at start of the week to meet customer’s demand during the week. If we assume that customer demand occurs at the rate units per week and that the demand process is completely random, the associated probability of having n items remaining in stock after time t is given by the following Truncated Poisson distribution.

Pn(t) = (t)N - n e-t where n = 1, 2, 3 … N.

(N – n)!

128

Pn(t) = 1 – N n-1 * Pn(t)

Question 2:At the beginning of each week, 15 units of an inventory item are stocked for use during the week. Withdrawals from stock occur only during the first 6 days (business is closed on Sundays) and follow a Poisson distribution with 3 units per day.When the stock level reaches 5 units, a new order of 15 units is placed for delivery at the beginning of next week. Because of the nature of the item all units left at the end of the week are discarded.

Solution:We can recognize that the consumption rate is = 3 units per day.Suppose that we are interested in computing the probability of having 5 units (reorder level) on day t then we compute it as:

Pn(t) = (t)N - n e-t

(N – n)!

P5(t) = (3t)15 - 5 e-3t

(15 – 5)!

NB: P5(t) represents the probability of reordering on day t. This probability peaks at t = 3 and then declines as we advance through the week. If we are interested in the probability of reordering by day t we must compute the cumulative probability of having 5 units or less on day t.i.e. Pn <= 5(t) = P0(t) + P1(t) + …..+ P5(t).

STEADY STATE MEASURES OF PERFORMANCE :

The steady state measures of performance is used to analyze the operation of the queuing situation for the purpose of making recommendations about the design of the system. Among these measures of performance are:

The expected number of customers waiting. The expected waiting time per customer. The expected utilization of the service facility.

The system comprises both the queue and the service facility.Let:

Ls = expected number of customers in system.Lq = expected number of customers in Queue.

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Ws = expected waiting time in system.Wq = expected waiting time in Queue.

Suppose that we are considering a Service Facility with c parallel servers then from the definition of Pn we get:

Ls = n=0 * npn

Lq = n=c + 1 * (n – c)pn

NB: A strong relationship exists between Ls and Ws (also Lq and Wq) so that either measure is automatically determined from the other.Let eff be the effective average arrival rate (independent of the number in the system n) then:

Ls = eff * Ws

Lq = eff * Wq The value of eff is determined from the state dependent n and the probabilities Pn is given as:

eff = n=0 * npn

NB: A direct relationship exists between Ws and Wq, which is equal to:

Expected waiting time in system = Expected waiting time in queue + expected service time.Ws = Wq + 1/.

Given that is the Service rate per busy server, the Expected Service time = 1/.

Ws = Wq + 1/.

Multiplying both sides by eff we obtain:

Ls = Lq + eff/ = expected number of customers in system.

The expected utilization of a service is defined as a function of the average number of busy servers.Since the difference between Ls and Lq must equal the Expected number of busy servers we obtain:

Expected number of busy servers = C = Ls – Lq = eff/.

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The Percent utilization of a service facility with c parallel servers is given by:

Percent utilization = C/C * 100

NB: In summary Given Pn we can compute the system’s measure of performance in the following order:

Pn Ls = n=npn Ws = Ls/eff Wq = Ws - 1/ Lq = eff * Wq C = Ls – Lq.

Question 3: Job orders arriving at a production Facility is divided into three groups. Group1 will take the highest priority for processing; Group3 will be processed only if there are no waiting orders from group 1 and 2. It is assumed that a job once admitted to the facility must be completed before any new job is taken in.Orders from groups 1,2 and 3 occur according to Poisson distribution with mean 4, 3 and 2 per day respectively. The service times for the three groups are constant with rates 10, 9 and 10 per day respectively.Find the following:

I. The expected waiting time in the system for each of the three queues.II. The expected waiting time in the system for any customer.

III. The expected number of waiting jobs in each of the three groups.IV. The expected number waiting in the system.

Solution:

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CODE: 710 / 04 /S02

UNIT 1 APPROXIMATIONS

TRAPEZIUM RULE

Any definite integral may be thought of as an area under which a given curve for a specific interval. We can evaluate a definite of when an indefinite integral is known.Some functions may not have a single integral hence an approximation method could be used to evaluate them.

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Trapezium divides the area under the curve into small trapeziums whose sum approximates the required integral.

Y-axis

Y =f(x)

A B X-axis

Area of trapezium = ½ (sum of 2 sides)*height

Consider the function y = f (x) to be the integral over the interval a, b i.e. limits are from a to b. We obtain the approximation of this definite integral by dividing the area under f (x) into any vertical trapezium width with delta x (δx)

TRAPEZIUM RULE

½ δx {f (x0)+2f(x1)+2f(x2)+2f(x3)+……….f (xn)}

Where δx = b - a nX0 = aX1 = x0 + δxX2 = x1 + δxX3 = x2 + δxXn = b

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