E M A L Z G 6 1 2 - M E T H O D S A N D T E C H N I Q U E S O F S Y S T E M E N G I N E E R I N G

PROF V.MURALIDHAR

WILP-BITS PILANI-BITS-TECHNIP COLLABORATIVE PROGRAM

MS IN ENGINEERING MANAGEMENT

Applications of Operations Research-Design

ImplementationOperations of large, humanly convinced soft systems

Linear programming

Queuing TheorySimulationSampling

TechniquesDecision Models

Integer programming

Inventory controlMaintenance

ModelsForecasting

TechniquesNetwork scheduling

Models

Mathematical Models (1)Operation Research Models (2)Statistical Models

(1)Operation Research Models :Decision VariablesObjective FunctionConstraintsFeasible SolutionOptimal Solution

OR-ORIGIN : World war-I & II Anti-submarine warfare-Thomas Alva Edison devised a war game to be used for simulating problems of

naval manoeuvre lpp model for United states economy

OR is a quantitative technique that deals with many management problems

OR involves the application of scientific methods in situations where executive requires description, prediction and comparison

OR is an experimental and applied science devoted to observe understanding and predicting the behavior of purposeful man-machine system

APPLICATIONS

AREAS OF Accounting facilities :Cash flow planning, credit policy

planning of delinquent account strategyManufacturing : max or min of objectives of products etc. Marketing : selections of product mix , productions scheduling

time , advertising allocation etcPurchasing : How much to order ? Depending on sale or profit

etc.Facility panning : warehouse location, Transportation loading

and un loading, Hospital planning Finance : Investment analysis , dividend policy etc.Production : OR is useful in designing,selecting and locating

sizes,scheduling and sequencing the production runs by proper allocation of machines etc.

TYPES OF OR MODELS

ANALOGUE MODEL (DIAGRAMATIC)SYMBOLIC MODEL (PHYSICAL MODEL)DESCRIPTIVE MODEL

METHODOLOGY OF OR MODEL

Identifying the problemFormulating the problemFinding out the key areas of the problemConstructing a mathematical modelDeriving the solution to the modelTesting and updating the model for feasibility

of the solutionEstablishing controls over the model and the

solutionImplementing the solution obtained

LINEAR PROGRAMING PROBLEM(LPP)

Optimization of linear function of several variables

Optimization-(maximization or minimization)Objective functionDecision variablesConstraintsExamples : maximization of profit or

production of any company/IndustryMinimization of expenditure/loss of any

company/Industry

Slack variables and surplus variables

CONSTRAINTS OF LESS THAN OR EQUAL TO TYPE IS CONVERTED INTO EQUALITY BY INTRODUCING SLACK VARIABLES.

CONSTRAINTS OF GREATER THAN OR EQUAL TO TYPE IS CONVERTED INTO EQUALITY BY INTRODUCING SURPLUS VARIABLES.

EXAMPLES :

FORMULATION OF AN LPP MODEL

(PRODUCTION-ALLOCATION PROBLEM) A Manufacturer produces two types of products X

and Y. Each X model requires 4 hours of grinding and 2 hours of polishing whereas each Y model requires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinders and 3 polishers. Each grinder works for 40 hours a week and each poisher works for 60 hours a week. Profit on X product is Rs3.00 and on Y product it is Rs4.00 . Whatever is produced in a week is sold in the market. How should the manufacturer allocate his production capacity to the two types of products so that he may make optimum profit in a week?

ANSWER

Max z= 3x+4y Subject to the constraints

0,0

18052

8024

yxnsrestrictioenonnegativwith

yx

yx

PRODUCT-MIX PROBLEM

The manager of an oil refinnery has to decide upon the optimal mix of two possible blending process of which the inputs and outputs per production run are as follows :

process input output

Crude-A Crude-B Crude-C Crude-D

12

54

35

54

84

The maximum amount available of crude-A and B are 200 units and 150 units respectively. Market requirements show that at least 100 units of gasoline X and 80units of gasoline y must be produced. The profit per

production run from process 1 and 2 are Rs3 and Rs4 respectively. Formulate the problem as a Lpp model.

Answer

0,0

8048

10045

15053

20045

int43

yxnsrestrictioenonnegativwith

yx

yx

yx

yx

sconstrathetosubjectyxzMax

METHODS OF OBTAINING OPTIMUM SOLUTIONS

GRAPHICAL METHODSIMPLEX METHODDUALITY METHODDUAL SIMPLEX METHODTWO-PHASE METHOD

GRAPHICAL METHOD

APPLIED ONLY WHEN THERE ARE TWO VARIABLES ONLY

CONVERTING INEQUALITY CONSTRAINTS TO EQUALITY CONSTRAINTS

DETERMINING THE POINTS BY PUTTING EACH VARIABLES EQUATED TO ZERO

PLOTTING OF GRAPHIDENTIFYING FEASIBLE REGIONFEASIBLE POINTS TO BE DETERMINEDFINDING THE VALUE OF OBJECTIVE FUNCTION AT

FEASIBLE POINTSFINDING OPTIMUM SOLUTIONS

Max z= 5x+4y

Subject to the constraints

0,

2

1

62

2446

yxwith

y

xy

yx

yx

OPTIMUM SOLUTION IS GIVEN BY

X=3 and y=3/2

Minimization Model

1.Find the minimum value of Z=4x+7y subject to the constraints

0,

5

55

6

yxwith

x

yx

yx

X=0,Y=1 AND MIN Z=7

Solution

2. A Farm is engaged in breeding pigs. The pigs are fed on various products grown on the form. Because of the need to ensure certain nutrient constituents, it is necessary to buy additional one or two

products which we shall call A and B.The nutrient constituents(vitamins and proteins) in each unit of the production are

given below :

Nutrient contents in the products

Nutrient contents in the products

Minimum amount of nutrient

Nutrient A B

1 36 6 108

2 3 12 36

3 20 10 100

Product A costs Rs20 per unit and Product B costs Rs40 per unit. How much of the products A and B should be purchased at the lowest possible cost so as to provide the

pigs nutrients not less than that the minimum required as given in the table.

Formulation and Solution :Minimize Z= 20x+40y

Solution : x=4 and y=2 and min Z = Rs160

0,

31001020

236123

)1(108636

yx

nutrientforyx

nutrientforyx

nutrientforyx

3.Find the minimum value of Z=20x+10y

Subject to the constraints

0,

303

6034

402

yx

yx

yx

yx

SOLUTION

X=6,Y=12 ; MIN z=240

Some exceptional cases

An unbounded solution :While solving lpp, there are situations when

the decision variables are permitted to increase infinitely without violating the feasibility condition, i.e., no upper bound. In such a case the objective function value can also be increased infinitely and hence there is an unbounded solution.

Example : Consider the lpp : Maximize Z= 3x+2y

Subject to the constraints

0,

3

1

yxwith

yx

yx

The following feasible region (solution space) shows that the maximum value of Z occurs at a point at infinity and thus the problem has an unbounded

solution

Graph :

No Feasible solution

If it is not possible to find a feasible region that satisfies all the constraints of the lpp, the given lpp is said to have feasible solution.This shows that the two inequalities that form the constraints set are inconsistent. As there is no set of points that satisfies all the constraints, there is no feasible solutionto the problem.

Example : Consider the lppMaximize Z =x+y

Subject to the constraints

0,

33

1

yx

yx

yx

2.Maximize Z=5x+3y

Subject to the constraints

0,

2

1243

yx

x

yx

Show that the following lpp has an unbounded solution

Minimize Z=6x-2y Subject to the constraints

0,

,4

22

yxwith

x

yx

Some Important types of solutions of a LPP

SOLUTIONSFEASIBLE

SOLUTIONBASIC FEASIBLE

SOLUTION

DEGENERATE SOLUTION

OPTIMAL SOLUTION UNBOUNDED

SOLUTION

SIMPLEX METHOD

STEP-1 : standard form of LPP :MAX OR MIN (Z)SUBJECT TO CONSTRAINTSAx< or > BSTEP-2 : Check whether the objective

function is to be max or min. If it is min then it can be converted to max type by using the result

Min(z)=Max(-z)

Continued …

STEP-3 :Determine a starting basic feasible solution by setting decision variables to at zero level and slack or surplus variables to nonzero level.

STEP-4 : Establish simplex tableau which exhibits the system of constraints :

Cj 3 4 9 0 0 0

C CV Q X1 X2 X3 S1 S2 S3 Ratio

0 S1 6 4 9 1 1 0 0 9/6

0 S2 3 7 1 9 0 1 0 1/3

0 S3 8 4 5 1 0 0 1 2/8

Zj 0 2 25 6 0 0 0

(Zj-Cj)

-1 21 -3 0 0 0

IF THE PROBLEM IS MINIMIZATION TYPE FOR OPTIMALITY SOLUTION (ZJ -CJ ) ROW

MUST BE <=0

If the problem is maximization type for optimality solution (Zj-Cj)

row must be >=0

Continued …

STEP-5 : Cj row --- The value of cost coefficients of the variables in the

objective function C column – Cost coefficients of non-basic variables in the objective

function CV column- Basic variables column X1,x2,…s1,s1,… ---Decision,slack or surplus or artificial variables Ratio—Ratio column Zj ---row—Multiplying key column(Incoming variable column) with C

column and adding (Zj-Cj) row—Net profit Bold row--- Key row(outgoing variable row) obtained by choosing

minimum element of ratio column Pivotal element--- Intersection element of key row and key column

Continued …

STEP-6 :To get new simplex table we proceed as

follows :If optimality is not obtained we move to next

simplex table for better solution :Choose (For Max)most negative value of (Zj-

Cj) row and (For Min)largest positive value of (Zj-Cj) row. Corresponding column denotes

key column which is Incoming (New Basic variable)variable column.

Continued …

STEP-7 :Divide each element of key column by the

corresponding element of Cost variable column to find ratio column.

Choose minimum value of ratio which indicates key row(Outgoing variable row(leaving the basic column))

Intersection element denote the pivotal element.STEP-8 : For new simplex table divide key row

element by pivotal element to make it unity(For new basic variable).

Continued …

STEP-9 :The remaining new elements in the simplex table can be obtained by using the following formula :

New Value = Old Value ---

elementPivotal

columnkeytoingcorrespondElement

XrowkeytoingcorrespondElement

Continued …

STEP-10 :Proceeding as in step-6 till we get optimal or

optimum solution.

Minimization problems

Use of Artificial variables :In the constraints of the > type , to obtain the initial basic

feasible solution we first put the given lpp in the standard form and then a nonnegative variable is added to the left side of each equation that lacks the much needed starting basic variables. This added variable is called artificial variable.

The artificial variables plays the same role as a slack variable in providing the initial basic feasible solution . The method will be valid only if we are able to force these artificial variables to be out or at zero level when the optimum solution is attained.

In other words, to get back to the original problem artificial variables must be driven to zero in the final solution; otherwise the resulting solution may be infeasible.

TWO METHODS FOR THE SOLUTION OF LPP HAVING ARTIFICIAL VARIABLES :

I. Big-M method or Method of Penalties II. Two-Phase Method

I .BIG-M Method OR Method of Penalties

STEP-I In this method we assign a very high penalty(say M) to the artificial variables in the objective function.

STEP-II Write the given L.P.P into its standard form and check whether there exists a starting basic solution : (a)If there is a ready starting basic feasible solution, move onto step-IV

(b)If there does not exists a ready starting basic feasible solution, move onto step-III .

STEP-III Add artificial variables to the left side of each equation that has no obvious starting basic variables.Assign a very high penalty (say M) to these variables in the objective function.

STEP-IV Substitute the values of artificial variables from the constraint equations into the modified objective function.

STEP-V Apply simplex method to the modified L.P.P. Following cases arise at the last iteration :

(a) At least one artificial variable is present in the basic solution with zero value. In such case the current optimum basic feasible solution is degenerate.

(b) At least one artificial variable is present in the basic solution with a positive value. In such a case, the given L.P.P does not posses an optimum basic feasible solution. The given problem is said to have a pseudo-optimum solution.

Minimize z=4x+ySUBJECT TO THE CONSTRAINTS

0,

42

634

33

yx

yx

yx

yx

Solution :

Using s1 as surplus variable and s2 as slack variable with zero cost coefficients in the objective function and adding artificial variables R1 and R2 with high penalty value M (MINIMIZING) we get

3x+y+R1=34x+3y-s1+R2=6X+2y+s1=4With x,y,s1,s2,R1,R2≥0

Continued …

Initially we let x=0,y=0,s2=0 (Non-Basic level)

Then we get R1=3,R2=6 and s2=4(Basic level)

Simplex table

Cj 4 1 100 100 0 0

C BV CV X Y R1 R2 s1 s2 ratio

0 R1 3 3 1 1 0 0 0 1

0 R2 6 4 3 0 1 -1 0 6/4

0 S2 4 1 2 0 0 0 1 4

Zj 0 0 0 0 0 0 0

Zj 900 700 400 100 100 -100 0

Zj-Cj

696 399 0 0 -100 0

Cj 4 1 100 0 0

C BV CV X Y R2 s1 s2 ratio

4 X 1 1 1/3 0 0 0 3

0 R2 2 0 5/3 1 -1 0 6/5

0 s2 3 0 5/3 0 0 1 9/5

Zj 4 4 4/3 0 0 0

Zj 204 4 504/3

100 -100 0

Zj-Cj 0 501/3

0 0 0

Cj 4 1 0 0

C BV CV X Y s1 s2 ratio

4 X 3/5 1 0 1/5 0

1 Y 6/5 0 1 -3/5 0

0 s2 0 0 0 1 1

Zj 18/5 4 1 1/5 0

Zj-Cj

0 0 1/5 0

Minimize z=12x+20ysubject to constraints

0,

120127

10086

yx

yx

yx

ANSWER :

X=15,Y=5/4 AND Min z=205

3.Max W=2x+y+3zsubject to the constraints

0,,

12432

52

zyx

zyx

zyx

SOLUTION

X=3,y=2,z=0and max w=8

DUALITY in LINEAR PROGRAMMING

Duality states that for every lpp of max(or min), there is a related unique problem of min(or max) based on the same data and the numerical values of the objective function to the two problems are same. The original problem is called PRIMAL PROBLEM while the other is called DUAL PROBLEM.

Dual problem we define in such a way that it is consistent with the standard form of the primal.

Examples

I Standard Primal :Max Z=C1X1+C2X2+C3X3+…..CnXnSubject to the constraints :A11X1+a12X2+…a1nXn=bi, (i=1,2,3,…m)

Dual :Min Z=b1y1+b2y2+b3y3+……bmymSubject to the constraints :A1jy1+a2jy2+…amjym≥Cj, (j=1,2,3,…n) yi unrestricted (i=1,2,3….m)

I Standard Primal :Min Z=C1X1+C2X2+C3X3+…..CnXnSubject to the constraints :ai1X1+ai2X2+…ain Xn=bi, (i=1,2,3,…n) Xj≥0 ; (j=1,2,3….m) Dual :Min Z=b1y1+b2y2+b3y3+……bmymSubject to the constraints :A1jy1+a2jy2+…amjyn≤Cj, (j=1,2,3,…n) yi unrestricted (i=1,2,3….m)

OPTIMAL DUAL SOLUTION :The following set of rules govern the derivation of the optimum dual solution

Rule-1 :If the primal(dual) variable corresponds to a slack and/or surplus variable in the dual(primal) problem, its optimum value is directly read off from the last row of the optimum dual(primal) simplex table, as the value corresponding to this slack and/or surplus variable.

Rule-2 :If the primal(dual) variable corresponds to artificial starting variable in the dual(primal) problem, its optimum value is directly read off from the last value row of the optimum dual(primal) simplex table, as the value corresponding to this artificial variable, after deleting the constant M, the penalty cost.

Rule-3 : If the primal(dual) problem is unbounded , then the dual(primal) problem does not have any feasible solution.

PROBLEMS

1.Formulate the dual of the following linear programming problem :

0,

,1025

,1553

int

35

21

21

21

21

xx

xx

xx

sconstrathetoSubject

xxzMaximize

Solution :

Introducing slack variables 0,0 21 ss

The reformulated linear programming problem is

0,,,

,10.025

,15.053

int

.0.035

2121

2121

2121

2121

ssxx

ssxx

ssxx

sconstrathetoSubject

ssxxzMaximize

Corresponding Dual is given by

)(

0,

,325

,553

int

1015

21

21

21

21

21

redundantedunrestrictyandy

yy

yy

yy

sconstrathetoSubject

yywMinimize

2.Find the dual of the following maximization problem :

0,

123

,204

,1532

int

3552

21

21

21

21

21

xxwhere

xx

xx

xx

sconstrathetoSubject

xxzMaximize

SOLUTION :

0,,,,

,20.0.04

,15.0.032

int

.0.0.03525

32121

32121

32121

32121

sssxx

sssxx

sssxx

sconstrathetoSubject

sssxxzMaximize

The corresponding duality is given by

)(

0,,

,3543

,2532

int

122015

21

321

321

321

321

redundantedunrestrictyandy

yyy

yyy

yyy

sconstrathetoSubject

yyywMinimize

3.Write the dual of the following linear programming problem :

0,,

3352

442

1027

,523

,9643

int

465

321

321

321

321

321

321

321

xxx

xxx

xxx

xxx

xxx

xxx

sconstrathetoSubject

xxxwMinimize

Reformulating in the standard form we get

0,,,,,,

,3.0.0.0.0352

,4.0.0.0.042

,10.0.0.0.027

,5.0.0.0.023

,9.0.0.0.0643

int

.0.0.0.0.0465

54321,321

54321321

54321321

54321321

54321321

54321321

54321321

sssssxxxwith

sssssxxx

sssssxxx

sssssxxx

sssssxxx

sssssxxx

sconstrathetoSubject

sssssxxxzMinimize

The corresponding duality is given by

)(,,,

0,,,,

,43426

,652234

,573

int

341059

54321

54321

54321

54321

54321

54321

redundantedunrestrictyandyyyy

yyyyy

yyyyy

yyyyy

yyyyy

sconstrathetoSubject

yyyyywMaximize

4.One unit of product A contributes Rs7 and requires 3 units of raw materials and 2 hours of labour. One unit of product B contributes Rs 5 and requires one unit of raw material and one unit of labour. Availability of raw material of present is 48 units and there are 40 hours of labour.

(i)Formulate it as a linear programming problem

(ii)Write its dual(iii)Solve the dual with simplex method and

find the optimal product mix and the shadow prices of the raw material and labour.

Solution :

0,

,402

,483

int

57

21

21

21

21

xx

xx

xx

sconstrathetoSubject

xxzMaximize

Dual problem

0,

,5

,723

int

4048

21

21

21

21

yy

yy

yy

sconstrathetoSubject

yywMinimize

SOLUTION

OPTIMUM SOLUTION OF DUAL VARIABLES ARE

50 21 yandy

ORIGINAL SOLUTION ARE

200)max()min(

400 21

zw

andxandx

Integer programming-An LPP wiTH ADDITIONAL REQUIREMENTS THAT THE VARIABLES CAN TAKE ON ONLY

INTEGER VALUES IS CALLED iNteger programming

Max or Min Z=C1X1+C2X2+C3X3+…..CnXnSubject to the constraints :ai1X1+ai2X2+…ain Xn=bi, (i=1,2,3,…m) and Xj≥0 ; (j=1,2,3….n)Where Xj are integer valued for j=1,2,3….p (p≤n)If p=n , the problem is called Pure Integer

programming problem , otherwise it is called Mixed Integer programming problem.

Also if all the variables of an integer programming problem are either 1 0r 0, the problem is called Zero-one programming problem.

CAPITAL BUDGETING PROBLEM

Five projects are being evaluated over a 3-year planning horizon .The following table gives the expected returns for each project and the associated yearly expenditures.Expenditures(mil

lions $) per year

Project 1 2 3 Returns(million$)

1 5 1 8 20

2 4 7 10 40

3 3 9 2 20

4 7 4 1 15

5 8 6 10 30

Available funds(million$)

25 25 25

solution

)1,0(,,,,

,25102108

,256497

,2587345

int

3015204020

54321

54321

54321

54321

54321

xxxxx

xxxxx

xxxxx

xxxxx

sconstrathetoSubject

xxxxxzMaximize

The Optimum integer solution is

$)(95max

05,14321

millionzand

xxxxx

THE SOLUTION SHOWS THAT ALL BUT PROJECT-5 MUST BE SELECTED

1,,,,0 54321 xxxxx

Solution is

$)(68.108max

7368.05,1432,5789.01

millionzand

xxxxx

INTEGER PROGRAMMING ALGORITHMS

Step-1 Relax the solution space of the ILP by deleting the integer restrictions on all integer variables and replacing any binary variable y with the continuous range 0≤y≤1.

The result of the relaxation is a regular LP.Step-2 Solve the LP and identify its

continuous optimum.Step-3 Starting from the continuous point ,

add special constraints that iteratively modify the LP solution space satisfying the integer requirements.

For generating special constraints there are two general methods :

1.Cutting Plane Method2.Branch and Bound method

1.Cutting Plane Method

Step-1Put the given LPP into its standard form and determine the optimum solution by using simplex methods ignoring integer value restrictions.

Step-2 Test for integerality of the optimum solution :(i) If the optimum solution admits all integer values, an

optimum basic feasible solution is attained.(ii) If the optimum solution does not include all integer

values then go to the step-3Step-3 Choose a row corresponding to the basic

variable which has largest fractional cut , say, fkAnd generate the constraint in the form

10

001

kj

n

jjkjk

fwhere

xffG

1010 0

001

kkj

n

jjkjk

fandfwhere

xffG

Continued ….

Step-4 Add the constraint to the optimum simplex table obtained in step-1. Apply dual simplex method to find an improved optimum solution.

Step-5 Go to step-2 and repeat the procedure until an optimum basic feasible all integer solution is obtained.

Solve the integer programming problem

egersareandxx

x

x

xx

sconstrathetoSubject

xxzMaximize

int0,

,35

25

,602

int

4060

21

2

1

21

21

Standard form

0,,,,

35

,25

,60.0.02

int

.0.0.04060

32121

32

21

32121

32121

sssxx

sx

sx

sssxx

sconstrathetoSubject

sssxxzMaximize

TRANSPORTATION MODEL

Definition :There are m sources and n destinations which are represented by a node.

ith source and jth destination are linked by the arc (i,j). The transporation cost from ith source to jth destination is represented by

The amount shipped by The amount of supply at source by

The amount of demand at destination j byThe objective of the model is to determine the unknown that will minimize the total transportation cost while

satisfying all the supply and demand restrictions.

ijc

ijx

ia

jb

ijx

GENERAL TRANSPORTATION MODEL

A B C D SUPPLY

X X11 (2) X12 (35) x13 x14 b1

Y x21 X22 (23) x23 x24 b2

Z X31 (18) x32 X33 (42) x34 b3

DEMAND a1 a2 a3 a4 Grand Total

PRODUCTION INVENTORY CONTROL

A company manufactures backpacks for serious hikers. The demand for its product occurs from March to June of each year. The company estimates the demand for the four months to be 100,200,180,and 300 units. The company uses part time labour to manufacture the backpacks and as such its production capacity varies monthly. It is estimated that company can produce 50,180,280, and 270 units for March to June, respectively.Because the production capacity and demand for different months do not match, a current month’s demand may be satisfied in one of the three ways :

1. Current month’s production2. Surplus production in earlier months3. Surplus production in later months

Continued …

In the first case, the production cost per backpack is $40.00. The second case incurs an additional holding cost of $50 per backpack per month. In the third case, an additional penalty cost of $2.00 is incurred per backpack per month delay. The company wishes to determine the optimal production schedule for the four months.

SOLUTION :TRANSPORTATION MODEL

1 2 3 4 CAPACITY

1 40 40.50 41.00

41.50 50

2 42.00 40.00 40.50

41.00 180

3 44.00 42.00 40.00

40.50 280

4 46.00 44.00 42.00

40.00 270

DEMAND

100 200 180 300

The unit transportation cost from period I to period j is computed as :

Cij = production cost in i, i=j production cost in I + holding cost from i to j , i<j production cost in i +penalty cost from i to j , i>j

TRANSPORTATION ALGORITHM

I NORTH WEST CORNER RULE METHODII LEAST COST METHODIII VOGEL’S APPROXIMATION METHOD

Transportation algorithm-vogel’s approximation method (vam)

VAM method is an improved version of the least cost method that generally produces better starting solution :

STEP-1 For each row(column), determine a penalty measure by subtracting the smallest

unit cost element in the row(column) from the next smallest unit cost element in the same row (column)

STEP-2 Identify the row or column with the largest penalty. Break ties arbitarily. Allocate

as much as possible to the variable with the least unit cost in the selected row or column. Adjust the supply and demand , and cross out the satisfied row or column. If a row and a column are satisfied simultaneously, only one of the two is crossed out, and the remaining row(column) is assigned zero supply(demand)

STEP-3 (a) If exactly one row or one column with zero supply or deman remains

uncrossed out, stop. (b)If one row(column) with positive supply(demand) remains uncrossed out,

determine the basic variables in the row(column) by the least cost method. (c) If all the uncrossed out rows and columns have (remaining) zero supply and

demand, determine the zero basic vaiables by the least cost method. Stop. (d) Otherwise go to step-1

Problems-1

Warehouse

Stores Availability

I II III IV

A 5 1 3 3 34

B 3 3 5 4 15

C 6 4 4 3 12

D 4 -1 4 2 19

Requirement

21 25 17 17 80

TRANSPORTATION COST IS GIVEN BY 6X5+6X1+17X3+5X3+15X3+12X3+19X-

1=rs 164

Warehouse

Stores Availability

I II III IV

A 5 (6) 1 (6) 3 (17)

3 (5) 34

B 3 (15)

3 5 4 15

C 6 4 4 3 (12)

12

D 4 -1 (19)

4 2 19

Requirement

21 25 17 17 80

INITIAL BASIC FEASIBLE SOLUTION USING VAM METHOD

Warehouse stores

Availability

Row penalties

I II III IV

A 5 1 3 3 34 2 2 0 0 0

B 3 3 5 4 15 0 0 1 1 -

C 6 4 4 3 12 1 1 1 - -

D 4 -1 4 2 19 3 - - - -

Requirement 21 25 17 17 80

1 2 1 1

Column penalties

2 2 1 1

2 - 1 0

2 - 1 0

Problem-1 ANSWER : COST=rs430

DESTINATION

Supply

ORIGIN D1 D2 D3 17

O1 13 15 (6) 16 (11)

12

O2 7 11 2 (12)

16

O3 10 (14)

20 (2) 9

Demand

14 8 23 45

Problem-2 ANSWER : COST=rs370

DESTINATION

Supply

ORIGIN

D1 D2 D3 D4

O1 5 (10)

8 3 (20)

6 30

O2 4 (20)

5 (20) 7 4 (10)

50

O3 6 2 (20) 4 6 20

Demand

30 40 20 10 100

Modi method (optimality solution)

Modi method helps to determine better transportation schedule by evaluating the non-basic cells. The Modi method selects a particular nonbasic cell that will yield maximum improvement through a set of index number for the row and column cells. The movement of maximum permissible units are then made for the non-basic cell. In a

Similar manner, revised index numbers indicate the next best non-basic cell. This is based on the following

principle : For a basic feasible solution of dual variables ui (i=1,2,3…

m) and vj(j=1,2,3…n) corresponding to capacity constraints i and requirement constraints j satisfying

Cij = ui+vj for each basic cell(occupied cell) then the index number (opportunity cost) Dij corresponding to each non-basic cell(I,j) is given by Dij = ui+vj-Cij

Problem-4 : find the minimum transportation cost using modi method

TO

I II III IV Availability

FROM A 15 10 17 18 2

B 16 13 12 13 6

C 12 17 20 11 7

Requirement

3 3 4 5 15

Answer : Initial transportation cost rs 320optimum cost rs174

ASSIGNMENT MODELS : The assignment problem is a special case of the transportation problem in which the objective is to assign a number of origins to the equal number of destinations at a minimum cost (or maximum profit). The assignment is to be made on a one-one basis.

Mathematical Formulation : Min Z= ∑∑CijXijSubject to the constraints∑Xij =1 and ∑Xij =1 ; xij = 0 or 1For all i=1,2,3,4,… and j=1,2,3,4…Where Cij is the cost associated with

assigning ith resource to jth activity.

REDUCTION THEOREM

TH-1 : In an assignment problem , if we add or subtract a constant to every element of any row(or any column) of the cost matrix [Cij], then an assignment that minimizes the total cost on one matrix also minimizes the total cost on the other matrix.

TH-2 :If Cij≥0 such that the minimum ∑∑Cij = 0, then Xij provides an optimum assignment.

HUNGARIAN METHOD (Hungarian mathematician D.KONIG)

Step-1 Determine the cost table from the given problem(i) If the number of sources is equal to the number of

destinations go to step-3.(ii) If the number of sources is not equal to the number of

destinations go to step-2.Step-2Add a dummy source or dummy destination, so that the

cost table becomes a square matrix. The cost entries of dummy source/destinations are always zero.

Step-3Locate the smallest element in each row of the given cost

matrix and then the same from each element of that row

Continued ….

Step-4 In the reduced matrix obtained in step-3, locate the smallest element of

each column and then subtract the same from each element of that column. Each column and row now have at least one zero.

Step-5 In the modified matrix obtained in step-4, search for an optimal

assignment as follows : (a) Examine the rows succesively until a row with single zero is found.

En rectangle this zer0 and cross off all other zeros in its column. Continue in this manner untill all the rows have been taken care of.

(b) Repeat the procedure for each column of the reduced matrix. (c) If a row and/ or column has two or more zeros and one cannot be

chosen by inspection then assign arbitary any one of these zeros and cross off all other zeros of that row/column.

(d) Repeat the above process until the chain of assigning rectangle or cross ends.

Continued ….

Step-6 If the number of assignments are equal to n an optimum solution is reached. If the number of assignments are less than n, go to next step. Step-7 Draw the minimum number of horizontal and/ vertical lines to cover all the zeros of the

reduced matrix as follows : (a)Mark (√) rows that do not have any assigned zero. (b) Mark(√) columns that have zeros in the marked rows. (c) Mark(√) rows that have assigned zeros in the marked columns (d) Repeat above until the chain of marking is completed. (e) Draw lines through all the unmarked rows and marked columns. This gives the

minimum number of lines. Step-8 Devel0p the new revised cost matrix as follows : (a) Find the smallest element of the reduced matrix not covered by any of the lines. (b) Subtract this element from all the uncovered elements and add the same to all the

elements lying at the intersection of any two lines. Step-9 Go to step-6 and repeat the procedure until an optimum solution is attained.

PROBLEMS :Answer : 1-I , 2-III , 3-II , 4-IV ;

Minimum Time = 42 hours

1.A department has four subordinates and four tasks are to be performed. The subordinates differ in efficiency and their tasks differ in their intrinsic difficulties. The estimate of time (in hour) each man would take to perform each task is given by Task

I II III IV

1 9 26 17 11

Subordinate

2 13 28 4 26

3 38 19 18 15

4 19 26 24 10

2.Shortest-Route problem

Find the shortest route between nodes 1 and 7 of the network in the following diagram by formulating the problem as a transshipment model. The distance between the different nodes are shown in the nettwork (Hint : Assume that node-1 has a net supply of 1 unit and node-7 has a net demand also of one unit.)

Network diagram

TYPES OF ASSIGNMENT MODELS

1.Maximum assignment model2.Restrictions on assignment3.Prohibited assignment model4.Travelling salesman problem