WILP-BITS PILANI-BITS-TECHNIP
COLLABORATIVE PROGRAM
MS IN ENGINEERING MANAGEMENT
EMAL ZG612-METHODS AND TECHNIQUES OF SYSTEM ENGINEERING
Prof V.MURALIDHAR
APPLICATIONS OF OPERATIONS RESEARCH-DESIGNIMPLEMENTATIONOPERATIONS OF LARGE, HUMANLY CONVINCED SOFT SYSTEMS
Linear programming
Queuing Theory
Simulation Sampling
Techniques Decision
Models
Integer programming
Inventory control
Maintenance Models
Forecasting Techniques
Network scheduling Models
MATHEMATICAL MODELS (1)OPERATION RESEARCH MODELS (2)STATISTICAL MODELS
(1)Operation Research Models : Decision Variables Objective Function Constraints Feasible Solution Optimal 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 strategy Manufacturing : max or min of objectives of products
etc. Marketing : selections of product mix , productions
scheduling time , advertising allocation etc Purchasing : 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 problem Formulating the problem Finding out the key areas of the problem Constructing a mathematical model Deriving the solution to the model Testing and updating the model for
feasibility of the solution Establishing controls over the model and
the solution Implementing the solution obtained
LINEAR PROGRAMING PROBLEM(LPP) Optimization of linear function of several
variables Optimization-(maximization or
minimization) Objective function Decision variables Constraints Examples : maximization of profit or
production of any company/Industry Minimization 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 METHOD SIMPLEX METHOD DUALITY METHOD DUAL SIMPLEX METHOD TWO-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 GRAPH IDENTIFYING FEASIBLE REGION FEASIBLE POINTS TO BE DETERMINED FINDING THE VALUE OF OBJECTIVE
FUNCTION AT FEASIBLE POINTS FINDING OPTIMUM SOLUTIONS
MAX Z= 5X+4Y Subject to the constraints
0,
2
1
62
2446
yxwith
y
xy
yx
yx
X=3 AND Y=3/2
Optimum solution is given by
MINIMIZATION MODEL 1.Find the minimum value of Z=4x+7y subject to the constraints
0,
5
55
6
yxwith
x
yx
yx
SOLUTIONX=0,y=1 and min z=7
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
X=6,Y=12 ; MIN Z=240
SOLUTION
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 LPPSOLUTIONSFEASIBLE
SOLUTIONBASIC FEASIBLE
SOLUTION
DEGENERATE SOLUTION
OPTIMAL SOLUTION
UNBOUNDED SOLUTION
SIMPLEX METHOD STEP-1 : standard form of LPP : MAX OR MIN (Z) SUBJECT TO CONSTRAINTS Ax< or > B STEP-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 MAXIMIZATION TYPE
FOR OPTIMALITY SOLUTION (ZJ-CJ)
ROW MUST BE >=0 If the problem is minimization
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=3 4x+3y-s1+R2=6 X+2y+s1=4 With 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
X=15,Y=5/4 AND MIN Z=205
ANSWER :
3.MAX W=2X+Y+3ZSUBJECT TO THE CONSTRAINTS
0,,
12432
52
zyx
zyx
zyx
X=3,Y=2,Z=0AND MAX W=8
SOLUTION
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+…..CnXn Subject to the constraints : A11X1+a12X2+…a1nXn=bi, (i=1,2,3,…m)
Dual : Min Z=b1y1+b2y2+b3y3+……bmym Subject 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+…..CnXn Subject 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+……bmym Subject 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 ARE200)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+…..CnXn Subject 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(millions $) per year
Project 1 2 3 Returns(million$)
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 Method 2.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-3
Step-3 Choose a row corresponding to the basic variable which has largest fractional cut , say, fk
And 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 by The 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 production 2. Surplus production in earlier months 3. 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 METHOD II LEAST COST METHOD III VOGEL’S APPROXIMATION METHOD
PROBLEMS