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