Goal Progrmming

Goal Programming

Presented by:Vivek kumar gupta

TABLE OF CONTENT

• Introduction -Theory - Steps involved in formulation of LGPP• Single Goal Programming• Multiple Goal Programming - Non premptive - Premptive

Mathematical model similar to Linear Programming, however it allows for multiple goals to be satisfied at the same time.

Lpp(Linear programming problem) is based on the assumption that the decision making has a single, qualifiable objective such as maximization of Profit/output/sale or minimization of cost/ time.

Often there are situations where instead of having single objective, multiple objectives or goals may be set. In such situations we need different technique that seek a compromise solution based on the relative importance of each objective.

Goal programming uses simplex method for finding the optimum solution to a multi dimensional linear objective function with linear constraints.

WHAT IS GOAL PROGRAMMING ?

Goal programming has been widely applied to decision problems in business organizations, government agencies, and nonprofit institutions. Some applications includes –

Accounting AnalysisCapital budgetingEconomic policy analysisEnergy resource planningFacility layout and location

CLASSIFICATION OF GOAL PROGRAMMING

1. Single goal programming: In this case only one goal has to be achieved

2. Multiple goal programming

(a) Non-premptive goal programming: In this case all the goals are equally ranked & are of roughly comparable importance.

(b) Premptive goal programming: In this case there is a hierarchy of priority levels for the goals so that goals of primary importance receive first priority attention, those secondary importance receive second priority attention, and so forth(if there are more than 2 priority levels)

Steps involved in formulation of linear Goal Programming Problem (LGPP)

Step 1: Find out the decision variable of the major decisionsStep 2: Formulate all the goals of the programme,these are generally fixed by the desire of the decision maker or resources or constraintsStep 3: Express each goal in terms of constraints equations by introducing negative and positive deviations i.e DU or d- & d+ or DoWhere Du or d- = underachievement of goal d+ or Do = over achievement of goal

Step 4: Fix the priority levels of each goals (only in case of multiple goals) & put these in achievement function

In general, once all goals & constraints are identified in a problem, management should analyze each goal to see if under or over-achievement of that goal is an acceptable situation.

If management seeks to attain a goal met exactly both d- and d+ must appear in an objective function.

A firm manufactures two products TV and DVD which must be processed through two processes, Assembly & Finishing. Assembly has 96 hours available and finishing has 78 hours available. For 1 TV set requires 6 hours in Assembly & 4 hours in finishing while 1 DVD set requires 5 hours in Assembly & 3 Hours in Finishing. If profit is Rs 90 per TV & 100 per DVD, find out the best combination of TV & DVD to realize a maximum profit of Rs 2500.

Q

Assembly(Hrs) Finishing(Hrs)

Profit(Rs)

TV 6 4 90

DVD 5 3 100

Total 96 78 2500

Let X1 & X2 be the no. of TV & DVD. The given problem can be first formulated as LPP

Max. Z = 90 X1 + 100 X2

Sub . to 6 X1 + 5 X2 ≤ 96 - Assembly constraints 4 X1 + 3 X2 ≤ 78 - Finishing constraints Where X1 & X2 ≥ 0 In case of goal programming, any goal stated can be :-• Exactly met• Under achieved • Over achieved Here the goal has to be set to as to make a profit of Rs 2500

Thus now the requirement can be expressed as 90 X1 + 100 X2 = 2500

• Now using Deviational Variables ..Let d- be the under achievement of profit goal & d+ be the over achievement of profit goal.

The given problem can be expressed as the following Goal Programming problem.

Minimize Z = d- (under achievement of profit goal)

Sub. to: 90 X1 + 100 X2 + (d-) – (d+) = 2500 (Goal constraint)

6 X1 + 5 X2 ≤ 96 (Assembly constraint) 4 X1 + 3 X2 ≤ 78 (Finishing constraint) Where X1 , X2, d+ & d- ≥ 0

We will now convert this into standard minimization problem.

Minimize Z = 0X1 + 0X2 + 0S1 + 0S2 +(d-) + 0(d+)Sub. to 90 X1 + 100 X2 + (d-) – (d+) = 2500 6X1 + 5X2 + S1 + 0S2 = 96 4X1 + 3X2 + 0S1 + S2 = 78 Where X1, X2, S1, S2, (d-) & (d+) ≥ 0

Basic

Variables

‘B’

Cj→ 0 0 0 0 1 0SOLUTIO

N VALUES

XB

Min.RatioCB

↓X1 X2 S1 S2 d- d+

d- 1 90 100 0 0 1 -1 2500 25

S1 0 6 5 1 0 0 0 96 19.20

S2 0 4 3 0 1 0 0 78 26

Zj90 100 0 0 1 -1 2500

Cj - Zj-90 -100 0 0 0 1

Table 1

Basic

Variables

‘B’

Cj→ 0 0 0 0 1 0SOLUTION VALUES

XBCB

↓X1 X2 S1 S2 d- d+

d- 1 -30 0 -20 0 1 -1 580

X2 0 6/5 1 1/5 0 0 0 96/5

S2 0 2/5 0 -3/5 1 0 0 102/5

Zj-30 0 -20 0 1 -1 580

Cj - Zj30 0 20 1 0 1

Table 2

Since Cj – Zj ≥ 0, thus solution is optimalX1 = 0, X2 = 96/5 , d- = 580 which means, target profit of 2500 is under achieved by Rs 580 i.e the profit actually earned is Rs 2500 – 580 = 1920

Example 7.11(N.D Vohra)• A firm is engaged in producing two products , A

and B. Each unit of product A requires 4 kg of raw material and 2 labor hours for processing, whereas each unit of product B requires 3 kg of raw material and 3 hours of labor , of the same type. Every week ,the firm has an availability of 60 kg raw material and 96 labor hours .One unit of product A sold yields Rs 40 and one unit of product B sold yields Rs 35 as profit. find out the best combination to realize a maximum profit of Rs.1400.

The LPP equations would be:Maximize Z = 40x1+ 35x2 (Profit)

Subject to 2x1 + 3x2 ≤ 60 (Labor hours constraint)

4x1 + 3x2 ≤ 96 (Raw material constraint)

x1, x2 ≥ 0

PRODUCTS Labor hours Raw Material Profit

A 2 4 Kg Rs. 40

B 3 3 Kg Rs. 35

Constraints 60 Hours 96 Kg

• The given problem can be expressed as the following Goal Programming problem-

• Minimize Z =d- Subject to 2x1 + 3x2 ≤ 60 4x1 + 3x2 ≤ 96 40x1 + 35 x2 +(d-) – (d+) = 1400 Where X1 , X2, d+ & d- ≥ 0

Now solving it via a Simplex Solution problem:-

Minimize Z = 0X1 + 0X2 + 0S1 + 0S2 +1(d-) + 0(d+)

Subject to 2x1 + 3x2 + S1 + 0S2 = 60

4x1 + 3x2 +0S1 + S2 = 96 40x1 + 35 x2 +(d-) – (d+) + S1 + S2 = 1400 Where X1, X2, S1, S2, (d-) & (d+) ≥ 0

Now solving it with help of Simplex Tables.

BasicVariab

les

‘B’

Cj→ 0 0 0 0 1 0

biMin.RatioCB

↓X1 X2 S1 S2 d- d+

S1 0 2 3 1 0 0 0 60 30

S2 0 4* 3 0 1 0 0 96 24

d- 1 40 35 0 0 1 -1 1400 35Zj 40 35 0 0 1 -1

Cj - Zj -40 -35 0 0 0 1

Simplex Tableau 1

BasicVariab

les

‘B’

Cj→

0 0 0 0 1 0

biMin.RatioCB

↓X1 X2 S1 S2 d- d+

S1 0 0 3/2* 1 -1/2 0 0 12 8

X1 0 1 3/4 0 1/4 0 0 24 32

d- 1 0 5 0 -10 1 -1 440 88Zj 0 5 0 -10 1 -1

Cj - Zj 0 -5 0 10 0 1

Simplex Tableau 2

Simplex Tableau 3Basic

Variables

‘B’

Cj→ 0 0 0 0 1 0

biCB

↓X1 X2 S1 S2 d- d+

X2 0 0 1 2/3 -1/3 0 0 8

X1 0 1 0 -1/2 1/2 0 0 18

d- 1 0 0 -10/3 -25/3 1 -1 400

Zj 0 0 -10/3 -25/3 1 -1

Cj - Zj 0 0 10/3 25/3 0 1

• So , the solution is • x1 =18• X2 = 8 • d+ = 0• d- = 400 , that means the profit goal is under

achieved by Rs. 400 and the maximum profit is Rs. 1000.

GOAL PROGRAMING WITH MULTIPLE GOALS

1.Non-preemptive Goal programming2.Preemptive Goal programmingNon-preemptive Goal ProgrammingAssumes the decision-maker has a linear utility function with

respect to the objectivesDifferential weights are used to various goals in line with their

relative significancePreemptive Goal ProgrammingUses prioritized goalsWhen optimal solution with respect to the higher priority goal is

obtained only then the next priority level is considered

NON-PREEMPTIVE GOAL PROGRAMMING

In non-preemptive goal programming, we first establish a goal for each objective and then seek a solution that minimizes the sum of deviations of these objectives from their respective goals

The goal programming assumes that the decision-maker has I linear utility function with respect to the objectives, that is to say, the marginal rate of substitution between the objectives is linear, regardless of the extent of deviation from the goal.

The deviations may be given different weights , called penalty weights, in accordance with the relative significance of the objectives, and the solution sought may be the one which minimizes the weighted sum of the derivatives.

the weight used in a goal programming model are indicative of the decision-maker’s utility for the various objectives.

Specifically, they measure the marginal rate of substitution between objectives and the degree of importance in attaining the goal of each objective relative to the others.

EXAMPLE 7.12 N.D VOHRA

The production manager of a company wants to schedule a weeks production run for two products P1 and P2, each of which requires the labour and materials as shown below. PRODUCT

P1 P2

Labour hours 2 4

Material M1 (kg)

4 5

Material M2 (kg)

5 4

The weekly availability of resources is limited to 600 labour hours, 1000 kg of material M1 and 1200 kg of M2. The unit profit for P1 and P2 is Rs.20 and Rs 32 respectively.

Products P1 and P2 are in fact new models and are replacements of the older ones which have been discontinued very recently .

The manager would like to maximise the profit but he is equally concerned with maintaining workforce of the division at the nearly constant level in the interest of employee morale. The workforce which consists of people engaged in production, sales, distribution, peons and other general staff, consisted of 108 persons in all. From a detailed study, it is known that production of one unit of P1 would maintain 0.3 person in the workforce while one unit of P2 would maintain 0.75 person.

SOLUTION:-

If we let x1 and x2 represent the number of units of P1 and P2 respectively, respectively, to be produced every week, the goals and constraints of the problem cab be stand as follows:

20x1+ 32x2 = 5400 goal-1 0.3x1+0.75x2 = 108 goal-2 2x1 + 4x2 <= 600 labour 4x1 + 5x2 <= 1000 material M1 5x1 + 4x2 <= 1200 material M2

This problem may not have a feasible solution which satisfies both the goals. Further, to be able to solve this problem using simplex method, we need to introduce deviational variables in each of the constraints involving goals. They may be defined as follows:

Let d1- = number of rupees below the profit goal of Rs 5400 d1+ = number of rupees above the profit goal of Rs 5400 d2- = number of people below the workforce goal of 108 d2+ = number of people above the workforce goal of 108

now, considering the nature of goals presumed that overachievement of either of them would attract no penalty, and we may seek a solution which would minimize the underachievement of both of them. Accordingly, the problem may be stated as follows

MINIMIZE Z = (d1-) + (d2-) Subject to 20x1 + 32x2 – (d1+) + (d1-) = 5400 0.3x1 + 0.75x2 + (d2-) - (d2+) = 108 2x1 + 4x2 + S1 = 600 4x1 + 5x2 + S2 = 1000 5x1 + 4x2 +S3 =1200x1,x2,S1,S2,S3,d1-,d1+,d2-and d2+ ≥ 0

SIMPLEX TABLE-1

BASIS

X1 X2 d1-

d1+

d2-

d2+

S1 S2 S3 bi Ratio

d1- 1 20 32 1 -1 0 0 0 0 0 5400

168.75

d-2 1 3/10 ¾* 0 0 1 -1 0 0 0 108 144

S1 0 2 4 0 0 0 0 1 0 0 600 150

S2 0 4 5 0 0 0 0 0 1 0 1000

200

S3 0 5 4 0 0 0 0 0 0 1 1200

300

Cj 0 0 1 0 1 0 0 0 0

Cj-Zj

-213/10

-131/4

0 0 1 0 0 0 0

SIMPLEX TABLE-2

BASIS

X1 X2 d1-

d1+

d2- d2+ S1 S2 S3 bi Ratio

d1_

1 36/5 0 1 -1 -128/3

128/3

0 0 0 792 299/16

X2 0 2/5 1 0 0 4/3 -4/3 0 0 0 144 -

S1 0 2/5 0 0 0 -16/3

16/3*

1 0 0 24 9/2

S2 0 2 0 0 0 -20/3

20/3

0 1 0 280 42

S3 0 17/5 0 0 0 -16/3

16/3

0 0 1 624 117

Cj 0 0 1 0 1 0 0 0 0

Cj-Zj

-36/5

0 0 1 131/3

-128/3

0 0 0

SIMPLEX TABLE-3

BASIS

X1 X2 d1-

d1+


d1_

1 4 0 1 -1 0 0 -8 0 0 600 150

X2 0 ½ 1 0 0 0 0 ¼ 0 0 150 300

d2+

0 3/40*

0 0 0 -1 1 3/16

0 0 9/2 60

S2 0 3/2 0 0 0 0 0 -5/4

1 0 250 500/3

S3 0 3 0 0 0 0 0 -1 0 1 600 200

Cj 0 0 1 0 1 o o o O

Cj-Zj

-4 0 0 1 0 0 8 0 0

SIMPLEX TABLE-4BASIS

X1 X2 d1-

d1+

d2- d2+ S1 S2 S3 Bi Ratio

d1_

1 0 0 1 -1 160/3

-160/3*

-18

0 0 360

27/4

X2 0 0 1 0 0 20/3 -20/3

-1 0 0 120

18

X1 0 1 0 0 0 -40/3

40/3

5/2

0 0 60 -

S2 0 0 0 0 0 20 -20 -5 1 0 160

8

S3 0 0 0 0 0 40 -40 -17/2

0 1 420

21/2

Cj 0 0 1 0 1 o o o O

Cj-Zj

o o o 1 -157/3

160/3

18 0 0


X1 X2 d1- d1+ d2-

d2+

S1 S2

S3 Bi

d2_

1 0 0 3/160 -3/160

1 -1 -27/80

0 0 27/4

X2 0 0 1 -1/8 1/8 0 0 5/4 0 0 75

X1 0 1 0 ¼ -1/4 0 0 -2 0 0 150

S2 0 0 0 -3/8 3/8 0 0 7/4 1 0 25

S3 0 0 0 -3/4 ¾ 0 0 5 0 1 150

Cj 0 0 1 0 1 o o o O

Cj-Zj

0 0 157/160

3/160 0 1 27/80

0 0

The optimal solution to the problem as obtained from simplex is x1 =150, x2 = 75, d2- = 27/4, s2 = 25, s3 =150, with the objective function value equal to 6.75 .

From this it is evident that workforce shall be 101.25, with the employment goal being underachieved to the extent of 6.75 people, while 150 kg of material M2 would remain unutilized. The other variables are non-basic so that all the available labour hours shall be used and the profit goal be met exactly.

DIFFERENTIAL WEIGHTING

In this case to avoid the problems associated with having large magnitudes of difference, we may standardize the weights so as to seek minimizing the percentage deviation from the goals in each objective.

For d1- : 1/5400 and for d2- : 1/108

Further, if the decision-maker feels that achieving the profit goal is twice as important as employment goal, he would give the weight as follows.

For d1-: 2x(1/5400)= 1/2700 and d2- : 1/108

Here we will do managerial assessment based on the trade of between different goals.It is given that one person underachieved of the unemployment goal is deemed equal to Rs100 in profit.By this we may rewrite the objective function as

Z = d1- + 100d2-

SIMPLEX TABLE-1

BASIS

X1 X2 d1-

d1+

d2-

d2+

S1 S2 S3 bi Ratio

d1- 1 20 32 1 -1 0 0 0 0 0 5400

168.75

d2- 100 3/10 ¾* 0 0 1 -1 0 0 0 108 144

S1 0 2 4 0 0 0 0 1 0 0 600 150

S2 0 4 5 0 0 0 0 0 1 0 1000

200

S3 0 5 4 0 0 0 0 0 0 1 1200

300

Cj 0 0 1 0 100 0 0 0 0

Cj-Zj

-50 -107

0 1 0 100 0 0 0

SIMPLEX TABLE-2

BASIS

X1 X2 d1-

d1+


d1_

1 36/5 0 1 -1 -128/3

128/3

0 0 0 792 299/16

X2 0 2/5 1 0 0 4/3 -4/3 0 0 0 144 -

S1 0 2/5 0 0 0 -16/3

16/3*

1 0 0 24 9/2

S2 0 2 0 0 0 -20/3

20/3

0 1 0 280 42

S3 0 17/5 0 0 0 -16/3

16/3

0 0 1 624 117

Cj 0 0 1 0 100 0 0 0 0

Cj-Zj

-36/5

0 0 1 428/3

-128/3

0 0 0

SIMPLEX TABLE-3

BASIS

X1 X2 d1-

d1+


d1_

1 4 0 1 -1 0 0 -8 0 0 600 150

X2 0 ½ 1 0 0 0 0 ¼ 0 0 500 500

d2+

0 3/40*

0 0 0 -1 1 3/16

0 0 9/2 60

S2 0 3/2 0 0 0 0 0 -5/4

1 0 250 500/3

S3 0 3 0 0 0 0 0 -1 0 1 600 200

Cj 0 0 1 0 100 o o o O

Cj-Zj

-4 0 0 1 100 0 8 0 0


X1 X2 d1-

d1+

d2- d2+ S1 S2 S3 Bi

d1_

1 0 0 1 -1 160/3

-160/3*

-18

0 0 360

X2 0 0 1 0 0 20/3 -20/3

-1 0 0 120

X1 0 1 0 0 0 -40/3

40/3

5/2

0 0 60

S2 0 0 0 0 0 20 -20 -5 1 0 160

S3 0 0 0 0 0 40 -40 -17/2

0 1 420

Cj 0 0 1 0 100 o o o O

Cj-Zj

o o o 1 140/3

160/3

18 0 0

It is given that from the optimal solution we have x1 =60 and x2 =120 , for a total profit 0f 20x 60 + 32x 120 = Rs 5040 , and d1- = 360, implying an underachievement of the profit goal of Rs 5400 by Rs 360. however, since d2- is equal to zero, it has the importance that the employment goal of 108 would be met . In terms of solution, the labour hours will be fully utilized while the raw materials would remain unutilized : 160 Kg of material M1 and 420 Kg of material M2.

QUES:-A firm producing two products radio and transistor which must be produced through two process I and Process II.90 hours are available in Process I & 72 hours are available in Process II.One radio requires 6 hours in Process I and 3 hours in process II. One transistor requires 3 hours in Process I and 6 hours in Process II. The profit is Rs.120 per radio and Rs 90 per transistor. Find the optimum solution in GPP, the firm has set up the following goals and has assigned them priorities P1,P2 & P3 where P1 is most important. Find the optimum solution.

PRIORITY GOAL

P1 Production of radio goal of 13.

P2 To achieve profit goal of Rs.1950

P3 Production of transistor goal of 5.

Let Dur = Under- achievement of radio goal Dor = Over- achievement of Radio goal Dup = Under-achievement of Profit goal Dop = Over-achievement of Profit goal Dut = Under-achievement of transistor goal Dot = Over-achievement of transistor goal

Objective function will be,

Minimize Z= P1 Dur +P2Dup + P3DutSubject To:- 120 X1 + 90X2 + Dup-Dop ( Profit Goal) X1 + Dur –Dor =13 (Radio Goal) X2 + Dut – Dot =5 (Transistor Goal) 6X1+3X2+S1+0S2=90 (Process I constraint) 3X1 + 6X2+0S1+S2=72 (Process II constraint)

Where X1,X2,S1,S2,Dur,Dor,Dup,Dop,Dut,Dot ≥0

P1 is more important than P2, it means P2 goal will not be achieved until P1 goal has been achieved. Similarly P3 goal will not be achieved till P2 goal is achieved.

Putting the above problem in the Simplex Table:-

SIMPLEX TABLE-1

BV SV X1 X2 S1 S2 Dup Dop Dur Dor Dut Dot ratio

P1 Dur 13 1* 0 0 0 0 0 1 -1 0 0 13

P2 DuP 1950

120 90 0 0 1 -1 0 0 0 0 65/4

P3 Dut 5 0 1 0 0 0 0 0 0 1 -1 ∞

0 S1 90 6 3 1 0 0 0 0 0 0 0 15

0 S2 72 3 6 0 1 0 0 0 0 0 0 24

Cj-Zj

P3 0 -1 - - - 0 - 0 - 1

P2 -120

-90

- - - 1 - 0 - 0

P1 -1 0 - - - 0 - 1 - 0

0 0 0 0 P2 0 P1 0 P3 0

C1-Z1 :-

P1= 0-1 = -1P2= 0-120= -120P3= 0-0 = 0

Similarly we have calculated the values for P1, P2,P3.

Each of the priorities P1, P2 and P3 has separate Zj and Cj-Zj in the order of their priority of goals. It is because the goals do not have the same dimension and cannot be just added.

Starting with the most important priority say P1, the most negative Cj-Zj is selected and entered. If there is no negative value in that row, we move to the next most important priority P2 and work the most negative Cj-Zj value.

SIMPLEX TABLE 2


0 X1 13 1 0 0 0 0 0 1 -1 0 0 -13

P2 DuP 390 0 90 0 0 1 -1 -120 120 0 0 3

P3 Dut 5 0 1 0 0 0 0 0 0 1 -1 ∞

0 S1 12 0 3 1 0 0 0 -6 6* 0 0 2

0 S2 13 0 6 0 1 0 0 -3 3 0 0 11

Cj-Zj

P3 - -1 - - - 0 0 0 - 1

P2 - -90

- - - 1 120 -120 - 0

P1 - 0 - - - 0 1 0 - 0

0 0 0 0 P2 0 P1 0 P3 0

SIMPLEX TABLE 3


0 X1 15 1 ½ 1/6

0 0 0 0 0 0 0 30

P2 DuP 150 0 30 -20

0 1 -1 0 0 0 0 5

P3 Dut 5 0 1 0 0 0 0 0 0 1 -1 50 Dor 2 0 ½ 1/

60 0 0 -1 1 O 0 4

0 S2 27 0 9/2

-1/2

1 0 0 0 0 0 0 6

Cj-Zj P3 - -1 0 - - 0 0 - - 1

P2 - -30

20 - - 1 0 - - 0

P1 - 0 0 - - 0 1 - - 0

0 0 0 0 P2 0 P1 0 P3 0

SIMPLEX TABLE 4

BV SV X1 X2 S1 S2

Dup Dop Dur Dor Dut Dot

0 X1 13 1 0 0 0 0 0 1 -1 0 0

P2 DuP 30 0 0 -30 0 1 -1 60 -60 0 0

P3 Dut 1 0 0 -1/3 0 0 0 2 -2 1 -1

0 X2 4 0 1 1/3 0 0 0 -2 2 0 0

0 S2 9 0 0 -2 1 0 0 9 -9 0 0

Cj-Zj P3 - - 1/3 - - 0 -2 2 - 1

P2 - - 30 - - 1 -60 60 - 0

P1 - - 0 - - 0 1 0 - 0

0 0 0 0 P2 0 P1 0 P3 0

There still exist -2 and -60 below Dur but it is to b disregarded as it will again lead to ass under achievement of radio goal.

So X1=13 X2=4

From the above table it is concluded that:-

The first goal P1 to produce 13 radio (goal) X1=13 has been achieved.

The second important goal P2= Rs 1950 has been under achieved by Rs 30 i.e. only Rs (1950-30) = Rs 1920 has been achieved.(120 * 3 + 90 * 4) =Rs 1920

The third goal P3 to produce 5 transistors is underachieved by 1. We still have 9 hours available in process II to manufacture 1 transistor but we have no time left in process I, so it cannot be produced.

Example 7.13 (N D Vohra)• A firm is engaged in producing two products A and B ,

which yield a unit profit of Rs. 400 and Rs. 300,resp.Both of these require a certain raw material whose supply is limited to 300 kg per month. The raw material requirement is one kg per unit of each product. Further , it is known that one unit of A requires 2 hours of processing while each unit of B needs one hour. The firm has a normal processing capacity of 400 hrs per month. The marketing department has stated that according to the prevailing conditions , and likely to continue in the forseeable future , the maximum number of units of A and B which can be sold each month is 150 and 350 , resp.

The manager of the firm has set the following goals in order of their importance – the priorities being preemptive.

• GOAL 1: Avoid under-utilisation of normal productive capacity.

• GOAL 2: Sell the maximum no. of units of products A& B .However ,since the two products yield the ratio of 400:300,the manager would give the weightage to the sales of the products in the same ratio.

• GOAL 3:Minimise overtime of processing capacity. You are required to formulate this as goal

programming problem and solve it graphically.

• Let x1 and x2 be the number of units of products A & B resp. to be produced every month.

• Now formulating this as goal programming problem:-• x1 + x2 ≤ 300

(Raw material constraint) 2x1 + x2 + d1

- + d1+ = 400 (Processing capacity constraint)

Where d1- = no. of hours unutilized processing capacity

d1+ = no. of hours overtime processing operation

Since the sales goals given are in terms of the max. possible sales volumes ,they an be expressed as :-

x1+ d2- =150 and x2 + d3

- = 350

(Here we are taking only deviational variables for under achievement and not for overachievement as the sales goals are the maximum ones.)

PRODUCTS Labor hours

Raw Material

Profit ProjectedMax. Sales

A(x1) 2 1 Kg Rs. 400

150 units

B(x2) 1 1 Kg Rs. 300

350 units

Constraints 400 Hours

300 Kg

• With P1 , P2 , P3 being the preemptive priority factors , the goal programming problem may be stated as follows:

• Minimize Z = P1d1- +P2(4d2

-+3d3-)+ P3d1

+

• Subject to X1+ X2 ≤ 300

2X1 + X2 + d1- + d1

+ = 400

X1 + d2- = 150

X2 + d3- = 350

X1 , X2 , d1- , d2

+ , d3- ≥ 0

Thank You…

Goal Progrmming

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