MBA, Semester 2 Operations Management Ms. Aarti Mehta Sharma

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AMITY GLOBALBUSINESS SCHOOL Bangalore

MBA, Semester 2

Operations Management

Ms. Aarti Mehta Sharma

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Linear programming

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An efficient method for determining an optimal decision from a large no. of decisions.

- Selection of product mix- Maximize profits, subject to constraints

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- Objective fn Z = c1x1 + c2x2 +…… cnxn

- c1,c2 are uncontrollable variables

- X1, x2 are different attributes

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AMITY GLOBALBUSINESS SCHOOL Bangalore General model

of LPP• Optimize (maximize or minimize) Z = c1x1 + c2x2 +…… cnxn

Subject to linear constraintsA11x1 + a12x2 + ….. + a1nxn (≤,=,>=)A21x1 + a22x2 + ….. + a2nxn (≤,=,>=).Am1x1 + am2x2 + ….. + amnxn (≤,=,>=)X1,x2….xn ≥ 0

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Graphical methodSimplex method

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AMITY GLOBALBUSINESS SCHOOL BangaloreGraphical

methodA small scale industry manufactures

electrical regulators, the assembly of which is being accomplished by a small group of skilled workers, both men and women. Due to the limitations of space and finance, the no. of workers employed cannot exceed 11 and their salary bill not exceed more than Rs 60,000 per month.

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The male members of the skilled workers are paid Rs 6000 p.m. and female workers Rs 5000 p.m. data collected on the performance of these workers indicate that a male member contributes Rs 10,000 pm to total return of the industry and a female worker contributes Rs 8,500 p.m. determine the no. of male and female workers to be employed to maximize the monthly total return.

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• X1 = male workers• X2 = female workers• Maximize Z = 10,000x1 + 8500 x2

X1 + x2 ≤ 116000 x1 + 5000 x2 ≤ 60,0006x1 + 5 x2 ≤ 60 x1≥ 0; x2 ≥ 0

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Turn inequalities into equalitiesX1 + x2 ≤ 11X1 + x2 = 11 when x1 =0, x2 = 11 when x1 =11, x2 = 06x1 + 5 x2 ≤ 606x1 + 5 x2 = 60 when x1 =0, x2 = 12 when x1 =10, x2 = 0

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x1

x2

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• O- 0,0• B – 10,0• P – 5,6• C – 0,11

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Maximize Z = 10,000x1 + 8500 x2

O ; 10,000 * 0 + 8500 * 0 = 0B ; 10,000 * 10 + 8500 * 0 = 1,00,000C ; 10,000 * 0 + 8500 * 11 = 93,500O ; 10,000 * 5 + 8500 * 6 = 1,01,000X1 = 5

X2 = 6

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AMITY GLOBALBUSINESS SCHOOL BangaloreQ

A company manufactures two types of products P1 and P2. each product uses lathe and milling machines. The processing time per unit of P1 on the lathe is 5 hours and on the milling machine is 4 hours. The processing time per unit of P2 on the lathe is 10 hours and on the milling machine is 4 hours. The maximum no. of hours available per week on the lathe and the milling machine are 60 hrs and 40 hrs. The profit per unit of P1 = Rs 6.00 and P2 = Rs 8.00. Formulate a LP model to determine the production volume of

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Each of the products such that the total profit is maximised.

Soln : maximize Z = 6x1 + 8 X2

subject to

5 x1 + 10 x2 ≤ 60 4 x1 + 4 x2 ≤ 40 x1 ≥ 0 ; x2 ≥0

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maximize Z = 6x1 + 8 X2 + 0s1 +0s2

subject to

5 x1 + 10 x2 = 60

4 x1 + 4 x2 = 40

x1 ≥ 0 ; x2 ≥0;

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AMITY GLOBALBUSINESS SCHOOL BangaloreQ

A company manufactures two types of products P1 and P2. each product uses lathe and milling machines. The processing time per unit of P1 on the lathe is 5 hours and on the milling machine is 4 hours. The processing time per unit of P2 on the lathe is 10 hours and on the milling machine is 4 hours. The maximum no. of hours available per week on the lathe and the milling machine are 60 hrs and 40 hrs. The profit per unit of P1 = Rs 6.00 and P2 = Rs 8.00. Formulate a LP model to determine the production volume of

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Each of the products such that the total profit is maximised.

Soln : maximize Z = 6x1 + 8 X2

subject to

5 x1 + 10 x2 ≤ 60 4 x1 + 4 x2 ≤ 40 x1 ≥ 0 ; x2 ≥0

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maximize Z = 6x1 + 8 X2 + 0s1 +0s2

subject to

5 x1 + 10 x2 + s1 = 60

4 x1 + 4 x2 + s2 = 40

x1 ≥ 0 ; x2 ≥0; s1,s2 ≥ 0

S1, s2 are called slack variables

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CBi Cj 6 8 0 0 solution

ratio

Basic variables

x1 x2 s1 s2

00

S1s2

54

104

10

01

6040

60/10 = 640/4 =10

Zj Cj - Zj

06

08

00

00

0

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• Continue till last row entries are all zero or negative• Choose largest value in last row = 8• Column with largest value is known as pivotal

column• Divide values of soln column with values of pivotal

colum• Choose min – pivotal row • 10 becomes pivot no. • Reduce 10 to 1; reduce other values in col to 0

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ratio

Basic variables

x1 x2 s1 s2

80

x2

s2

1/22

10

1/10-2/5

01

616

6/1/2 = 1216/2 = 8

Zj Cj - Zj

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80

4/5-4/5

00

48

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Basic variables

x1 x2 s1 s2

86

x2x1

01

10

1/5-1/5

-1/41/2

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Zj Cj - Zj

60

80

2/5-2/5

1-1

64

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• Stop when all values in last row are 0 or negative

• The corresponding optimal solution is : X1(production volume of P1) = 8 units X2(production volume of P2) = 2 units

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AMITY GLOBALBUSINESS SCHOOL BangaloreCase

Maximize Z= 10x1 + 15 x2 + 20 x 3

Subject to

2x1 + 4x2 + 6x 3<= 24

3x1 + 9x2 + 6x3<= 30

x1,x2, x3 >=0

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AMITY GLOBALBUSINESS SCHOOL BangaloreSolution

Maximize Z= 10x1 + 15 x2 + 20 x3 +0s1 + 0s2 Subject to2x1 + 4x2 + 6x 3 + s1 = 24 3x1 + 9x2 + 6x3 + s2 = 30 x1,x2, x3 s1, s2,s3 >=0

Where s1 & s2 are slack variables

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CBi Cj 10 15 20 0 0 solution

ratio

Basic variables

x1 x2 x3 s1 s2

00

S1s2

23

49

66

10

01

2430

24/6 = 45

Zj Cj - Zj

010

015

020

00

00

0

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CBi

Cj 10 15 20 0 0 soln ratio

Basic var

x1 x2 x3 s1 s2

20

0

X3

s2

1/3

1

2/3

5

1

0

1/6

-1

0

1

4

6

12

6

Zj

Cj - Zj

20/3

10/3

40/3

5/3

20

0

10/3

-10/3

0

0

80 (20*4)

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CBi Cj 10 15 20 0 0 SolutionBasic variables

x1 x2 x3 s1 s2

20

10

X3

X1

0

1

-1

5

1

0

1/2

-1

-1/3

1

2

6

Zj

Cj - Zj

10

0

30

-15

20

0

0

0

10/3

-10/3

100

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Since all the values of Cj-Zj are less than or equal to zero, the optimality is there and

X1 = 6, X2 = 0; X3 = 2 and Zoptimum = 100

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AMITY GLOBALBUSINESS SCHOOL BangaloreCase

A company makes two kinds of leather bags bag A and bag B. Bag A is a high quality bag and bag B is of lower quality. The respective profits are Rs. 4 and Rs. 3 per bag. The production of each of type A requires twice as much time as a bag of type B. if all bags were of type B, the company could make 1000 bags per day.

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The supply of leather is sufficient for only 800 bags per day (Both A & B combined). Bag A requires a fancy buckle and only 400 of these are available . There are only 700 buckles a day available for belt B.

What should be the daily production of each type of belt ? Formulate this problem as an LP problem and solve it using the simplex method.

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Maximize Z = 4x1 +3x2

Subject to the constraints2x1 + x2 <=1000

X1 +x2 <= 800

X1<=400

X2<= 700 and x1,x2 >=0

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Maximize Z = 4x1 +3x2 +0s1 + 0s2 +0s3 + 0s4

Subject to the constraints2x1 + x2 + s1 =1000

X1 +x2 + s2= 800

X1+ s3 =400

X2 + s4 = 700 and x1,x2,s1,s2,s3,s4 >=0

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Cj 4 3 0 0 0 0Profit per unit Cb

Basic variables

Soln variables

x1 x2 s1 s2 s3 s4 Min ratio

0 S1 1000 2 1 1 0 0 0 5000 S2 800 1 1 0 1 0 0 8000 S3 400 1 0 0 0 1 0 4000 S4 700 0 1 0 0 0 1 Not

defined zj 0 0 0 0 0 0

Cj - Zj 4 3 0 0 0 0

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AMITY GLOBALBUSINESS SCHOOL BangaloreCj 4 3 0 0 0 0Profit per unit Cb

Basic variables

Soln variables


0 S1 200 0 1 1 0 -2 0 200/10 S2 400 0 1 0 1 -1 0 400/14 x1 400 1 0 0 0 1 0 -0 S4 700 0 1 0 0 0 1 700/1

zj 4 0 0 0 4 0

Cj - Zj 0 3 0 0 -4 0

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R2(new) R2(old)-R1(new)R4(new)R4(old)-R1(new)

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AMITY GLOBALBUSINESS SCHOOL BangaloreCj 4 3 0 0 0 0Profit per unit Cb

Basic variables

Soln variables


3 X2 200 0 1 1 0 -2 00 S2 200 0 0 -1 1 1 0 200/14 X1 400 1 0 0 0 1 0 400/10 S4 500 0 0 -1 0 2 1 500/2 zj 4 3 3 0 -2 0

Cj - Zj 0 0 -3 0 2 0

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R1(new) R1(old) + 2R2(new)

R3(new) R3(old) – R2(new)

R4(new) R4(old) – 2R2

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Cj 4 3 0 0 0 0Profit per unit Cb

Basic variables

Soln variables

x1 x2 s1 s2 s3 s4

3 x2 600 0 1 -1 2 0 00 S3 200 0 0 -1 1 1 04 x1 200 1 0 1 -1 0 00 S4 100 0 0 1 -2 0 1 zj 4 3 1 2 0 0

Cj - Zj 0 0 -1 -2 0 0

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Since Cj – Zj <= 0, the current solution cannot be improved upon. Thus , the company must manufacture x1 = 200 bags of type A and x2 = 600 bags of type B to obtain the max profit of Rs. 2,600.

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MBA, Semester 2 Operations Management Ms. Aarti Mehta Sharma

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