MBA, Semester 2 Operations Management Ms. Aarti Mehta Sharma Amity Business School MBA, Semester 2 Operations Management Ms. Aarti Mehta Sharma

Linear programming

An efficient method for determining an optimal decision from a large no. of decisions. Selection of product mix Maximize profits, subject to constraints

Objective fn Z = c1x1 + c2x2 +…… cnxn c1,c2 are uncontrollable variables X1, x2 are different attributes

General model of LPP Optimize (maximize or minimize) Z = c1x1 + c2x2 +…… cnxn Subject to linear constraints A11x1 + a12x2 + ….. + a1nxn (≤,=,>=) A21x1 + a22x2 + ….. + a2nxn (≤,=,>=) . Am1x1 + am2x2 + ….. + amnxn (≤,=,>=) X1,x2….xn ≥ 0

Graphical method Simplex method

Graphical method A 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.

The male members of the skilled workers are paid Rs 6000 p. m 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.

X1 = male workers X2 = female workers Maximize Z = 10,000x1 + 8500 x2 X1 + x2 ≤ 11 6000 x1 + 5000 x2 ≤ 60,000 6x1 + 5 x2 ≤ 60 x1≥ 0; x2 ≥ 0

Turn inequalities into equalities X1 + x2 ≤ 11 X1 + x2 = 11 when x1 =0, x2 = 11 when x1 =11, x2 = 0 6x1 + 5 x2 ≤ 60 6x1 + 5 x2 = 60 when x1 =0, x2 = 12 when x1 =10, x2 = 0

O- 0,0 B – 10,0 P – 5,6 C – 0,11

Maximize Z = 10,000x1 + 8500 x2 O ; 10,000 * 0 + 8500 * 0 = 0 B ; 10,000 * 10 + 8500 * 0 = 1,00,000 C ; 10,000 * 0 + 8500 * 11 = 93,500 O ; 10,000 * 5 + 8500 * 6 = 1,01,000 X1 = 5 X2 = 6

Q 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

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

maximize Z = 6x1 + 8 X2 + 0s1 +0s2 subject to 5 x1 + 10 x2 = 60 4 x1 + 4 x2 = 40 x1 ≥ 0 ; x2 ≥0;

Q 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

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

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

CBi Cj 6 8 solution ratio Basic variables x1 x2 s1 s2 S1 5 4 10 1 60 40 60/10 = 6 40/4 =10 Zj Cj - Zj

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

CBi Cj 6 8 solution ratio Basic variables x1 x2 s1 s2 1/2 2 1 1/10 -2/5 16 6/1/2 = 12 16/2 = 8 Zj Cj - Zj 4 4/5 -4/5 48

CBi Cj 6 8 solution Basic variables x1 x2 s1 s2 1 1/5 -1/5 -1/4 1/2 2 Zj Cj - Zj 2/5 -2/5 -1 64

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

Case Maximize Z= 10x1 + 15 x2 + 20 x 3 Subject to

Solution Maximize Z= 10x1 + 15 x2 + 20 x3 +0s1 + 0s2 Subject to Where s1 & s2 are slack variables

CBi Cj 10 15 20 solution ratio Basic variables x1 x2 x3 s1 s2 S1 2 3 4 9 6 1 24 30 24/6 = 4 5 Zj Cj - Zj

CBi Cj 10 15 20 soln ratio Basic var x1 x2 x3 s1 s2 X3 1/3 1 2/3 5 1/6 -1 4 6 12 Zj Cj - Zj 20/3 10/3 40/3 5/3 -10/3 80 (20*4)

CBi Cj 10 15 20 Solution Basic variables x1 x2 x3 s1 s2 X3 X1 1 -1 5 1/2 -1/3 2 6 Zj Cj - Zj 30 -15 10/3 -10/3 100

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

Case 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.

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.

Maximize Z = 4x1 +3x2 Subject to the constraints 2x1 + x2 <=1000 X1 +x2 <= 800 X1<=400 X2<= 700 and x1,x2 >=0

Maximize Z = 4x1 +3x2 +0s1 + 0s2 +0s3 + 0s4 Subject to the constraints 2x1 + x2 + s1 =1000 X1 +x2 + s2= 800 X1+ s3 =400 X2 + s4 = 700 and x1,x2,s1,s2,s3,s4 >=0

Cj 4 3 S1 1000 2 1 500 S2 800 S3 400 S4 700 Not defined zj Cj - Zj Profit per unit Cb Basic variables Soln variables x1 x2 s1 s2 s3 s4 Min ratio S1 1000 2 1 500 S2 800 S3 400 S4 700 Not defined zj Cj - Zj

Cj 4 3 Profit per unit Cb Basic variables Soln variables x1 x2 s1 s2 s3 s4 Min ratio S1 200 1 -2 200/1 S2 400 -1 400/1 - S4 700 700/1 zj Cj - Zj -4

R2(new) R2(old)-R1(new)

Cj 4 3 Profit per unit Cb Basic variables Soln variables x1 x2 s1 s2 s3 s4 Min ratio X2 200 1 -2 S2 -1 200/1 X1 400 400/1 S4 500 2 500/2 zj Cj - Zj -3

R1(new) R1(old) + 2R2(new)

Cj 4 3 600 1 -1 2 S3 200 S4 100 -2 zj Cj - Zj Profit per unit Cb Profit per unit Cb Basic variables Soln variables x1 x2 s1 s2 s3 s4 600 1 -1 2 S3 200 S4 100 -2 zj Cj - Zj

Since Cj – Zj <= 0, the current solution cannot be improved upon 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.