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1) After several years working as a corporate lawyer, a woman decides that she would prefer to start her own business. Her hobby is making pottery and she decides to turn her hobby into a business. After some experiments she decides to specialize in making cups and bowls. She sets her prices so that she will make a S2 profit on each cup and a profit of S1.50 on each bowl. You can assume that she sells every cup and every bowl that she makes. The woman has 40 hours per week to make the cups and bowls. It takes her 12 minutes to make a cup and 6 minutes to make a bowl. There are 60 minutes in an hour. Each week, the woman orders 125 kg of clay. (This is the raw material that pottery is made from.) Each cup uses 375g of clay. Each bowl uses 500g of clay. There are 1000g of clay in lkg of clay. Although she has many reasons for pursuing a new business, one of the reasons is to make money, and the woman would like the business to be as profitable as possible. a) (2 points) Use the letter C to represent the number of the cups and B to represent the number of bowls. Use the given information to write down the linear programming problem to maximize the weekly profit. b) (2 points) Solve the above programming problem by geometric solution method c) (3 points) By using the Simplex Algorithm find the maximum weekly profit.

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Answer #1

(a)

From question, Number of cups = C and Number of bowls = B

Required to MAXIMIZE WEEKLY PROFITS, Z = 2*C + 1.5*B

Subject to Contraints:

  • 12*C + 6*B = 2400 ................ Time constraint of 2400 minutes per week
  • 375*C + 500*B = 125000....... Total clay available per week in grams

(b)

y2400 900yc125000 250 200 150 100 50 x1 50 100 150 200 250 3:00 350.

The value of the objective function at each of these extreme points is as follows Extreme Point Coordinates (x1- 2) Objective function value Lines through Extreme Point 0(0, 0) 2(0) + 1.5(0) = 0 112r + 62400 A(200, 0) 2(200+1.5(0) 400 1 → 12x1 +6x2 2400 125000 B(120, 160) 2(120)+ 1.5(160) 480 2 → 375x, + 500x2 2375x+500x125000 C(0. 250) 2(0)1.5(250) 375

The maximum value of the objective function Z = 480, occurs at the extreme point (120,160).

Hence, the optimal solution to the given LP problem is : C =120, B =160 and max Z = 480.

(c)

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of types we should add slack variable S 2. As the constraint-2 is of types we should add slack variable S2 After introducing slack variables Max Z = 2x1 + 1.5x2 + 0S1 +0S2 subject to 2400 375 x1 500 x2 S2-125000 and x1,x2.S1.S2 2 0

Iteration-1 1.5 t) MinRatic Ху *1 2400 12 125000 375 S1 2400 (12) 200 125000 375 500 0 - 333.3333 1.5 0 0)

Negative minimum Z, - C, is -2 and its column index is 1. So, the entering variable is x Minimum ratio is 200 and its row index is 1. So, the leaving basis variable is S . The pivot element is 12. Enteringx,Departing S,Key Element -12 + R1(new) R1(old) 12 t R2(new) - R2(old) 375R (new)

Iteration-2 MinRatio 0.0833 - 400 0.5 50000 312.5 (312.5) 31.25 160 Z-400 0.1667 -0.5 T 0.1667

Negative minimum Z, - C, is -0.5 and its column index is 2. So, the entering variable is x2 Minimum ratio is 160 and its row index is 2. So, the leaving basis variable is S2. .. The pivot element is 312.5 Entering x2,DepartingS2, Key Element + R2(new)-R2(old) x 0.0032 312.5 + R(new)- R1(old)- 0.5R2(new)

Iteration-3 0 MinRatio 0.1333 -0.1 0.1167 0.1167 -0.0016 0.0032 0.0016 0.0016 160 0 Z = 480 1.5 Z, - C 0 0

Since all Z,-C,20 Hence, optimal solution is arrived with value of variables as: x1 = 120, x2-160 Max Z = 480

Hence, optimal solution is arrived with value of variables as :
C = 120, B = 160

Max weekly profits, Z=480

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