In: Math
Weekly production schedules are needed for the manufacture of two products X and Y. Each unit of X uses one component made in the factory, while each unit of Y uses two of the components, and the factory has a maximum output of 80 components a week. Each unit of X and Y requires 10 hours of subcontracted work and agreements have been signed with subcontractors for a minimum weekly usage of 200 hours and a maximum weekly usage of 600 hours. The marketing department says that all production of Y can be sold but there is a maximum demand of 50 units of X, despite a long-term contract to supply 10 units of X to one customer. The net profit on each unit of X and Y is $200 and $300 respectively.
Linnear Programming step by step, show work manually without using excel, Include a graph, and corner. I would like to try to understand how the question is done step by step. Thank you so much
Let the weekly production are needed for the manufacture of two products X and Y be x units and y units respectively.
Then by the given conditions we have,
Maximize Z = 200x+300y
Subject to x+2y 80
10x+10y 200
10x+10y 600
x 50
x,y 0
i.e., Maximize Z = 200x+300y
Subject to x+2y 80
x+y 20
x+y 60
x 50
x,y 0
Introducing slack variables a,c,d and surplus variable b the problem becomes,
Maximize Z = 200x+300y+0a+0b+0c+0d
Subject to x+2y+a = 80
x+y-b = 20
x+y+c = 60
x+d = 50
x,y,a,b,c,d 0
COMPUTATION TABLE :
cj | 200 | 300 | 0 | 0 | 0 | 0 | ||
cB | B | b | a1 | a2 | a3 | a4 | a5 | a6 |
0 | a3 | 80 | 1 | 2 | 1 | 0 | 0 | 0 |
0 | a4 | 20 | 1 | 1 | 0 | -1 | 0 | 0 |
0 | a5 | 60 | 1 | 1 | 0 | 0 | 1 | 0 |
0 | a6 | 50 | 1 | 0 | 0 | 0 | 0 | 1 |
zj-cj | -200 | -300 | 0 | 0 | 0 | 0 | ||
0 | a3 | 40 | -1 | 0 | 1 | 2 | 0 | 0 |
300 | a2 | 20 | 1 | 1 | 0 | -1 | 0 | 0 |
0 | a5 | 40 | 0 | 0 | 0 | 1 | 1 | 0 |
0 | a6 | 50 | 1 | 0 | 0 | 0 | 0 | 1 |
zj-cj | 100 | 0 | 150 | -300 | 0 | 0 | ||
0 | a3 | 20 | -1/2 | 0 | 1/2 | 1 | 0 | 0 |
300 | a2 | 40 | 1/2 | 1 | 1/2 | 0 | 0 | 0 |
0 | a4 | 20 | 1/2 | 0 | -1/2 | 0 | 1 | 0 |
0 | a6 | 50 | 1 | 0 | 0 | 0 | 0 | 1 |
zj-cj | -50 | 0 | 150 | 0 | 0 | 0 | ||
0 | a3 | 40 | 0 | 0 | 0 | 1 | 1 | 0 |
300 | a2 | 20 | 0 | 1 | 1 | 0 | -1 | 0 |
200 | a1 | 40 | 1 | 0 | -1 | 0 | 2 | 0 |
0 | a6 | 10 | 0 | 0 | 1 | 0 | 2 | 1 |
zj-cj | 0 | 0 | 100 | 0 | 100 | 0 |
In the last tableau all zj-cj 0.
The optimal solution is : x = 40, y = 20 and Zmax = (200*40)+(300*20) = 14000
Therefore, the weekly production are needed for the manufacture of two products X and Y are 40 units and 20 units respectively and total profit is $14000.