Quantitative Methods
Statistics and Quantitative Methods
ASSIGNMENT: 2
PREPARED FOR:
CHOWDHURY
PREPARED BY:
VALLEY STATE UNIVERSITY
DATE:
13. The Electrotech Corporation manufactures two industrial-sized electrical devices: generators and alternators. Both of these products require wiring and testing during the assembly process. Each generator requires 2 hours of wiring and 1 hour of testing and can be sold for a $250 profit. Each alternator requires 3 hours of wiring and 2 hours of testing and can be sold for a $150 profit. There are 260 hours of wiring time and 140 hours of testing time available in the next production period and electrotech wants to maximize profit.
(A) Formulate an LP model for this problem.
Decision Variable X1 = Generator
X2 = Alternator
Obj. = Maximum Profit X1 | X2 | Time | Activity | 2 | 3 | 260 | Wiring | 1 | 2 | 140 | Testing |
Max. Profit = 250X1 + 150X2
Subject to: 2x1 + 3x2 < 260 X1 + 2x2 < 140 X,y > 0
2x1 = 3x2 < 260
X1 =0 x2 = 0
2(0) = 3x2 = 260 2x1 = 3(0) =260 3x2 = 260 2x1 = 260 X2 = 86.7 x1 = 130
X1 + 2x2 = 140
X1 = 0 x2 =0 2x2 = 140 x1 = 2(0) 140 X2 = 70 x1 = 140
250x1 + 150x2 = Profit
Profit – 250x1 = 150x2
150x2 = Profit -250
150 150 150
X2 = profit/150 – 5/3 Rise/Run
(B) Sketch the feasible region for this problem.
See Attachment
(C) Determine the optimal solution to this problem using level curves.
The optimal solution to this problem using level curves is 130.00
Total Profit $32,500.00 or 41,175.00
See Attachment
(D) Interpret your findings.
In order for the Electrotech Corporation to maximize their profit they will have to produce 130 Generators 0 alternators for a maximum profit of 32,500. Any deviation from this will lower the company’s
ASSIGNMENT: 2
PREPARED FOR:
CHOWDHURY
PREPARED BY:
VALLEY STATE UNIVERSITY
DATE:
13. The Electrotech Corporation manufactures two industrial-sized electrical devices: generators and alternators. Both of these products require wiring and testing during the assembly process. Each generator requires 2 hours of wiring and 1 hour of testing and can be sold for a $250 profit. Each alternator requires 3 hours of wiring and 2 hours of testing and can be sold for a $150 profit. There are 260 hours of wiring time and 140 hours of testing time available in the next production period and electrotech wants to maximize profit.
(A) Formulate an LP model for this problem.
Decision Variable X1 = Generator
X2 = Alternator
Obj. = Maximum Profit X1 | X2 | Time | Activity | 2 | 3 | 260 | Wiring | 1 | 2 | 140 | Testing |
Max. Profit = 250X1 + 150X2
Subject to: 2x1 + 3x2 < 260 X1 + 2x2 < 140 X,y > 0
2x1 = 3x2 < 260
X1 =0 x2 = 0
2(0) = 3x2 = 260 2x1 = 3(0) =260 3x2 = 260 2x1 = 260 X2 = 86.7 x1 = 130
X1 + 2x2 = 140
X1 = 0 x2 =0 2x2 = 140 x1 = 2(0) 140 X2 = 70 x1 = 140
250x1 + 150x2 = Profit
Profit – 250x1 = 150x2
150x2 = Profit -250
150 150 150
X2 = profit/150 – 5/3 Rise/Run
(B) Sketch the feasible region for this problem.
See Attachment
(C) Determine the optimal solution to this problem using level curves.
The optimal solution to this problem using level curves is 130.00
Total Profit $32,500.00 or 41,175.00
See Attachment
(D) Interpret your findings.
In order for the Electrotech Corporation to maximize their profit they will have to produce 130 Generators 0 alternators for a maximum profit of 32,500. Any deviation from this will lower the company’s