Chapter 2 An Introduction to Linear Programming
Chapter 2
An Introduction to Linear Programming
18. a.
Max 4A + 1B + 0S1 + 0S2 + 0S3
s.t.
10A + 2B + 1S1 = 30 3A + 2B + 1S2 = 12 2A + 2B + 1S3 = 10
A, B, S1, S2, S3 0
b.
c. S1 = 0, S2 = 0, S3 = 4/7
23. a. Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced
Max 2400E + 1800L
s.t.
6E + 3L 2100 Engine time L 280 Lady-Sport maximum 2E + 2.5L 1000 Assembly and testing E, L 0
b.
c. The binding constraints are the manufacturing time and the assembly and testing time.
38. a. Let S = yards of the standard grade material per frame P = yards of the professional grade material per frame
Min 7.50S + 9.00P
s.t.
0.10S + 0.30P 6 carbon fiber (at least 20% of 30 yards) 0.06S + 0.12P 3 kevlar (no more than 10% of 30 yards) S + P = 30 total (30 yards) S, P 0
b.
c.
Extreme Point Cost
(15, 15) 7.50(15) + 9.00(15) = 247.50
(10, 20) 7.50(10) + 9.00(20) = 255.00
The optimal solution is S = 15, P = 15
d. Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is reduced to 7.50(15) + 8(15) = $232.50.
e. At $7.40 per yard, the optimal solution is S = 10, P = 20. The value of the optimal solution is reduced to 7.50(10) + 7.40(20) = $223.00. A lower price for the professional grade will not change the S = 10, P = 20 solution because of the requirement for the maximum percentage of kevlar (10%).
39. a. Let S = number of units purchased in the stock fund M = number of units purchased in the money market fund
Min 8S + 3M
s.t.
50S + 100M 1,200,000 Funds available 5S + 4M 60,000 Annual income M 3,000 Minimum units in money market S, M, 0
Optimal Solution: S = 4000, M =
An Introduction to Linear Programming
18. a.
Max 4A + 1B + 0S1 + 0S2 + 0S3
s.t.
10A + 2B + 1S1 = 30 3A + 2B + 1S2 = 12 2A + 2B + 1S3 = 10
A, B, S1, S2, S3 0
b.
c. S1 = 0, S2 = 0, S3 = 4/7
23. a. Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced
Max 2400E + 1800L
s.t.
6E + 3L 2100 Engine time L 280 Lady-Sport maximum 2E + 2.5L 1000 Assembly and testing E, L 0
b.
c. The binding constraints are the manufacturing time and the assembly and testing time.
38. a. Let S = yards of the standard grade material per frame P = yards of the professional grade material per frame
Min 7.50S + 9.00P
s.t.
0.10S + 0.30P 6 carbon fiber (at least 20% of 30 yards) 0.06S + 0.12P 3 kevlar (no more than 10% of 30 yards) S + P = 30 total (30 yards) S, P 0
b.
c.
Extreme Point Cost
(15, 15) 7.50(15) + 9.00(15) = 247.50
(10, 20) 7.50(10) + 9.00(20) = 255.00
The optimal solution is S = 15, P = 15
d. Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is reduced to 7.50(15) + 8(15) = $232.50.
e. At $7.40 per yard, the optimal solution is S = 10, P = 20. The value of the optimal solution is reduced to 7.50(10) + 7.40(20) = $223.00. A lower price for the professional grade will not change the S = 10, P = 20 solution because of the requirement for the maximum percentage of kevlar (10%).
39. a. Let S = number of units purchased in the stock fund M = number of units purchased in the money market fund
Min 8S + 3M
s.t.
50S + 100M 1,200,000 Funds available 5S + 4M 60,000 Annual income M 3,000 Minimum units in money market S, M, 0
Optimal Solution: S = 4000, M =