Assn 3
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711.a. 1 Boston
7 100
1
11
Denver
8 13
20
100
2 Atlanta
2 Dallas
70
17 12 10
8
150
50
3
3 Los Angeles
60
18 13
Chicago 16 4 St. Paul
b.
80
There are alternative optimal solutions. Solution #1 Denver to St. Paul: Atlanta to Boston: Atlanta to Dallas: Chicago to Dallas: Chicago to Los Angeles: Chicago to St. Paul:
10 50 50 20 60 70
Solution # 2 Denver to St. Paul: Atlanta to Boston: Atlanta to Los Angeles: Chicago to Dallas: Chicago to Los Angeles: Chicago to St. Paul:
10 50 50 70 10 70
Total Profit: $4240 If solution #1 is used, Forbelt should produce 10 motors at Denver, 100 motors at Atlanta, and 150 motors at Chicago. There will be idle capacity for 90 motors at Denver. If solution #2 is used, Forbelt should adopt the same production schedule but a modified shipping schedule.
720
A linear programming formulation of this problem can be developed as follows. Let the first letter of each variable name represent the professor and the second two the course. Note that a DPH variable is not created because the assignment is unacceptable. Max 2.8AUN + 2.2AMB +
3.3AMS +
3.0APH + 3.2BUN +
∙ ∙ ∙
s.t. AUN +
AUN +
AMB BUN
+ +
AMS + BMB + CUN +
BUN AMB
+ +
CUN + BMB + AMS +
All Variables ≥ 0 Optimal Solution: A to MS course B to Ph.D. course C to MBA course D to Undergraduate course Max Total Rating
Rating 3.3 3.6 3.2 3.2 13.3
APH BMS CMB DUN DUN CMB BMS APH
+ + +
BPH CMS + CPH DMB + DMS
+ + +
DMB CMS + DMS BPH + CPH
+ 2.5DMS ≤ ≤ ≤ ≤ = = = =
1 1 1 1 1 1 1 1
a.
7-26.
1 Augusta
300
2 Tupper Lake
100
5
5 4 Portsmouth
4
6 NewYork
100
7 Philadelphia
150
5 7
3
150
8
3 Albany
7
5 Boston
6 10
b. Min s.t.
7x13 + 5x14 + 3x23 + 4x24 + 8x35 + 5x36 + 7x37 + 5x45 + 6x46 + 10x47 x13 + x14 x13
x23 x23 x14
+ x24 x24
+ x35 x35
+ x36
+ x37
+ x36
+ x45 + x45
x37 xij ≥ 0 for all i and j c.
Optimal Solution:
Variable Value x13 x14 x23 x24 x35 x36 x37 x45 x46 x47
Objective Function: 4300
50 250 100 0 0 0 150 150 100 0
+ x46 + x46
+ x47
+ x47
≤ ≤ = = = = =
300 100 0 0 150 100 150
8-16.
a.
min 105x9 + x9 x9 + x9 + x9 + x9 x9 + x9 +
105x10 + 105x11 + 32y9 + 32y10 + 32y11 + 32y12 + 32y1 + 32y2 + 32y3 ≥ 6 + y9 ≥ 4 + y9 + x10 y10 ≥ 8 x10 + x11 + y9 + y10 + y11 ≥ 10 x10 + x11 + y9 + y10 + y11 + y12 ≥ 9 + x10 + x11 y10 + y11 + y12 + y1 + ≥ 6 x11 y11 + y12 + y1 + y2 + x10 y12 + y1 + y2 + y3 ≥ 4 + x10 + x11 y1 + y2 + y3 ≥ 7 x10 +
x11
+
x11
y2 + +
y3 ≥ 6 y3 ≥ 6
xi, yj ≥ 0 and integer for i = 9, 10, 11 and j = 9, 10, 11, 12, 1, 2, 3 b.
c.
Solution to LP Relaxation obtained using LINDO/PC: y9 = 6
y12 = 6
y11 = 2
y1 = 1
y3 = 6
All other variables = 0. Cost: $672.
The solution to the LP Relaxation is integral therefore it is the optimal solution to the integer program. A difficulty with this solution is that only parttime employees are used; this may cause problems with supervision, etc. The large surpluses from 5, 121 (4 employees), and 34 (9 employees) indicate times when the tellers are not needed for customer service and may be reassigned to other tasks.
d.
Add the following constraints to the formulation in part (a). x9 ≥ 1 x11 ≥ 1 x9 +x10 + x11 ≥ 5 The new optimal solution, which has a daily cost of $909 is x9 = 1
y9 = 5
x11 = 4
y12 = 5 y3 = 2
There is now much less reliance on parttime employees. The new solution uses 5 fulltime employees and 12 part time employees; the previous solution used no fulltime employees and 21 parttime employees.
7 100
1
11
Denver
8 13
20
100
2 Atlanta
2 Dallas
70
17 12 10
8
150
50
3
3 Los Angeles
60
18 13
Chicago 16 4 St. Paul
b.
80
There are alternative optimal solutions. Solution #1 Denver to St. Paul: Atlanta to Boston: Atlanta to Dallas: Chicago to Dallas: Chicago to Los Angeles: Chicago to St. Paul:
10 50 50 20 60 70
Solution # 2 Denver to St. Paul: Atlanta to Boston: Atlanta to Los Angeles: Chicago to Dallas: Chicago to Los Angeles: Chicago to St. Paul:
10 50 50 70 10 70
Total Profit: $4240 If solution #1 is used, Forbelt should produce 10 motors at Denver, 100 motors at Atlanta, and 150 motors at Chicago. There will be idle capacity for 90 motors at Denver. If solution #2 is used, Forbelt should adopt the same production schedule but a modified shipping schedule.
720
A linear programming formulation of this problem can be developed as follows. Let the first letter of each variable name represent the professor and the second two the course. Note that a DPH variable is not created because the assignment is unacceptable. Max 2.8AUN + 2.2AMB +
3.3AMS +
3.0APH + 3.2BUN +
∙ ∙ ∙
s.t. AUN +
AUN +
AMB BUN
+ +
AMS + BMB + CUN +
BUN AMB
+ +
CUN + BMB + AMS +
All Variables ≥ 0 Optimal Solution: A to MS course B to Ph.D. course C to MBA course D to Undergraduate course Max Total Rating
Rating 3.3 3.6 3.2 3.2 13.3
APH BMS CMB DUN DUN CMB BMS APH
+ + +
BPH CMS + CPH DMB + DMS
+ + +
DMB CMS + DMS BPH + CPH
+ 2.5DMS ≤ ≤ ≤ ≤ = = = =
1 1 1 1 1 1 1 1
a.
7-26.
1 Augusta
300
2 Tupper Lake
100
5
5 4 Portsmouth
4
6 NewYork
100
7 Philadelphia
150
5 7
3
150
8
3 Albany
7
5 Boston
6 10
b. Min s.t.
7x13 + 5x14 + 3x23 + 4x24 + 8x35 + 5x36 + 7x37 + 5x45 + 6x46 + 10x47 x13 + x14 x13
x23 x23 x14
+ x24 x24
+ x35 x35
+ x36
+ x37
+ x36
+ x45 + x45
x37 xij ≥ 0 for all i and j c.
Optimal Solution:
Variable Value x13 x14 x23 x24 x35 x36 x37 x45 x46 x47
Objective Function: 4300
50 250 100 0 0 0 150 150 100 0
+ x46 + x46
+ x47
+ x47
≤ ≤ = = = = =
300 100 0 0 150 100 150
8-16.
a.
min 105x9 + x9 x9 + x9 + x9 + x9 x9 + x9 +
105x10 + 105x11 + 32y9 + 32y10 + 32y11 + 32y12 + 32y1 + 32y2 + 32y3 ≥ 6 + y9 ≥ 4 + y9 + x10 y10 ≥ 8 x10 + x11 + y9 + y10 + y11 ≥ 10 x10 + x11 + y9 + y10 + y11 + y12 ≥ 9 + x10 + x11 y10 + y11 + y12 + y1 + ≥ 6 x11 y11 + y12 + y1 + y2 + x10 y12 + y1 + y2 + y3 ≥ 4 + x10 + x11 y1 + y2 + y3 ≥ 7 x10 +
x11
+
x11
y2 + +
y3 ≥ 6 y3 ≥ 6
xi, yj ≥ 0 and integer for i = 9, 10, 11 and j = 9, 10, 11, 12, 1, 2, 3 b.
c.
Solution to LP Relaxation obtained using LINDO/PC: y9 = 6
y12 = 6
y11 = 2
y1 = 1
y3 = 6
All other variables = 0. Cost: $672.
The solution to the LP Relaxation is integral therefore it is the optimal solution to the integer program. A difficulty with this solution is that only parttime employees are used; this may cause problems with supervision, etc. The large surpluses from 5, 121 (4 employees), and 34 (9 employees) indicate times when the tellers are not needed for customer service and may be reassigned to other tasks.
d.
Add the following constraints to the formulation in part (a). x9 ≥ 1 x11 ≥ 1 x9 +x10 + x11 ≥ 5 The new optimal solution, which has a daily cost of $909 is x9 = 1
y9 = 5
x11 = 4
y12 = 5 y3 = 2
There is now much less reliance on parttime employees. The new solution uses 5 fulltime employees and 12 part time employees; the previous solution used no fulltime employees and 21 parttime employees.
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