76
Chapter 7
Integer Linear Programming
7.15
UB = 380.5 (X
1
= 22.167, X
2
= 0.167)
LB = 374 (X
1
= 22, X
2
= 0)
1
Z=380.5
X
1
<
22
X
1
>
23
2
X
1
= 22
X
2
= 0.29
3
Infeasible
Solution
UB = 380.5
LB = 380.29
5
X
1
= 21
X
2
= 1
Stop
4
X
1
= 22
X
2
= 0
Stop
*Inferior*
X
2
<
0
X
2
>
1
UB = 380.5
LB = 379
UB=380.5
LB = 374
Stop
*Optimal*
7.16
X
1
= 2; X
2
= 3; X
3
= 2; Z
MAX
= 1260.
7.17
X
1
= 2; X
2
= 4; Z
MAX
= 48
7.18
X
1
= 0; X
2
= 1; X
3
= 1; X
4
= 1; X
5
= 1; X
6
= 0; Z
MAX
= 72000.
7.19
X
1
= 1; X
2
= 0; X
3
= 0; X
4
= 1; X
5
= 1; X
6
= 1: Z
MAX
= 500.
7.20
X
1
= 15; X
2
= 15; X
3
= 25: Z
MAX
= 822500.
7.21
X
1
= 6; X
2
= 3; X
3
= 118; Z
MAX
= 18625.
7.22
X
1
= 7; X
2
= 1; X
3
= 1; Z
MAX
= 85.
7.23
X
1
= 6; X
2
= 0.667; Z
MAX
= 32.
7.24
X
1
= 6; X
2
= 1.6; Z
MIN
= 55.2.
7.25
Z = 360000.
7.26
Z = 314.
7.27
Z = 318.
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Solutions Manual and Workbook
77
Chapter 8
The Transportation Model
8.1
(Initial Solution)
4
Note:
Using the Northwest corner
method, we do not need the costs to
obtain an initial feasible solution.
If we use them, we can compute the
total cost as:
35(10)+35(8)+15(8)+15(3)+65(6)+10(4)+
55(3) = $1,390
8.2
Note:
At one point there is a tie between Row A, Row B, and Column X.
It
was broken arbitrarily in favor of Row A.
Total Cost by VAM:
35(4)+50(5)+70(4)+10(3)+
20(5)+25(4)+20(3) = $960
.
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 Spring '10
 Koslov
 Harshad number, Hebrew numerals, BMW Sports Activity Series, Pentagonal number, northwest corner

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