Chapter 5 - Network Modeling : S-41
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Chapter 5
Network Modeling
16. a.
1.012
A1
A2
1.012
1.012
A3
A4
1.012
1.012
1.012
A5
1.035
A6
1.012
1.035
B3
1.035
B1
BB
1.035
1.012
1.035
1.012
1.035
P3
-9999
+50
1.012
B5
1.035
P5
+50
1.058
EB
+200
C1
C4
1.058
1.11
D1
b.
c.
Minimize the total flow from BB (beginning balance) to A1, B1, C1 and D1.
See file: PRB5_16.XLS
Flow
From
To
$94.985 1 BB
3 B1
$178.673 1 BB
4 C1
$46.676 3 B1
8 B3
$48.309 3 B1
9 P3
$178.673 4 C1
11 C4
$48.309 8 B3
14 P5
$189.036 11 C4
16 EB
Total cash required = $273,658
33.
Iteration Node Added
1
1
5
2
4
3
3
4
7
5
6
6
2
7
8
8
9
9
Total Cost
Cost
$0
$85
$20
$25
$30
$20
$30
$35
$25
$270
S-42 : Chapter 5 - Network Modeling
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5.
a.
NY
17
Chic.
7
6
+1
18
24
18
95
32
50
LA
Denver
27
2
13
30
3
19
5
25
Mem p
24
14
5
35
14
St Lou
35
105
4
San Diego
45
1
-1
b.
MIN
5 X12 + 13 X13 + 45 X15 + 105 X17 + 27 X23 + 19 X24 + 50 X25 + 95 X27 +
14 X34 + 30 X35 + 32 X36 + 14 X43 + 35 X45 + 24 X46 + 35 X54 + 18 X56 + 25 X57 +
24 X64 + 18 X65 + 17 X67
ST
-X12 - X13 - X15 - X17 = -1
+X12 - X23 - X24 - X25 - X27 = 0
+X13 + X23 + X43- X34 - X35 - X36 = 0
+X24 + X34 + X54 + X64 - X43 - X45 - X46 = 0
+ X15 + X25 + X35 + X45 + X65 - X54 - X56 - X57 = 0
+ X36 + X46 + X56 - X64 - X65 - X67 = 0
+ X17 + X27 + X57 + X67 = +1
Xij  0
c.
See file: PRB5_5.XLS
The solution is: X13=X36=X67=1 with a minimum total cost of $62.
Chapter 5 - Network Modeling : S-43
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12. a.
35
Cincinnati
5
15
Boston
1
25
Chicago
7
20
Pittsburgh
3
30
5
15
20
5
Memphis
6
15
15
Baltimore
2
25
30
10
Atlanta
4
10
Dallas
8
999
b.
c.
See file: PRB5_12.XLS
The maximum flow is 55 tons.
13. a.
+410
Beg
Inv
1
+310
+580
$0
D1
$1.50
lb = 0
3
lb = 50
D2
5
+540
+50
$1.50
D3
$1.50
D4
$1.50
lb = 50
7
lb = 50
9
lb = 50
Fin
Inv
10
-120
$49 lb = 400
$45
lb = 400
$46 lb = 400
$47 lb = 400
P1
P2
P3
P4
2
4
6
8
-500
-520
-450
-550
note: lb =lower bound
b.
c.
d.
See file PRB5_13.XLS
Produce 420 in month 1, 520 in month 2, 400 in month 3, 450 in month 4, carry 110 in inventory from
month 1 to 2, 50 from month 2 to 3, 140 from month 3 to 4, and 50 at the end of month 4. Total Cost =
$83,565.
Not much, only $45.
S-44 : Chapter 5 - Network Modeling
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28. a.
This is a transportation problem. Note that demand exceeds supply by 20 units.
Stores
Warehouses
1
+20
2
+25
3
+30
4
+35
5
-30
1
4
6
5
3
6
-30
2
4
4
4
-30
b.
3
3
2
MIN
5 X11 + 4 X12 + 6 X13 + 5 X14
+3 X21 + 6 X22 + 4 X23 + 4 X24
+4 X31 + 3 X32 + 3 X33 + 2 X34
ST
-X11 - X12 - X13 - X14 = -30
-X21 - X22 - X23 - X24 = -30
-X31 - X32 - X33 - X34 = -30
+X11 + X21 + X31 + XD1  +20
+X12 + X22 + X32 + XD2  +25
+X13 + X23+ X33 + XD3  +30
+X14 + X24+ X34 + XD4  +35
Xij  0
c.
See file: PRB5_28.XLS
The optimal solution is: X12 = 25, X14 = 5, X21 = 20, X23 = 10, X34 = 30, XD3 = 20.
Minimum total cost = $285.
Note that store 3 receives 20 units less than demanded.
d.
Assign arbitrarily large costs (such as $999) to the arcs representing these flows.
The optimal solution is then: X13 = 30, X21 = 20, X24 = 10, X32 = 5, X34 = 25.
Minimum total cost = $345.
Note that store 2 receives 20 units less than demanded.