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Sample Questions Posted Below
ANSWER: a
ANSWER: c
ANSWER: b
ANSWER: d
ANSWER: a
ANSWER: b
ANSWER: a
ANSWER: a
ANSWER: c
ANSWER: b
ANSWER: d
ANSWER: a
ANSWER: b
A  B  C  D  E  F  G  H  I  J  K  L  
1  
2  
3  
4  Supply/  
5  Ship  From  To  Unit Cost  Nodes  Net Flow  Demand  
6  55  1  LAV  2  PHO  60  1  LAV  −100  −100  
7  45  1  LAV  4  REN  120  2  PHO  50  50  
8  5  2  PHO  3  LAX  160  3  LAX  30  30  
9  0  3  LAX  5  SAN  70  4  REN  45  45  
10  25  5  SAN  3  LAX  90  5  SAN  90  90  
11  0  5  SAN  4  REN  70  6  DEN  35  35  
12  0  5  SAN  6  DEN  90  7  SLC  −150  −150  
13  0  6  DEN  5  SAN  50  
14  0  7  SLC  4  REN  190  
15  115  7  SLC  5  SAN  90  
16  35  7  SLC  6  DEN  100  
17  
18  Total  25600 
ANSWER: b
ANSWER: c
ANSWER: a
ANSWER: c
ANSWER: a
ANSWER: a
ANSWER: c
ANSWER: d
ANSWER: b
ANSWER: c
ANSWER: c
ANSWER: c
A  B  C  D  E  F  G  H  I  J  K  L  
1  
2  
3  
4  Supply/  
5  Ship  From  To  Unit Cost  Nodes  Net Flow  Demand  
6  55  1  LAV  2  PHO  60  1  LAV  −100  −100  
7  45  1  LAV  4  REN  120  2  PHO  50  50  
8  5  2  PHO  3  LAX  160  3  LAX  30  30  
9  0  3  LAX  5  SAN  70  4  REN  45  45  
10  25  5  SAN  3  LAX  90  5  SAN  90  90  
11  0  5  SAN  4  REN  70  6  DEN  35  35  
12  0  5  SAN  6  DEN  90  7  SLC  −150  −150  
13  0  6  DEN  5  SAN  50  
14  0  7  SLC  4  REN  190  
15  115  7  SLC  5  SAN  90  
16  35  7  SLC  6  DEN  100  
17  
18  Total  25600 
ANSWER: c
ANSWER: d
ANSWER: d
ANSWER: a
ANSWER: c
ANSWER: b
ANSWER: a
ANSWER: c
ANSWER: a
ANSWER: d
ANSWER: a
ANSWER: c
ANSWER: a
ANSWER: b
ANSWER: d
ANSWER: b
ANSWER: a
MIN: 5 X12 + 3 X13 + 2 X14 + 3 X24 + 2 X34
Subject to: −X12 − X13 − X14 = −10
X12 − X24 = 2 X13 − X34 = 3
X14 + X24 + X34 = 5
Xij ≥ 0 for all i and j
ANSWER:
Age in years
01  12  
Operating Cost  $15,000  $16,500 
Tradein Value  $20,000  $16,000 
Draw the network representation of this problem
ANSWER:
Write out the LP formulation for this problem.
ANSWER: MIN: 26 X12 + 50 X13 + 25 X23 + 50 X24 + 28 X34 + 40 X35 + 17 X45
Subject to: −X12 − X13 = −1
X12 − X23 − X24 = 0
X13 + X23 − X34 − X35 = 0 X24 + X34 − X45 = 0
X45 + X35 = 1
To
From  Lexington  Washington  Charlottesville 
Roanoke  50  −  80 
Lexington  −  50  40 
Charlottesville  −  30 
Draw the network representation of this problem.
ANSWER:
To
From  Portland  Spokane  Salt Lake City  Denver 
Seattle Portland Spokane  100 (X12)
− − 
500 (X13)
350 (X23) − 
600 (X14)
300 (X24) 250 (X34) 
−
− 200 (X35) 
Salt Lake
City 
−  −  −  200 (X45) 
Write out the LP formulation for this problem.
ANSWER: MIN: 100 X12 + 500 X13 + 600 X14 + 350 X23 + 300 X24 + 250 X34 + 200 X35 + 200 X45
Subject to: −X12 − X13 − X14 ≥ −100
X12 − X23 − X24 = 0 X13 − X34 − X35 = 0
X14 + X24 + X34 − X45 = 0 X35 + X45 ≥ 100
Xij ≥ 0
To  
From  Portland  Spokane  Salt Lake City  Denver 
Seattle  100  500  600  − 
Portland  −  350  300  − 
Spokane  −  −  250  200 
Salt Lake City  −  −  −  200 
What values should go into cells G6:L13 in the following Excel spreadsheet?
A  B  C  D  E  F  G  H  I  J  K  L  
1  
2  
3  
4  Supply/  
5  Ship  From  To  Unit Cost  Nodes  Net Flow  Demand  
6  1  SEA  2  POR  1  SEA  
7  1  SEA  3  SPO  2  POR  
8  1  SEA  4  SLC  3  SPO  
9  2  POR  3  SPO  4  SLC  
10  2  POR  4  SLC  5  DEN  
11  3  SPO  4  SLC  
12  3  SPO  5  DEN  
13  4  SLC  5  DEN  
14  
15  Total cost 
ANSWER:
A  B  C  D  E  F  G  H  I  J  K  L  
1  
2  
3  
4  Supply/  
5  Ship  From  To  Unit Cost  Nodes  Net Flow  Demand  
6  1  SEA  2  POR  100  1  SEA  −100  −100  
7  1  SEA  3  SPO  500  2  POR  0  0  
8  1  SEA  4  SLC  600  3  SPO  0  0  
9  2  POR  3  SPO  350  4  SLC  0  0  
10  2  POR  4  SLC  300  5  DEN  100  100  
11  3  SPO  4  SLC  250  
12  3  SPO  5  DEN  200  
13  4  SLC  5  DEN  200  
14  
15  Total cost 
To
From  Portland  Spokane  Salt Lake City  Denver 
Seattle  100  500  600  − 
Portland  −  350  300  − 
Spokane  −  −  250  200 
Salt Lake City  −  −  −  200 
What values would you enter in the Analytic Solver Platform task pane for the following Excel spreadsheet? Objective Cell:
Variables Cells: Constraints Cells:
A  B  C  D  E  F  G  H  I  J  K  L  
1  
2  
3  
4  Supply/  
5  Ship  From  To  Unit Cost  Nodes  Net Flow  Demand  
6  1  SEA  2  POR  100  1  SEA  −100  −100  
7  1  SEA  3  SPO  500  2  POR  0  0  
8  1  SEA  4  SLC  600  3  SPO  0  0  
9  2  POR  3  SPO  350  4  SLC  0  0  
10  2  POR  4  SLC  300  5  DEN  100  100  
11  3  SPO  4  SLC  250  
12  3  SPO  5  DEN  200  
13  4  SLC  5  DEN  200  
14  
15  Total cost 
ANSWER: Objective Cell:
G15
Variables Cells:
B6:B13
Constraints Cells:
B6:B13 ≥ 0 K6:K10 = L6:L10
A  B  C  D  E  F  G  H  I  J  K  L  
1  
2  
3  
4  Select  Driving  Supply/  
5  Route  From  To  Time  Nodes  Net Flow  Demand  
6  0  1  SEA  2  POR  3  1  SEA  −1  
7  0  1  SEA  3  SPO  4  2  POR  0  
8  1  1  SEA  4  SLC  12  3  SPO  0  
9  0  1  SEA  5  DEN  18  4  SLC  0  
10  0  2  POR  3  SPO  9  5  DEN  1  
11  0  2  POR  4  SLC  12  
12  0  2  POR  5  DEN  16  
13  0  3  SPO  4  SLC  10  
14  0  3  SPO  5  DEN  15  
15  1  4  SLC  5  DEN  5  
16  
17  Total Driving Time  17 
ANSWER:
A  B  C  D  E  F  G  H  I  J  K  L  
1  
2  
3  
4  Select  Driving  Supply/  
5  Route  From  To  Time  Nodes  Net Flow  Demand  
6  0  1  SEA  2  POR  3  1  SEA  −1  −1  
7  0  1  SEA  3  SPO  4  2  POR  0  0  
8  1  1  SEA  4  SLC  12  3  SPO  0  0  
9  0  1  SEA  5  DEN  18  4  SLC  0  0  
10  0  2  POR  3  SPO  9  5  DEN  1  1  
11  0  2  POR  4  SLC  12  
12  0  2  POR  5  DEN  16  
13  0  3  SPO  4  SLC  10  
14  0  3  SPO  5  DEN  15  
15  1  4  SLC  5  DEN  5  
16  
17  Total Driving
Time 
17 
Write out the LP formulation for this problem.
ANSWER: MIN: 15X13 + 13X14 + 9X23 + 11X24 + 4X35 + 7X36 + 8X37 + 3X45 + 9X46 + 6X47
Subject to: −X13 − X14 = −100
−X23 − X24 = −50
0.80X13 + 0.95X23 − X35 − X36 − X37 = 0
0.85X14 + 0.85X24 − X45 − X46 − X47 = 0
0.95X35 + 0.90X45 = 50
0.90X36 + 0.95X46 = 25
0.90X37 + 0.95X47 = 75
Xij ≥ 0
A  B  C  D  E  F  G  H  I  J  K  L  M  
1  
2  
3  
4  Unit  Net  Supply/  
5  Flow from Node  Yield  Flow into Node  Cost  Nodes  Flow  Demand  
6  1  Crude A  0.90  3  Refinery 1  15  1  Crude A  −120  
7  1  Crude A  0.85  4  Refinery 2  13  2  Crude B  −60  
8  2  Crude B  0.80  3  Refinery 1  9  3  Refinery 1  0  
9  2  Crude B  0.85  4  Refinery 2  11  4  Refinery 2  0  
10  3  Refinery 1  0.95  5  Lube Oil  4  5  Lube Oil  75  
11  3  Refinery 1  0.90  6  Gasoline  7  6  Gasoline  50  
12  3  Refinery 1  0.90  7  Diesel  8  7  Diesel  25  
13  4  Refinery 2  0.90  5  Lube Oil  3  
14  4  Refinery 2  0.95  6  Gasoline  9  
15  4  Refinery 2  0.95  7  Diesel  6  
16  
17  Total cost 
ANSWER: D6*A6, copied to E7:E15
What values would you enter in the Analytic Solver Platform task pane for the following Excel spreadsheet? Objective Cell:
Variables Cells: Constraints Cells:
A  B  C  D  E  F  G  H  I  J  K  L  M  
1  
2  
3  
4  Unit  Net  Supply/  
5  Flow from Node  Yield  Flow into Node  Cost  Nodes  Flow  Demand  
6  1  Crude A  0.90  3  Refinery 1  15  1  Crude A  −120  
7  1  Crude A  0.85  4  Refinery 2  13  2  Crude B  −60  
8  2  Crude B  0.80  3  Refinery 1  9  3  Refinery 1  0  
9  2  Crude B  0.85  4  Refinery 2  11  4  Refinery 2  0  
10  3  Refinery 1  0.95  5  Lube Oil  4  5  Lube Oil  75  
11  3  Refinery 1  0.90  6  Gasoline  7  6  Gasoline  50  
12  3  Refinery 1  0.90  7  Diesel  8  7  Diesel  25  
13  4  Refinery 2  0.90  5  Lube Oil  3  
14  4  Refinery 2  0.95  6  Gasoline  9  
15  4  Refinery 2  0.95  7  Diesel  6  
16  
17  Total cost 
ANSWER: Objective Cell:
H17
Variables Cells:
A6:A15
Constraints Cells:
A6:A15 ≥ 0 L6:L12 ≥ M6:M12
Distance  Center 1  Center 2  Center 3  Center 4  Supply 
Plant A  45  60  53  75  500 
Plant B  81  27  49  62  700 
Plant C  55  40  35  60  650 
Demand  350  325  400  375 
Draw the transportation network for Clifton’s distribution problem.
ANSWER:
ANSWER: Let Xij = flow from plant i (A, B, or C) to distribution center j (Center 1, 2, 3, or 4).
MIN: 45 X14 + 60X15 + 53X16 + 75X17 + 81X24 + 27X25 + 49X26 + 62X27
+ 55X34 + 40X35 + 35X36 + 60X37 Subject to: −X14 − X15 − X16 − X17 ≥ −500
−X24 − X25 − X26 − X27 ≥ −700
−X34 − X35 − X36 − X37 ≥ −650 X14 + X24 + X34 ≥ 350
X15 + X25 + X35 ≥ 325 X16 + X26 + X36 ≥ 400 X17 + X27 + X37 ≥ 375
All Xij ≥ 0
Distance  Center 1  Center 2  Center 3  Center 4  Supply 
Plant A  45  60  53  75  500 
Plant B  81  27  49  62  700 
Plant C  55  40  35  60  650 
Demand  350  325  400  375 
Draw the balanced transportation network for Clifton’s distribution problem.
ANSWER:
ANSWER: Let Xij = flow from plant i (A, B, or C) to distribution center j (Center 1, 2, 3, or 4).
MIN: 45X14 + 60X15 + 53X16 + 75X17 + 81X24 + 27X25 + 49X26 + 62X27
+ 55X34 + 40X35 + 35X36 + 60X37 Subject to: −X14 − X15 − X16 − X17 − X18 = −500
−X24 − X25 − X26 − X27 − X28 = −700
−X34 − X35 − X36 − X37 − X38 = −650 X14 + X24 + X34 = 350
X15 + X25 + X35 = 325 X16 + X26 + X36 = 400 X17 + X27 + X37 = 375 X18 + X28 + X38 = 400
All Xij ≥ 0
Plane 1  Plane 2  Plane 3  
Repair Person 1  11  9  21 
Repair Person 2  17  7  13 
Repair Person 3  9  12  17 
Repair Person 4  14  8  28 
Repair Person 5  12  5  12 
Draw the network flow for this assignment problem assuming Joe would like to maximize the total preference in his workertoaircraft schedule.
ANSWER:
ANSWER: Let Xij = assignment of repairperson i (1, 2, 3, 4, or 5) to plane j (1, 2, or 3).
MIN: 11X16 + 9X17 + 21X18 + 17X26 + 7X27 + 13X28 + 9X36 + 12X37
+ 17X38 + 14X46 + 8X47 + 28X48 + 12X56 + 5X57 + 12X58
Subject to: −X16 − X17 − X18 ≥ −1
−X26 − X27 − X28 ≥ −1
−X36 − X37 − X38 ≥ −1
−X46 − X47 − X48 ≥ −1
−X56 − X57 − X58 ≥ −1
X16 + X26 + X36 + X46 + X56 = 1 X17 + X27 + X37 + X47 + X57 = 1 X18 + X28 + X38 + X48 + X58 = 2
All Xij ≥ 0
Plane 1  Plane 2  Plane 3  
Repair Person 1  11  9  21 
Repair Person 2  17  7  13 
Repair Person 3  9  12  17 
Repair Person 4  14  8  28 
Repair Person 5  12  5  12 
Draw the balanced network flow for this assignment problem assuming Joe would like to maximize the total preference in his workertoaircraft schedule.
ANSWER:
ANSWER: Let Xij = assignment of repairperson i (1, 2, 3, 4, or 5) to plane j (1, 2, or 3).
MIN: 11X16 + 9X17 + 21X18 + 17X26 + 7X27 + 13X28 + 9X36 + 12X37
+ 17X38 + 14X46 + 8X47 + 28X48 + 12X56 + 5X57 + 12X58 Subject to: −X16 − X17 − X18 − X19 = −1
−X26 − X27 − X28 − X29 = −1
−X36 − X37 − X38 − X39 = −1
−X46 − X47 − X48 − X49 = −1
−X56 − X57 − X58 − X59 = −1 X16 + X26 + X36 + X46 + X56 = 1 X17 + X27 + X37 + X47 + X57 = 1 X18 + X28 + X38 + X48 + X58 = 2 X19 + X29 + X39 + X49 + X59 = 1
All Xij ≥ 0
ANSWER:
Write out the LP formulation for this problem.
ANSWER: MINIMIZE 25X13 + 25X14 + 45X23 + 45X24
Subject to: −X13 − X14 ≥ −50
−X23 − X24 ≥ −50 0.50X13 + 0.50X23 ≥ 15
0.33X14 + 0.33X24 ≥ 20
Xij ≥ 0 for i = 1,2; j = 3,4
ANSWER: Supply 1
Demand 6
Transsshipment 2, 3, 4, 5
Units  Unit  Net  Supply/  
of Flow  From  To  Cost  Nodes  Flow  Demand  
5  1  A  2  B  20  1  A  −40  −40  
35  1  A  3  C  15  2  B  5  5  
0  2  B  4  D  30  3  C  5  5  
25  3  C  4  D  10  4  D  10  10  
5  3  C  5  E  25  5  E  5  5  
15  4  D  6  F  10  6  F  15  15  
0  5  E  6  F  30  
Total  1150 
ANSWER:
ANSWER: MAX:
Subject 
X61
to: X61 − X12 − X13 = 0 

X12 − X25 = 0  
X13 − X35 − X34 = 0  
X34 − X46 = 0  
X25 + X35 − X56 = 0  
X56 + X46 − X61 = 0  
0 ≤ X12 ≤ 5  0 ≤ X13 ≤ 4  
0 ≤ X25 ≤ 6  0 ≤ X35 ≤ 6  
0 ≤ X34 ≤ 6  0 ≤ X56 ≤ 8  
0 ≤ X46 ≤ 2  0 ≤ X61 ≤ ∞ 
Units  Upper  Net  Supply/  
of Flow  From  To  Bound  Nodes  Flow  Demand  
4  1  A  2  B  4  1  A  0  0  
8  1  A  3  C  8  2  B  0  0  
4  2  B  4  D  6  3  C  0  0  
0  2  B  5  E  2  4  D  0  0  
4  3  C  4  D  4  5  E  0  0  
4  3  C  5  E  5  
8  4  D  5  E  9  
12  5  E  1  A  999  
12  Maximal flow 

ANSWER:
MAX X41
Subject to: X41 − X12 − X13 = 0
X12 − X24 = 0 X13 − X34 = 0
X24 + X34 − X41 = 0 0 ≤ X12 ≤ 5,
0 ≤ X13 ≤ 4,
0 ≤ X24 ≤ 3,
0 ≤ X34 ≤ 2,
0 ≤ X41 ≤ ∞
ANSWER: It is a maximal flow problem.
ANSWER: Arc Value
1 − 2  4 
2 − 4  3 
3 − 5  2 
4 − 5  4 
5 − 6  6 
Total  19 
ANSWER: Arc Value
1 − 2  4 
2 − 5  2 
3 − 5  5 
3 − 4  4 
Total  15 
ANSWER: Arc Value
1 − 2  5 
1 − 3  4 
2 − 4  6 
3 − 5  4 
5 − 6  2 
Total  21 

You are a military training analyst in charge of initial training for the XXX career field and must decide how to best train the new recruits to satisfy the requirements for skilled recruits. There are six different courses (A, B, C, D, E,
Course  Cost Per Student  Min. Num. of Trainees  Max. Num. of Trainees 
A  25  15  40 
B  55  10  50 
C  30  15  50 
D  10  15  50 
E  20  10  50 
F  15  10  50 
There are 100 recruits available for training and a demand for 100 skilled recruits. Assume all recruits pass each course and that you are trying to put students in classes in order to minimize the total cost of training. Assume non integer solutions are acceptable. Further, assume each course will be held.
What is the expected student load for each course? Should any course be expanded?
Should any course or sequence be considered for elimination?
Next, assume that not all students pass each course. In fact only 90% of the students pass courses A, E, and F and only 95% of the students pass courses B, C, and D. Each course is considered independent. The requirement for 100 skilled recruits remains. Your job is now to determine the number of recruits to place into the training program to obtain the 100 trained recruits while continuing to minimize the total cost of training.
Implement a spreadsheet model of this changed model and use Risk Solver Platform (RSP)
ANSWER: a. Draw a network flow diagram describing the problem.
b.  Formulate the associated network flow linear program.  
Minimize  25 X_{1A} + 55 X_{1B} + 30 X_{1C} + 10 X_{1D} + 20 X_{1E} + 15 X_{1F}  
Subject to:  X_{1A} – X_{1B} – X_{1C} = 100  
X_{1A} – X_{1E} – X_{1D} = 0  
X_{1B} – X_{B,TP} = 0  
X_{1C} – X_{1F} = 0  
X_{AD} – X_{DF} = 0  
X_{AE} – X_{E.TP} = 0  
X_{CF} + X_{DF} – X_{F,TP} = 0  
X_{B,TP} + X_{E,TP} + X_{F,TP} = 100  
10 £ X_{1A} £ 50  
15 £ X_{1A} £ 40  
10 £ X_{1B} £ 50  
15 £ X_{1C} £ 50  
10 £ X_{AE} £ 50  
15 £ X_{AD} £ 50  
10 £ X_{B,TP} £ 50  
10 £ X_{CF} £ 50  
10 £ X_{DF} £ 50  
10 £ X_{E,TP} £ 50  
10 £ X_{F,TP} £ 50 
c.  Implement your model in Excel and solve the model. Answer each of the following:  
Load for each course:  
A – 40  D – 15  
B – 25  E – 25  
C – 35  F – 50  
Should any course be expanded:  
Courses A and F are running at capacity  
Should any course or sequence be considered for elimination:  
Sequence ADF. Course D is at a minimum level. This minimum forces underutilization of course B. 
d.  (5 points) Redraw and properly label the network flow diagram of part (a) to accommodate the above changes. 
e.  Formulate this modified model.  
Minimize  25 X_{1A} + 55 X_{1B} + 30 X_{1C} + 10 X_{1D} + 20 X_{1E} + 15 X_{1F}  
Subject to:  X_{1A} – X_{1B} – X_{1C} ³ 0  
0.90 X_{1A} – X_{1E} – X_{1D} = 0  
0.95 X_{1B} – X_{B,TP} = 0  
0.95 X_{1C} – X_{1F} = 0  
0.95 X_{AD} – X_{DF} = 0  
0.90 X_{AE} – X_{E.TP} = 0  
0.90 X_{CF} + X_{DF} – X_{F,TP} = 0  
X_{B,TP} + X_{E,TP} + X_{F,TP} = 100  
15 £ X_{1A} £ 40  
10 £ X_{1B} £ 50  
15 £ X_{1C} £ 50  
10 £ X_{AE} £ 50  
15 £ X_{AD} £ 50  
10 £ X_{CF} + X_{DF} £ 50  
f.  Implement this changed model in Excel and solve. How many recruits are needed and what is the change in total training cost?  
A total of 116 recruits are needed, increasing costs to $5,538.95. 
(Fixed Charge Problem via Network Flow with Side Constraints)
Jack Small Enterprises runs two factories in Ohio, one in Toledo and one in Centerville. His factories produce a variety of products. Two of his product lines are polished wood clocks which he adorns with a regional theme. Naturally, clocks popular in the southwest are not as popular in the northeast, and vice versa. Each plant makes both of the clocks. These clocks are shipped to St Louis for distribution to the southeast and western states and to Pittsburgh for distribution to the south and northeast.
Jack is considering streamlining his plants by removing certain production lines from certain plants. Among his options is potentially eliminating the clock production line at either the Toledo or the Centerville plant. Each plant carries a fixed operating cost for setting up the line and a unit production cost, both in terms of money and factory worker hours. This information is summarized in the table below.


Production Cost
per Clock 
Clocks Produced
per Hour 
Available 

Plant 
Fixed Cost for
Line 
Southwest
Clock 
Northeast
Clock 
Southwest
Clock 
Northwest
Clock 
Hours per
Month 

Toledo  $20,000  $10  $12  5  6  500  
Centerville  $24,000  $ 9  $13  5.5  6.2  675  
The Southwest clocks are sold for $23 each and the Northwest clocks are sold for $25 each. Demand rates used for production planning are 1875 Southwest clocks for sale out of the St Louis distribution center and 2000 Northeast clocks for sale out of the Pittsburgh distribution center. Assume all these units are sold. The per clock transportation costs from plant to distribution center is given in the following table.
(cost per clock shipped) Cost to Ship to Distribution Center
Plant  St Louis  Pittsburgh 
Toledo  $2  $4 
Centerville  $3  $2 
Develop a generalized network flow model for this problem and implement this model in solver. Use the model to answer the following questions.
ANSWER: The following network diagram captures the Small Production problem:
The decision whether to keep open or close a plant is captured in the early arcs, which are modeled as binary, flow equal zero or flow equal one. Once opened, the flow opening that plant is transformed into the hours available at that plant. These hours are then used to produce either the Southwest Clock (product 1) or the Northeast Clock (product 2). Any unused hours flow into the collection node, Unused Hours. These products has a per unit production cost captured by the next set of flows. Distribution costs and the distribution plan are captured in the arcs between the Prod # Cost nodes and the St Louis and Pittsburgh nodes. Finally, sales of the products are captured in the flows from the warehouses to the Prod # Sales nodes. These final nodes indicate the product demand set for the problem.
This problem is essentially a generalized network flow problem with side constraints. Analysis of this problem indicates closing Toledo. Both clock products are produced at Centerville. The 675 hours available at Centerville are allocated as follows: 340.9 hours for the Southwest Clock, 322.6 hours for the Northeast Clock, and the remaining 11.5 hours unused. This yields an expected monthly profit of $16,625 on these clock products. This is a savings of $20,000. With both plants open, the expected loss per month is $3,375 as Toledo will produce the Southwest Clocks leaving an excess of 125 hours and Centerville will produce the Northeast Clocks leaving an excess of 352.4 hours.
ANSWER: a
ANSWER: a
ANSWER: b
ANSWER: a
ANSWER: a
ANSWER: a
ANSWER: a
ANSWER: a
ANSWER: d
ANSWER: a
ANSWER: d
ANSWER: a
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