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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
a. X12 + X23 = 100 b. X12 − X23 ≤ 100
c. −X12 + X23 ≥ −100 d. X12 − X23 ≥ 100
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
a. −X12 − X13 = 0 b. −X12 − X24 = 1 c. X12 + X13 = 0 d. −X12 + X24 = 0
ANSWER: d
a. X13 + X23 − .95 X35 − .90 X36 − .90 X37 = 0 b. .80 X13 + .95 X23 − X35 − X36 − X37 = 0
c. .80 X13 + .95 X23 − .90 X36 − .90 X37 ≥ 0 d. X13 + X23 − X35 − X36 − X37 ≥ 0
ANSWER: b
a. X35 + X36 + X37 = 75 b. X37 + X47 ≥ 75
c. .90 X37 + .95 X47 = 75
d. X37 + X47 −X36 − X35 − X45 − X46 ≥ 75
ANSWER: c
ANSWER: c
a. −X12 − X13 = 0
b. MIN −50 X12 − 200 X13 + 100 X24 + 35 X34
c. MIN 50 X12 + 200 X13 + 100 X24 + 35 X34
d. MAX −50 X12 − 200 X13 + 100 X24 + 35 X34
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 |
a. SUMIF($C$6:$C$16,I6,$B$6:$B$16)−SUMIF($E$6:$E$16,I6,$B$6:$B$16) b. SUMIF($I$6:$I$12,B6,$B$6:$B$16)−SUMIF($I$6:$I$12,I6,$B$6:$B$16) c. SUMIF($E$6:$E$16,I6,$B$6:$B$16)−SUMIF($C$6:$C$16,I6,$B$6:$B$16)
d. SUMPRODUCT(B6:B16,G6:G16)
ANSWER: c
ANSWER: d
ANSWER: d
ANSWER: a
ANSWER: c
ANSWER: b
ANSWER: a
ANSWER: c
ANSWER: a
ANSWER: d
a. X12 − X23 − X24 = 0 b. X12 + X23 + X24 = 0 c. X12 ≤ 4
d. X12 + X13 − X23 = 0
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
0-1 | 1-2 | |
Operating Cost | $15,000 | $16,500 |
Trade-in 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 worker-to-aircraft 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 worker-to-aircraft 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,
F) used for training in the XXX career field and four different sequences of courses that can be taken to achieve the required skill level. These sequences are A-E, B, C-F, and A-D-F. The table below provides information on the six courses.
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 X1A + 55 X1B + 30 X1C + 10 X1D + 20 X1E + 15 X1F | |||
Subject to: | X1A X1B X1C = 100 | |||
X1A X1E X1D = 0 | ||||
X1B XB,TP = 0 | ||||
X1C X1F = 0 | ||||
XAD XDF = 0 | ||||
XAE XE.TP = 0 | ||||
XCF + XDF XF,TP = 0 | ||||
XB,TP + XE,TP + XF,TP = 100 | ||||
10 X1A 50 | ||||
15 X1A 40 | ||||
10 X1B 50 | ||||
15 X1C 50 | ||||
10 XAE 50 | ||||
15 XAD 50 | ||||
10 XB,TP 50 | ||||
10 XCF 50 | ||||
10 XDF 50 | ||||
10 XE,TP 50 | ||||
10 XF,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 A-D-F. Course D is at a minimum level. This minimum forces underutilization of course B. |
d. | (5 points) Re-draw and properly label the network flow diagram of part (a) to accommodate the above changes. |
e. | Formulate this modified model. | |||
Minimize | 25 X1A + 55 X1B + 30 X1C + 10 X1D + 20 X1E + 15 X1F | |||
Subject to: | X1A X1B X1C 0 | |||
0.90 X1A X1E X1D = 0 | ||||
0.95 X1B XB,TP = 0 | ||||
0.95 X1C X1F = 0 | ||||
0.95 XAD XDF = 0 | ||||
0.90 XAE XE.TP = 0 | ||||
0.90 XCF + XDF XF,TP = 0 | ||||
XB,TP + XE,TP + XF,TP = 100 | ||||
15 X1A 40 | ||||
10 X1B 50 | ||||
15 X1C 50 | ||||
10 XAE 50 | ||||
15 XAD 50 | ||||
10 XCF + XDF 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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