network problems: a review of some fundamental algorithms · introduction to operations research...
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Introduction to Operations Research
ANSWERS
Network Problems: A Review of Some Fundamental Algorithms
Prof. dr. E-H. Aghezzafir. Rodrigo Rezende Amaral
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Exercise 6
(e.g., if a truck carries three packages, they should be from three different types)
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Exercise 61
2
7
so
1
2
5
si
3
4
5
6
3
4
33
3
3
3
3
3
6
4
5
4
31
1
1
1
1
*
*
*
*
*
* * 5 arcs goes from each node on the left to the 5 nodes on the right
If max flow= 21 then all packages can be loaded.
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Exercise 7
(Consider that the number of cars in each city is >> 300)
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Exercise 7
C10
C11
Each node cit represents city i in time t
∞
C21
C31
C41
C42
∞300
300
300
300
300
300
t = 0 t = 1 t = 2
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Exercise 8
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Exercise 8
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0* ∞ ∞ ∞ ∞ ∞ ∞ ∞
N.Y. Cleveland St. Louis Nashville Phoenix Dallas Salt Lake L.A.
d[v]π[v]
: path length from the source (NY) to v
: previous node which should be visited in order to achieve d[v]
v
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N.Y. N.Y. N.Y. St. Louis Cleveland Nashville Phoenix
0* 400* 950 800 ∞ ∞ ∞ ∞
0* 400* 950 800* 2200 1300 ∞ ∞
0* 400* 950* 800* 2200 1300 2000 ∞
0* 400* 950* 800* 2050 1300* 2000 ∞
0* 400* 950* 800* 2050* 1300* 2000* 2600
0* 400* 950* 800* 2050* 1300* 2000* 2600
0* 400* 950* 800* 2050* 1300* 2000* 2450*
N.Y. Cleveland St. Louis Nashville Phoenix Dallas Salt Lake L.A.
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Exercise 8
b) Balanced transportation matrix:
Cleveland St. Louis Nashville Phoenix Dallas Salt Lake L.A.
N.Y.400
1950 800 ∞ ∞ ∞ ∞
1
Cleveland 10 ∞ ∞ 1800 900 ∞ ∞
1
St. Louis∞ 0 ∞
11100 600 ∞ ∞
1
Nashville∞ ∞
10 ∞ 600 1200 ∞
1
Phoenix∞ ∞ ∞ 0 ∞ ∞
1400
1
Dallas∞ ∞ ∞ 900
10 1000 1300
1
Salt Lake∞ ∞ ∞ ∞ ∞
10 600
1
1 1 1 1 1 1 1
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Exercise 8
c) MCNFP
Cleveland St. Louis Nashville Phoenix Dallas Salt Lake L.A.
N.Y. x400
x950
x800
…∞ ∞ ∞
x∞
111 12 13 17
Cleveland x0 ∞ ∞ 1800 900 ∞
…∞
121
St. Louis x∞ 0 ∞ 1100 600 ∞ ∞
131
Nashville …∞ ∞ 0 ∞ 600 1200 ∞
1
Phoenix∞ ∞ ∞ 0 ∞ ∞
x400
157
Dallas∞ ∞ ∞ 900 0 1000
x1300
167
Salt Lake x∞
…∞ ∞ ∞
x∞
x0
x600
171 75 76 77
1 1 1 1 1 1 1
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Exercise 8
c) MCNFP
400 950 800 ∞ ∞ ∞ ∞
0 ∞ ∞ 1800 900 ∞ ∞
∞ 0 ∞ 1100 600 ∞ ∞
∞ ∞ 0 ∞ 600 1200 ∞
∞ ∞ ∞ 0 ∞ ∞ 400
∞ ∞ ∞ 900 0 1000 1300
∞ ∞ ∞ ∞ ∞ 0 600
min 𝑧 =
𝑖=1
7
𝑗=1
7
𝑐𝑖𝑗𝑥𝑖𝑗
𝑠. 𝑡.
𝑖=1
7
𝑥𝑖𝑗 = 1 , 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑗
𝑗=1
7
𝑥𝑖𝑗 = 1 , 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑖
𝐴𝑙𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 ≥ 0
𝐶 =
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Exercise 9
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Exercise 9
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Exercise 9
Graphical representation:
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Exercise 9
a) LP formulation:
x12
x13x35
x34
x25
x24
x46
x56
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b) Ford-Fulkerson method
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b) Ford-Fulkerson method
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b) Ford-Fulkerson method
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Exercise 10
of Exercise 8 using Prim's Algorithm.
(Consider the arcs as non-directed)
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Exercise 10
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Exercise 11
DeptShift
1 2 3
1 (6, 8) (11, 12) (7, 12)
2 (4, 6) (11, 12) (7, 12)
3 (2, 4) (10, 12) (5, 7)
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Exercise 11
DeptShift
1 2 3 ∑j
1 6 ≤ x11 ≤ 8 11 ≤ x12 ≤ 12 7 ≤ x13 ≤ 12 ∑j=1,2,3 x1j≥ 26
2 4 ≤ x21 ≤ 6 11 ≤ x22 ≤ 12 7 ≤ x23 ≤ 12 ∑j=1,2,3 x2j≥ 24
3 2 ≤ x31 ≤ 4 10 ≤ x32 ≤ 12 5 ≤ x33 ≤ 7 ∑j=1,2,3 x3j≥ 19
∑i∑i=1,2,3 xi1 ∑i=1,2,3 xi2 ∑i=1,2,3 xi3
≥ 13 ≥ 32 ≥ 22
Min ∑∑ xij …
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Exercise 11
We start with the maximum possible number of nurses for every department and every shift. Then maximize the number of nurses which can be unassigned.
DeptShift
1 2 3 ∑j=1,2,3
1 6 ≤ 8 ≤ 8 11 ≤ 12 ≤ 12 7 ≤ 12 ≤ 12 32 ≥ 26
2 4 ≤ 6 ≤ 6 11 ≤ 12 ≤ 12 7 ≤ 12 ≤ 12 30 ≥ 24
3 2 ≤ 4 ≤ 4 10 ≤ 12 ≤ 12 5 ≤ 7 ≤ 7 23 ≥ 19
∑i=1,2,3 18 36 31
≥ 13 ≥ 32 ≥ 22
∑∑ ẋij = 85
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Exercise 11
DeptShift
1 2 3 ∑j=1,2,3
1 y11 ≤ 2 y12 ≤ 1 y13 ≤ 5 ∑j=1,2,3 y1j≤ 6
2 y21 ≤ 2 y22 ≤ 1 y23 ≤ 5 ∑j=1,2,3 y2j≤ 6
3 y31 ≤ 2 y32 ≤ 2 y33 ≤ 2 ∑j=1,2,3 y3j≤ 4
∑i=1,2,3 ∑i=1,2,3 yi1 ∑i=1,2,3 yi2 ∑i=1,2,3 yi3
≤ 5 ≤ 4 ≤ 9
Min ∑∑ xij = 85 - Max ∑∑ yij
We start with the maximum possible number of nurses for every department and every shift. Then maximize the number of nurses which can be unassigned.
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Exercise 11
S
source of “unassigned
nurses”
Dept j Shift i t
∑i yij ∑j yijyij
# of unassigned nurses currently in dept j
# of unassigned nurses currently in dept j and shift i
# of unassigned nurses currently in shift i
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Exercise 11
S S2 t
5
D1 S1
S3
D2
D3
6
2
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Exercise 10
- First solve the problem with only 2 experient contractors.
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Exercise 10
s
c2
c1
d3
d1
d2
(0, u1)
(c11, ∞) b(d1) = -r1
(cij, uij)
b(s) = ∑ ri
i j
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Exercise 10
s
c2
c1
d3
d1
d2
(0, u1)
(c11, ∞)
b(d1’) = -(r1-1)
b(s) =
∑ ri
c4
c3
(exp.)
d1’
d2’
d3’(inexp.)
(c11, ∞)
(c12, ∞)
(c12, ∞)
(c13, ∞)
(c13, ∞)
(0, u3)(c31, ∞)
(c32, ∞)
(c33, ∞)
b(d1) = -1
b(d2’) = -(r2-1)
b(d2) = -1
(exp.)
(inexp.)
b(d3’) = -(r3-1)
b(d3) = -1
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Minimum Cost Flow Problem - Exercise 1
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• Negative Cycle Algorithm
– Find a feasible flow
– While there exist a negative cycle C in the residual graph
• Find the minimum Δ of the residual capacities in C
• Augment the flow in C with Δ units
• Update the residual graph
Minimum Cost Flow Problem - Exercise 1
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2222
10
10
Cost: 110+50+330+110 = 600
Minimum Cost Flow Problem - Exercise 1
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2222
1010
38
2
∞
Original Graph Residual Graph
∞ ∞12 20
10
22 22
10
Minimum Cost Flow Problem - Exercise 1
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1410
4
5
-5-15
15
5
-5
5
11
-11
Residual Graph (Costs)
Minimum Cost Flow Problem - Exercise 1
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14-10
4
5
-5-15
15
5
-5
5
11
-11
Negative cost cycle
Minimum Cost Flow Problem - Exercise 1
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38
2
∞
∞ ∞12 20
10
22 22
10
Negative cost cycle
2 units can be send through this cycle
Minimum Cost Flow Problem - Exercise 1
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2020
12
10
2
Previous Cost: 110+50+330+110 = 600
Current Cost: 132+50+8+300+100 = 590 = 600 + 2 * (-5)
Minimum Cost Flow Problem - Exercise 1
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-5-15
5
54
-4
10
5
-5
15
-11
5
No more negative cycles can be found
Minimum Cost Flow Problem - Exercise 1
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2020
12
10
2
Minimum Cost Flow Problem - Exercise 1
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Minimum Cost Flow Problem - Exercise 2
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50
2525
50
25
50 75
Flow
Minimum Cost Flow Problem - Exercise 2
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25
575
15
10 5
40
50
50
50 75
25
25 25
60
50
40
40
50
100
40
Residual Forward Flow Residual Backward Flow
Minimum Cost Flow Problem - Exercise 2
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15
10 45
5
8 15
5
-10
-15
-8 -15
-5
-10 -45
10
30
10
10
30
45
25
X Cost of using the arc
Minimum Cost Flow Problem - Exercise 2
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15
10 45
5
8 15
5
-10
-15
-8 -15
-5
-10 -45
10
30
10
10
30
45
25
2525
50 40
Minimum Cost Flow Problem - Exercise 2Negative
cost cycle
25 units can be send through this cycle
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50 2525
50
25
50 75
Flow
Minimum Cost Flow Problem - Exercise 2
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25
30 100
15
10 5
40
50
50
50 75
25
25 25
60
25
15
40
50
100
40
Residual Forward Flow Residual Backward Flow
Minimum Cost Flow Problem - Exercise 2
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Residual Forward Flow Residual Backward Flow
5
-10
-15
-8 -15
-5
10 10
10
30
45
25
15
-10
10 45
-1030
8
15
5
Minimum Cost Flow Problem - Exercise 2
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Residual Forward Flow Residual Backward Flow
5
-10
-15
-8 -15
-5
10 10
10
30
45
25
15
-10
10 45
-1030
8
15
5
No more negative cycle can be found
Minimum Cost Flow Problem - Exercise 2
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Solution
Minimum Cost Flow Problem - Exercise 2