1. Outline LP formulation of minimal cost flow problem
useful results from shortest path problem
optimality condition
residual network
potentials of nodes
reduced costs of arcs (with respect to a given set of potentials)
the optimality conditions for networks
network simplex and existence of negative cycle
transportation simplex method
transportation by eliminating negative cycles
determining an initial feasible solution for a minimal flow problem 1

2. Minimum Cost Flow Models
G: (connected) network, i.e., G = (N, A)
N: the collection of nodes in G, i.e., N = {j}
A: the collection of arcs in G, i.e., A = {(i, j)}
cij: the cost of arc (i, j)
uij: the capacity of arc (i, j)
b(i): the amount of flow out of node i
b(i) > (resp. <) 0 for out (resp. in) flow
How to deal with an undirected arc?
Assume direct arcs. 2

3. Minimum Cost Flow Models1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9
(cij, uij)
N = {1, 2, 3, 4, 5, 6}
A = {(1, 2), (1, 3), (2, 3), (2, 4), (2, 5), (3, 5), (4, 6), (5, 4), (5, 6)}
b(1) = 9, b(6) = -9, and b(i) = 0 for i = 2 to 5 3

4. A Minimum Cost Flow LP Model1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9 4

5. Useful Results from Shortest-Path Algorithms5

6. Useful Results from Shortest-Path Algorithms
optimality condition for shortest distances
residual network
potentials of nodes
reduced costs of arcs with respect to a given set of potentials 6

7. Optimality Condition for Shortest Distances
Theorem. Let G = (N, A) be a directed graph; d() be a function defined on N. Then d() is the shortest distance of node () from node 1 iff
d(1) = 0;
d(i) is the length of a path from node 1 to node i; and
d(j)  d(i) + cij for all (i, j)  A. 7

8. Optimality Condition for Shortest Distances1 2 3 4 7 6
d(1) = 0
d(2) = 2
d(3) = 3
d(4) = 7 1 2 3 4 7 6
d(1) = 0
d(2) = 2
d(3) = 3
d(4) = 8
LHS figure: {d(i)} = {0, 2, 3, 7}
for all i, d(i) being shortest distance of node i from node 1
satisfying the three conditions
RHS figure: {d(i)} = {0, 2, 3, 8}
d(4) = the distance of a path from node 1 to node 4
d(3) + c34 < d(4), i.e., this set cannot be the collection of shortest distances from nodes 8

9. Residual Network (6,4) (8,5) s t 1 (2,7)
Figure 1. Costs and Capacities of Arcs t
(6, 4)
(8, 5) s 1
(2, 7)
Figure 3. The Residual Network Corresponding to Figure 1 s t 1
(-8,3)
(-2, 3)
(6, 4)
(8, 2)
(2, 4) 3
Figure 2. Flows on arcs s t 1
Figure 4. The Residual Network Corresponding to Figure 2 9

10. A Cycle in a Residual Network
actual meaning of sending flow in cycle s-t-1-s (Figure 4)
a different way to send flow from s to t
flow along t-1-s  reduction of flow in s-1-t
(-8,3)
(-2, 3)
(6, 4)
(8, 2)
(2, 4)
Figure 4. The Residual Network Corresponding to Figure 2 s t 1
Figure 5. The New Residual Network After Adding a Flow of 3 units in cycle s-t-1-s s t 1
(6, 1)
(8, 5)
(2, 7)
(-6, 3) s t 1 3
Figure 6. Flows on arcs 10

11. Potentials of Nodes and the Corresponding Reduced Costs of Arcs
first the idea
call any arbitrary set of numbers {i}, one for a each node, a set of potentials of nodes
define the reduced cost with respect to this set of potentials {i} by
interesting properties for this set of reduced costs 11

12. Potentials of Nodes and the Corresponding Reduced Costs of Arcs
C(P) be the total cost of path P
C(P) be the total reduced cost of path P
i = potential of node i
P = a path from node s to node t
then C(P) = C(P)  s + t 12

13. Potentials of Nodes and the Corresponding Reduced Costs of Arcs
C(P) be the total cost of path P
C(P) be the total reduced cost of path P
i = potential of node i
P = a path from node s to node t, then
C(P) = C(P)  s + t
a shortest path from s to t for C() is also a shortest path from s to t for C() 1 2 3 4 5 6 8 7
1 = 3
2 = -5
3 = 2
4 = 6
5 = -3
6 = 5
C(P)
C(P) 1 2 3 4 5 6 10 18 -2 13 14 13

14. Potentials of Nodes and the Corresponding Reduced Costs of Arcs
C(P) be the total cost of path P
C(P) be the total reduced cost of path P
i = potential of node i
P = a path from node s to node t, then
if j = -d(j), then for all (i, j)  A
arcs with forming a tree
j = -d(j)
1 = 0
2 = -8
3 = -3
4 = -10
5 = -6
6 = -12 4 8 5 3 1 2 3 4 5 6 1 2 3 4 5 6 8 7
1 = 3
2 = -5
3 = 2
4 = 6
5 = -3
6 = 5 14

15. Spanning Tree and BFS
in a minimal cost flow network problem, a BFS  a spanning tree of the network
 # of constraints = # of nodes
# of basic variables = # of nodes - 1 15

16. The optimality Conditions for Networks16

17. The Optimality Conditions for Networks
Theorem 1. (Theorem 9.1, Negative Cycle Optimality Conditions) A feasible solution x* is optimal iff there is no negative cost (direct) cycle in the corresponding residual network G(x*).
Theorem 2. (Theorem 9.3, Reduced Cost Optimality Conditions) A feasible solution x* is optimal iff there exists dual variable  such that for every arc (i, j) in the residual network G(x*).
Theorem 3. (Theorem 9.4, Complementary Slackness Optimality Conditions) A feasible solution x* is optimal iff there exists dual variables  such that
if then = 0;
if 0 < < uij, then
if then = uij. 17

18. Ideas for Theorem 1
a cycle in a residual network: a different way to send flow across a network
existing of a negative cycle  existence of another flow pattern with lower cost
no negative cycle  no other flow pattern with lower cost 18

19. Ideas for Theorem 3
Theorem 3. (Theorem 9.4, Complementary Slackness Optimality Conditions) A feasible solution x* is optimal iff there exists dual variables  such that
optimality conditions of simplex method for bounded variables 19

20. Ideas for Theorem 2
contradictory statements of Theorem 2 & Theorem 3?
Theorem 2: all reduced costs non-negative (for minimization)
Theorem 3: some reduced costs positive, some negative
Theorem 2: on reduced costs for (variables representing) arcs of a residual network
Theorem 3: on reduced costs of all variables 20

21. Ideas for Theorem 2
Theorem 2. (Theorem 9.3, Reduced Cost Optimality Conditions) A feasible solution x* is optimal iff there exists dual variable  such that for every arc (i, j) in the residual network G(x*).
Theorem 2  minimum
for any potential  and cycle: C(P) = C(P)
for every arc in G(x)  no negative cycle in G  minimal 21

22. Ideas for Theorem 2
Theorem 2. (Theorem 9.3, Reduced Cost Optimality Conditions) A feasible solution x* is optimal iff there exists dual variable  such that for every arc (i, j) in the residual network G(x*).
minimum  Theorem 2
conditions of Theorem 3 satisfied by a minimum solution
a forward arc (i, j) with by Theorem 3
no reversed arc in G(x*)
a forward arc (i, j) with = uij, in the original network
no such forward arc in G(x*)
for the reversed arc in G(x*):
a forward arc (i, j) with in the original network
for the forward arc in G(x*): 22

23. Equivalence Between Network Simplex and Negative Cost Cycle23

24. Equivalence Between Network Simplex Method and Existence of a Negative Cost Cycle1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9
(cij, uij)
at a certain point, an extended basic feasible solution is shown below, where the dotted lines show arcs at upper bounds. 1 2 3 4 5 6 7 9 -9
Flow 24

25. An Iteration of the Network Simplex1 2 3 4 5 6 7 9 -9
Flow
to find the entering variable
x46 entering
reduction of flow  from upper bound
dual variables
1 = 0
2 = 1 – c12 = -3
4 =-8
3 =-2
5 =-6
6 =-10 1 2 3 4 5 6
How about (2, 3), (2, 5), (5, 4)?
beneficial to reduce flow in (4, 6) 25

26. An Iteration of the Network Simplex1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9
(cij, uij)
to find the leaving variable
reduction of flow  from x46
leaving variable, x13 or x35, for  = 1 1 2 3 4 5 6
Change of Flow
0  2+  3
0  7  8
0  5  7
0  5  5
0  4+  6 1 2 3 4 5 6 7 9 -9
Current Flow Pattern 26

27. An Iteration of the Network Simplex1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9
(cij, uij)
 = 1
arbitrarily making x35 leaving 1 2 3 4 5 6
0  7  8
Change of Flow
0  5  7
0  5  5
0  2+  3
0  4+  6 1 2 3 4 5 6 9 -9
New Flow Pattern 27

28. An Iteration of the Negative Cycle Algorithm
cost = -1
maximum allowable flow = 1 unit 1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9
(cij, uij) 1 2 3 4 5 6 7 9 -9
Current Flow Pattern
Residual Network
(cij, xij) 28

29. An Iteration of the Negative Cycle Algorithm
cost = -1
maximum allowable flow = 1 unit 1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9
(cij, uij)
(cij, xij)
(cij, xij) 29

30. Same Result from Network Simplex Method and Negative Cost Cycle1 2 3 4 5 6
(3, 8)
(2, 3)
(2, 2)
(5, 7)
(4, 3)
(5, 4)
(3, 5)
(4, 6) 9 -9
(cij, uij)
residual network
(cij, xij) 1 2 3 4 5 6 9 -9
New Flow Pattern 30

31. Transportation Simplex Method31

32. Transportation Simplex Method
balanced transportation problem
2 sources, 3 destinations
5 constraints, with one degree of redundancy
4 basic variables in a BFS 32

33. Transportation Simplex Method
four basic variables in a BFS 33

34. Transportation Simplex Method
reduced cost (i, j) = cij – ui – vj 34

35. Transportation Simplex Method35

36. Transportation Problem by Eliminating Negative Cycles36

37. Transportation Problem by the Negative Cycle Approach37

38. Transportation Problem by the Negative Cycle Approach38

39. Transportation Problem by the Negative Cycle Approach39

40. Transportation Problem by the Negative Cycle Approach
no more negative cycle
optimal flow: x13 = 100, x14 = 500, x15 = 200, x23 = 300, x24 = 0, x25= 0,
minimum cost = 7600 40

41. To Determine an Initial Feasible Solution41

42. To Find an Initial Feasible Solution
to solve a maximal flow problem 42

43. To Find an Initial Feasible Solution
to solve a minimum cost flow problem
with x1t = 800, x2t = 300, xts = 1100, xs3 = 400, xs4 = 500, xs5 = 200 as the initial solution of the minimal cost flow problem 43