Complex networks, the minimum cost of single-source single point of maximum flow algorithm (MCMF) is widely used.
In the actual network problems, not only from the viewpoint of V S to V t maximum flow rate, but also consider the costs feasible flow in the network transmission process, which is the minimum cost network of maximum flow problem.
General reference minimum-cost flow problem: Given a network capacity D = (V, A, C ), each arc (V I, V J ) are given in addition to the capacity C ij of , but also gives the transmission unit traffic cost B ij of ≧ 0, referred to as D = (V, A, C , B), wherein B ij of ∈B. To find V at the expense of the capacity in the network D S → V T of the maximum flow f = {fij}, and the total cost of the transmission streams:
B (F) = sB ij of F ij of the minimum
, Method for finding maximum flow is from the last lecture can be seen on the network capacity from a feasible flow, try to find a path from V S → V t augmented chain, then augmented along this chain to adjust the flow and make new traffic increased feasible flow. In this new feasible flow basis and then look for its augmented chain. This is repeated until then could not find the augmented chain, you get the maximum flow of the network.
Now seeks to minimize the cost of the maximum flow, first look, when to adjust the f θ = 1 obtained along an augmenting μ chain on new feasible flow f feasible flow f ', the total cost of B (f') than b (f) how much?
Because the arc to the front of f + [mu] ij of '= f ij of + 1'd,
Examples: Given cost, capacity of the network of FIG. (B ij of , C ij of ), Find the maximum flow of minimum cost network.
solution:
(1), the initial flow rate takes 0 total cost of f (0) = 0.
(2), (right in FIG Each heavy network line charges b is constructed from the original cost of the network of FIG network I J , b ij of as a cost per unit flow rate).
(3), with the current cost to find a minimum cost network of FIG path, i.e. vs-> v3-> v2-> vt. Path drawn by the capacity of each segment of the line by the maximum flow, i.e., the adjustment amount θ = min {8,5,7} = 5 , i.e. the flow rate is 5, the current minimum cost is f (1) = 1 + 5 * 5 * 5 * 2 + 1 = 20. The following figure shows the network traffic FIG .
(4), the base (3) on the network configuration of FIG new cost, the method is configured to: when no flow through the line, the flow direction is not changed, the original cost of the cost. The vs-> v2;
When the line is equal to the maximum flow capacity of the line, the flow direction is changed, the original cost of the negative charges. The v3-> v2 becomes v2-> v3;
When the flow is less than the maximum capacity of the line of the line, the reverse flow increases, the cost of the original cost is negative. As vs-> v3 new v3-> vs;
(5), the cost of the network map obtained in (4), find the least cost path, i.e. vs-> v2-> v1-vt, each segment obtained by the capacity of the line path through the maximum flow rate, i.e., the adjustment amount θ = min {10,7-5} = 2, the minimum cost flow stream 7 is obtained, the minimum cost of the current f (2) = 2 * 4 + 5 * 1 + 5 * 2 + 7 * 1 = 30.
（6）、构造可行流 f2的费用网络图。
（7）、在（6）中得到的费用网络图上，找到费用最少的路径，即vs->v3->v4->vt，通过该路径上每条线段的容量得出该线路能通过的最大流量，即调整量θ = min{8-5,10,4} = 3，得到的最小费用流的流量为10，当前的最小费用为 f(3) = 2*4+8*1+5*2+7*1+3*3+3*2 = 42
（8）、构造可行流 f3 的费用网络图。
（9）、在（8）的费用网络图上，找到费用最少的路径，即vs->v2->v3->v4->vt，通过该路径上每条线段的容量得出该线路能通过的最大流量，即调整量θ = min{10-2, 10-3, 4-3, 5} = 1，得到的最小费用流的流量为11，当前最小费用为 f(4) = 3*4+8*1+4*2+7*1+4*3+4*2 = 55
（10）、构造可行流 f4 的费用网络图。
由于无法找到从 vs->vt 的最短路，所以 f(4) 就是该网络的最小费用最大流。