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Transportation and Network Flow Problems From Linear & Nonlinear Programming ch.5 Presented by Wei Shiou Information Management Dept. National Taiwan University 2002.11.7 Introduction Part A The Transportation Problem The Assignment Problem Part B Minimum Cost Flow Maximal Flow 問題描述 There are m origins that contain various amounts of a commodity that must be shipped to n destinations to meet demand requirements It is assumed that the system is balanced m n ai bj i 1 j 1 For Example A specific transportation problem with four origins and five destinations is defined by a 30,80,10,60 b 10,50,20,80,20 3 4 6 8 9 2 2 4 5 5 C 2 2 2 3 2 3 3 2 4 2 Standard Form x11 x12 ....x1n a1 x 21 x 22 ....x 2 n a2 .......... .......... .......... .......... .......... .......... .......... .......... .......... .. .......... x m1 x m 2 ....x mn a m x11 x 21 x m1 b1 x12 x 21 x m1 b2 .......... .......... .......... .......... .......... .......... .......... .......... .......... .. .......... x1n x2n x mn bn 可行解 We can let xij = aibj / S for i=1 to m and j=1 to n then it is clearly a feasible solution 相當於將supply依照比例分配給每個 destination 必然存在可行解!? Theorem. A transportation problem always has a solution , but there is exactly one redundant equality constraint . When any one of the equality constraints is dropped , the remaining system of n+m-1 equality constraints is linearly independent . 完整證明於 page.120 Simplex Method的應用 Step 1 用西北角法找出 initial basic feasible solution Step 2 計算 simplex multipliers and the relative cost coefficients , if all relative cost coefficients are nonnegative then stop. otherwise, go to Step.3 Step 3 選擇一個nonbasic variable corresponding to enter the basis。Compute the cycle of change and update the solution. Go to Step.2 請參照page.130 The Northwest Corner Rule Solution array： 10 20 30 30 20 30 80 10 10 40 20 60 10 50 20 80 20 The Northwest Corner Rule Step 1 Start with the cell in the upper left-hand corner Step 2 Allocate the maximum feasible amount consistent with row and column sum requirements involving that cell Step 3 Move one cell to the right if there is any remaining row requirement. Otherwise move one cell down till the end. 名詞定義 A nonsingular square matrix M is said to be triangular if by a permutation of its rows and columns it can be put in the form of a lower triangular matrix 判斷 M 是否 triangular 的過程叫做 back substitution, 可參照page.123 Basis Triangularity Basis Triangularity Theorem: Every basis of the transportation problem is triangular. 完整證明於 page.124 Corollary: If the row and column sums of a transportation problem are integers, then the basic variables in any basic solution are integers. Simplex Multipliers We partition the vector of multipliers as (u, v) Where ui represents the multiplier associated with the ith row sum constraint and vj represents the multiplier associated with the jth column sum constraint Simplex Multipliers In transportation problem, the relative cost coefficients are rij cij ui v j for i 1,2,....,m j 1,2,....,n 任意給定一個multiplier值，可算出所有其他值， 由此計算rij，找出下一個enter variable Cycle of Change Corollary: If the unit costs cij of a transportation problem are all integers, then the simplex multipliers associated with iany mbasis are integers 1,2,...., Theorem: Let B be a basis from A (ignoring one row), and let d be another column. Then the components of the vector y=B-1d are either 0,+1,-1 Cycle of Change From the Theorem mentioned before, we know that if a new enter variable has a value S, then S is equal to the value subtracted from the old basic variable We will set S equal to the smallest magnitude of these basic variables in the stepping-stone path Example 如page.131 3 4 6 8 9 5 10 20 30 2 2 4 5 5 3 30 20 30 80 2 2 2 3 2 1 10 10 3 3 2 4 2 2 + 40 20 60 -2 -1 1 2 0 10 50 20 80 20 Coefficient array Transportation tableau Degeneracy In a transportation tableau with m rows and n columns, there must be m+n-1 cells with allocations; otherwise, it is degenerate. 必須引入特殊符號來代替basic variable的 「零」，否則可能找不到stepping-stone path The Assignment Problem 問題描述: Assigning n workers to n jobs, there is a benefit(or cost) for each assignment. Each worker must be assigned to exactly one job and each job must have one assigned worker General formulation: n n Minimize c x j 1 i 1 ij ij n Subject to x 1 for i 1 to ij n j 1 n x i 1 ij 1 for j 1 to n xij 0 for i 1 to n , j 1 to n Minimum Cost Flow 問題描述: (以下簡稱MCF) Consider a network having n nodes. Corresponding to each node I, there is a number bi representing the available supply at the node(using bi<0 to represent a required demand) To determine flows xij>=0 in each arc of the network so that the net flow into each node I is bi while minimizing the total cost Still assume that the net work is balanced Problem Structure (MCF) n n Minimize c j 1 i 1 ij x ij n n Subject to x x j 1 ij k 1 ki bi for i 1 to n xij 0 for i 1 to n , j 1 to n The transportation problem is a special case of this problem! The more general problem is often termed the transshipment problem Simplex Method (MCF) Step 1 Start with a given basic feasible solution Step 2 Compute simplex multipliers and relative cost coefficients. If all rij are nonnegative, stop; Otherwise, go Step 3 Step 3 Select a nonbasic flow with negative relative cost coefficient to enter the basis. And move out the flow from the cycle, go Step 2 Maximal Flow 問題描述: (以下簡稱MF) To determine the maximal flow possible from one given source node to a sink node under arc capacity constraints. Reachable 自行參考page.141或Ford-Fulkerson’s Algorithm Capacitated Networks A capacitated network is a network in which some arcs are assigned nonnegative capacities The capacities of an arc(I,j) is denoted kij and this capacity is indicated on the graph by placing the number kij adjacent to the arc General Form (MF) maxinize f n n subject to x j 1 1j x j1 f 0 j 1 n n xj 1 ij x ji j 1 0 i 1, m n n xj 1 mj x jm f 0 j 1 i, j 0 xij k ij Min Cut Theorem In a network the maximal flow between a source and a sink is equal to the minimal cut capacity of all cuts separating the source and sink The capacity of the cut is the sum of the capacities of the arcs in the cut 證明於page.147 Primal-Dual Algorithm n n Minimize c j 1 i 1 ij x ij n n Subject to x x j 1 ij k 1 ki bi for i 1 to n xij 0 for i 1 to n , j 1 to n m n Maximize u a v b i i j j i 1 j 1 Subject to ui + vj <= cij Primal-Dual Algorithm m n minimize yi 1 i zj j 1 n subject to x j1 ij y i ai i 1 to m m x i 1 ij zi b j j 1 to n for all i , j xij 0 y i 0 z j 0 for i , j S xij 0 Primal-Dual Algorithm (MF) m n Since y i z in any solution j i 1 j 1 Alternatively, the problem can be cast as a maximal flow problem. The objective can be rewritten as m n n m ai xij b j xij i 1 j 1 j 1 i 1 Dual Feasible Solution With Primal-Dual Algorithm , it is unnecessary to construct the network explicitly! Hungarian’s Method This is The End Thank You All