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The transhipment problem is to minimise the cost of transporting goods between various sources and destinations.

Python, 164 lines
 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164``` ```''' This program uses the stepping stone algorithum to solve the transhipment problem. That is how to transport various quuantities of material to various destinations minimising overall cost, given the various costs of sending a unit from each source to each destination. The sum of supply and demand must equal.''' def PrintOut(): GetDual() nCost = 0 print print ' DEMAND' + ' ' * ( m * 10) + 'SUPPLY' for y in aDemand: print '%10i' % y, print for x in range( n): for y in range( m): nCost += aCost[ x][ y] * aRoute[ x][ y] if aRoute[ x][ y] == 0: print '[<%2i>%4i]' %( aCost[ x][ y], aDual[ x][ y]), else: print '[<%2i>(%2i)]' %( aCost[ x][ y], aRoute[ x][ y] + 0.5), print ' : %i' % aSupply[ x] print 'Cost: ', nCost print 'Press ENTER to continue' raw_input() def NorthWest(): ''' The simplest method to get an initial solution. Not the most efficient''' global aRoute u = 0 v = 0 aS = [ 0] * m aD = [ 0] * n while u <= n - 1 and v <= m - 1: if aDemand[ v] - aS[ v] < aSupply[ u] - aD[ u]: z = aDemand[ v] - aS[ v] aRoute[ u][ v] = z aS[ v] += z aD[ u] += z v += 1 else: z = aSupply[ u] - aD[ u] aRoute[ u][ v] = z aS[ v] += z aD[ u] += z u += 1 def NotOptimal(): global PivotN global PivotM nMax = -nVeryLargeNumber GetDual() for u in range( 0, n): for v in range( 0, m): x = aDual[ u][ v] if x > nMax: nMax = x PivotN = u PivotM = v return ( nMax > 0) def GetDual(): global aDual for u in range( 0, n): for v in range( 0, m): aDual[ u][ v] = -0.5 # null value if aRoute[ u][ v] == 0: aPath = FindPath( u, v) z = -1 x = 0 for w in aPath: x += z * aCost[ w[ 0]][ w[ 1]] z *= -1 aDual[ u][ v] = x def FindPath( u, v): aPath = [[ u, v]] if not LookHorizontaly( aPath, u, v, u, v): print 'Path error, press key', u, v raw_input() return aPath def LookHorizontaly( aPath, u, v, u1, v1): for i in range( 0, m): if i != v and aRoute[ u][ i] != 0: if i == v1: aPath.append( [ u, i]) return True # complete circuit if LookVerticaly( aPath, u, i, u1, v1): aPath.append( [ u, i]) return True return False # not found def LookVerticaly( aPath, u, v, u1, v1): for i in range( 0, n): if i != u and aRoute[ i][ v] != 0: if LookHorizontaly( aPath, i, v, u1, v1): aPath.append([ i, v]) return True return False # not found def BetterOptimal(): global aRoute aPath = FindPath( PivotN, PivotM) nMin = nVeryLargeNumber for w in range( 1, len( aPath), 2): t = aRoute[ aPath[ w][ 0]][ aPath[ w][ 1]] if t < nMin: nMin = t for w in range( 1 , len( aPath), 2): aRoute[ aPath[ w][ 0]][ aPath[ w][ 1]] -= nMin aRoute[ aPath[ w - 1][ 0]][ aPath[ w - 1][ 1]] += nMin # example 1 aCost = [[ 2, 1, 3, 3, 2, 5] ,[ 3, 2, 2, 4, 3, 4] ,[ 3, 5, 4, 2, 4, 1] ,[ 4, 2, 2, 1, 2, 2]] aDemand = [ 30, 50, 20, 40, 30, 11] aSupply = [ 50, 40, 60, 31] ''' example 2 aCost = [[ 1, 2, 1, 4, 5, 2] ,[ 3, 3, 2, 1, 4, 3] ,[ 4, 2, 5, 9, 6, 2] ,[ 3, 1, 7, 3, 4, 6]] aDemand = [ 20, 40, 30, 10, 50, 25] aSupply = [ 30, 50, 75, 20] ''' ''' example3 aCost = [[ 5, 3, 6, 2] ,[ 4, 7, 9, 1] ,[ 3, 4, 7, 5]] aDemand = [ 16, 18, 30, 25] aSupply = [ 19, 37, 34] ''' n = len( aSupply) m = len( aDemand) nVeryLargeNumber = 99999999999 # add a small amount to prevent degeneracy # degeneracy can occur when the sums of subsets of supply and demand equal elipsis = 0.001 for k in aDemand: k += elipsis / len( aDemand) aSupply[ 1] += elipsis # initialisation aRoute = [] for x in range( n): aRoute.append( [ 0] * m) aDual = [] for x in range( n): aDual.append( [ -1] * m) NorthWest() PivotN = -1 PivotM = -1 PrintOut() # MAIN while NotOptimal(): print 'PIVOTING ON', PivotN, PivotM BetterOptimal() PrintOut() print "FINISHED" ```
 Created by James Coliins on Sat, 29 Nov 2008 (MIT)

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