''' 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"