= This is because, for each of the links x, and X;ke the value is 1 (the person uses this link) and for each of the links Xgj, xy, X/1 and X;m the value is zero (the person does not use these links). For the network shown in Fig. 2.1, we get for node 2, *12-X24 - X25 = 0, node 3, X13-X35 = 0, node 4, X 24-X45 - X46 = 0, node 5, X25 + x3 + x4s - X56 - XS = 0, node 6, X46 + X56 - X68 = 0 and node 7 X37-X78 0 Further, since the person has to start from node 1, along one of the paths (links) starting from it, we have - X12 - 13 = -1 Similarly, as the person has to reach node 8 along one of the paths, we get *68 + X78 = 1 The above eight are, then, the constraints (equality type) that must be satisfied. The objective is to minimize the total cost of travelling from node 1 to 8, given by Minimize Z = 5x12 + 7x13 + 10x24 + 3x25 + 8x35 + x4 + 6X46 + 7x56 + 4xs7 + 5x68 + 3x78 This is, then the linear programming model for the network problem wherein every variable has a value 1 or 0. EXAMPLE 2.6-21 (Product Mix Problem) The Delhi Florist Company is planning to make up floral arrangements for the upcoming festival. The company has available the following supply of flowers at the costs shown: Type Number available Cost per flower ) Red roses 800 0.20 Gardenias 456 0.25 Carnations 4.000 0.15 White roses 920 0.20 Yellow roses 422 0.22 These flowers can be used in any of the four popular arrangements whose make up and selling prices are as follows: Arrangement Requirements Selling price Economy 4 red roses 6 2 gardenias 8 carnations Maytime 8 white roses 8 5 gardenias 10 carnations 4 yellow roses Spring-colour 9 red roses 10 f14