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Unformatted text preview: Problem 1 Lets first label the options Tv, Trade magazine, Newspaper, Radio, Popular magazine and promotional campaign as by number 1 , 2 , 3 , 4 , 5 , 6 respectively. Define x i be the decision variables. Let x i = 1 if advertisement option i is used 0 otherwise The objective is to maximize the number of customers reached. max1000000 x 1 + 200000 x 2 + 300000 x 3 + 400000 x 4 + 450000 x 5 + 450000 x 6 500000 x 1 + 150000 x 2 + 300000 x 3 + 250000 x 4 + 250000 x 5 + 100000 x 6 1800000 (cost constrain) 700 x 1 + 250 x 2 + 200 x 3 + 200 x 4 + 300 x 5 + 400 x 6 1500 (designers constrain) 200 x 1 + 100 x 2 + 100 x 3 + 100 x 4 + 100 x 5 + 1000 x 6 1200 (salesman constrain) x 4 + x 5 x 6 ( x 4 or x 5 = 1 when x 6 = 1) x 2 + x 5 1 (2,4 cannot both be 1) x i 1 , integer i 1 Problem 2 a) Since the cost for the first 20kg and the rest 30kg of commodities is different. We express the total amount of commodity x as the sum of two variables, 1 , 2 . Hence: x = 1 + 2 1 20 2 30 We are also interested in finding if whether the courier travel to Europe or not, therefore, we use a binary variable y to indicate it: y = 1 if the courier will travel 0 if the courier will not travel Therefore, we have 1 + 2 50 y . The objective is to maximize the net profit: 40 1 +40 2 450 y 5 2 max 450 y + 40 1 + 35 2 . Notice that when 2 > 0, it must be true that 1 = 20. Therefore, we introduce a binary variable : = 1 if 1 is at its upper bound 0 otherwise The constraints can be then replaced by 20 1 20 2 30 To conclude: max 450 y + 40 1 + 35 2 20 1 20 2 30 1 + 2 50 y ,y are binaries 1 Since this is a mixed IP problem, we can only use branch and bound algorithm. b) Let the output x = 1 + 2 + 3 . In this case, i s are integers. Using similar idea of part a) to express the constrains of s. We have 4 1 1 4 4 2 2 4 1 3 992 2 where 1 and 2 are binaries: 1 = 1 if 1 is at its upper bound at 4 0 otherwise 2 = 1 if 2 is at its upper bound at 4 0 otherwise The operating cost is a piecewise linear function: 0 . 1 1 + 0 . 05 2 + 0 . 025 3 . Therefore, the objective is maximize 0 . 5 1 + 0 . 5 2 + 0 . 5 3 . 1 1 . 05 2 . 025 3 max 0 . 4 1 + 0 . 45 2 + 0...
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This note was uploaded on 10/04/2010 for the course ORIE 3300 taught by Professor Todd during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 TODD

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