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HW4solution

# HW4solution - Problem 1 Lets rst label the options Tv Trade...

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Problem 1 Let’s 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. max 1000000 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) 0 x i 1 , integer i 1

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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, δ 1 , δ 2 . Hence: x = δ 1 + δ 2 0 δ 1 20 0 δ 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 δ 1 + δ 2 50 y . The objective is to maximize the net profit: 40 δ 1 +40 δ 2 - 450 y - 5 δ 2 max - 450 y + 40 δ 1 + 35 δ 2 . Notice that when δ 2 > 0, it must be true that δ 1 = 20. Therefore, we introduce a binary variable ω : ω = 1 if δ 1 is at its upper bound 0 otherwise The constraints can be then replaced by 20 ω δ 1 20 0 δ 2 30 ω To conclude: max - 450 y + 40 δ 1 + 35 δ 2 20 ω δ 1 20 0 δ 2 30 ω δ 1 + δ 2 50 y ω, y are binaries 1
Since this is a mixed IP problem, we can only use branch and bound algorithm. b) Let the output x = δ 1 + δ 2 + δ 3 . In this case, δ 0 i s are integers. Using similar idea of part a) to express the constrains of δ ’s. We have 4 ω 1 δ 1 4 4 ω 2 δ 2 4 ω 1 0 δ 3 992 ω 2 where ω 1 and ω 2 are binaries: ω 1 = 1 if δ 1 is at its upper bound at 4 0 otherwise ω 2 = 1 if δ 2 is at its upper bound at 4 0 otherwise The operating cost is a piecewise linear function: 0 . 1 δ 1 + 0 . 05 δ 2 + 0 . 025 δ 3 . Therefore, the objective is maximize 0 . 5 δ 1 + 0 . 5 δ 2 + 0 . 5 δ 3 - 0 . 1 δ 1 - 0 . 05 δ 2 - 0 . 025 δ 3 max 0 . 4 δ 1 + 0 . 45 δ 2 + 0 . 475 δ 3 To conclude; max 0 . 4 δ 1 + 0 . 45 δ 2 + 0 . 475 δ 3 4 ω 1 δ 1 4 4 ω 2 δ 2 4 ω 1 0 δ 3 992 ω 2 δ 1 , δ 2 , δ 3

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