Homework3solution - ORIE3310 Spring 2009 Homework 4 2. (a)...

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ORIE3310 Spring 2009 Homework 4 2. (a) It would almost seem intuitive to think that since we are shipping more units than originally that we would have to incur more cost since the costs are the same but the amount shipped has increased. However, let us see in parts (b) and (c) how an additional unit really affects our allocation and subsequently, cost. (b) Let us use the northwest corner rule to set up an initial allocation; that gives us: further, we can choose to make u1 = 0 to get the simplex multipliers (I'm not sure if you've used this method in class, but you can read the BHM explanation of why it works. ..it makes things much easier): 10 5 5 10 15 5 15 10 10 10 10 2 1 2 3 1 4 2 8 4 v1=4 v2=2 v3=-2 u1=0 u2=6 2 2 1 3 1
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ORIE3310 Spring 2009 Homework 4 This gives us the following reduced costs: ĉ 11 = 4 – 0 – 0 = 0 ĉ 12 = 2 – 0 – 2 = 0 ĉ 13 = 4 – 0 – (-2) = 6 ĉ 21 = 12 – 6 – 2 = 4 ĉ 22 = 8 – 6 – 2 = 0 ĉ 23 = 4 – 6 – (-2) = 0 and shipping cost of: cost = 10*4 + 5*2 + 5*8 + 10*4 = 130 We observe that each reduced cost for is equal to 0 when we ship from node i to j and non-negative when we do not ship from i to j. We now have found an optimal solution. (c) Let us again use the northwest corner rule to obtain an initial allocation. We see that the same edges remain, thus we can use the same simplex multipliers as in part (b) since the costs of shipping on these edges has not changed. That means that our cost looks like: cost = 10*4 + 6*2 + 4*8 + 11*4 = 128 This is easy to see from the graph of our multipliers. When we add a supply to 1 and a demand to 3 we see that we push an extra unit on x 12 , detract a unit from x 22 , and push an extra unit onto x 23 . This means we incur a cost of 2, detract a cost of 8, and incur an extra cost of 4. Our new cost function is exactly 2 – 8 + 4 = -2 the amount of the original cost. On the other hand, it is not always the case that the cost is less, for instance adding a unit of supply to 2 and adding a unit of demand to 1 would make us incur a cost of 4 – 2 + 8 – 4 = 6 more than our original solution. 10 6 4 11 15 6 15 11 10 10 11 2 1 1 3 2
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ORIE3310 Spring 2009 Homework 4 4(a) There are 10 equations, but they are linearly dependent (more precisely, the sum of first 5 equations is the same as the sum of last 5 equations). _ _ |1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | |0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | |0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 | |0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 | |0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 | |1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 | |0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 | |0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 | |0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 | |0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 | These equations correspond to the equations limiting the number of jobs an individual may work on (top 5), as well as the constraints that limit the number of people that may work on
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Homework3solution - ORIE3310 Spring 2009 Homework 4 2. (a)...

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