Assignment6Solution - Fall 2010 Optimization I(ORIE...

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Fall 2010 Optimization I (ORIE 3300/5300) Assignment 6 Solution Problem 1 In this problem, we have five people (supply) and only four stroke styles (demand). Since the supply quantity is greater than the demand quantity, we need to add a “dummy” demand: we add a new stroke style called “bench” that everyone can swim in 0 seconds. The person swimming this style will stay at the bench. After this, we have five styles, five swimmers, and the problem is clearly an assignment problem of assigning swimmers to different strokes. To fit it into a transportation model file, we have to make sure that all demands and supplies are equal to one, while the costs are clearly the times of swimmers in different strokes. We use the model file transp.mod from the amplcml’s MODELS folder. The data file is (there are more than one way to do this): set ORIG:= Barbara Carol Dianne Elizabeth Olympia; set DEST:= breaststroke butterfly backstroke freestyle bench; param supply default 1; param demand default 1; param cost (tr): Barbara Carol Dianne Elizabeth Olympia:= breaststroke 45.4 42.8 36.1 43.2 36.7 butterfly 34.3 36.6 29.5 39.9 33.4 backstroke 39.7 37.4 33.9 36.8 39.0 freestyle 31.2 33.1 29.4 31.6 32.5 bench 0.00 0 0 0 0; The output is: ampl: reset; model transp.mod; data transpdata.dat; ampl: solve; MINOS 5.5: optimal solution found. 14 iterations, objective 134.2 ampl: display Trans; Trans [*,*] : backstroke bench breaststroke butterfly freestyle := Barbara 0 0 0 -7.91731e-17 1 Carol 1.80525e-17 1 0 0 0 Dianne 0 0 -7.91731e-17 1 0 Elizabeth 1 0 0 0 9.02627e-18 Olympia 0 0 1 0 0 ; To get a nicer output, we can type: 1
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Fall 2010 Optimization I (ORIE 3300/5300) ampl: option omit_zero_rows 1; ampl: display Trans; Trans := Barbara butterfly -7.91731e-17 Barbara freestyle 1 Carol backstroke 1.80525e-17 Carol bench 1 Dianne breaststroke -7.91731e-17 Dianne butterfly 1 Elizabeth backstroke 1 Elizabeth freestyle 9.02627e-18 Olympia breaststroke 1 ; (Note: here, 7.91731e-17 is the same as zero. It is not displayed as zero due to round-off error.) Grading scheme: 2 points for adding a dummy demand (the new stroke called “bench”) such that the supply quantity is equal to the demand quantity; 2 points for using the transp.mod file provided; 6 points for correct data file and output. Problem 2 (a) Clearly, the problem is feasible. The point ( x 1 ,x 2 ) = (0 , 0) is feasible: x 1 - 2 x 2 = 0, so the first and second inequality constraints are both satisfied. It is also easy to see that the problem is bounded, since the first constraint provides an upper bound and the second constraint provides a lower bound to the objective value of any feasible solution. (b) The feasible region: Now, we prove that this region does not have any extreme point. In general, to show that a feasible region does not have any extreme point, we do the following. Consider an arbitrary point in the region (that is, an arbitrary feasible solution), call it x , and show that we can find two other feasible solutions, call them y and z such that
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Assignment6Solution - Fall 2010 Optimization I(ORIE...

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