Assignment6Solution

# 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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