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![Let Xi = 1 if Swimmer j is assigned to stroke i, = 0 if Swimmer j is not assigned to stroke i, Since there are only four stro](//img.homeworklib.com/questions/6013aa10-11ee-11ec-b21a-2f4140315194.png?x-oss-process=image/resize,w_560)
![X31 + X32 + X33 +X34 + X35 <1 X41 + X42 + X43 +X44 + X45 S1 X51 + X52 + X53 +X54 + X55 <1 X11 + X21 + X31 +X41 + X51 S1 X12 +](//img.homeworklib.com/questions/608ebcd0-11ee-11ec-87b3-31f08f429674.png?x-oss-process=image/resize,w_560)
![0 7.1 8.8 Minimum no. of lines covering maximum zeros = 3, so optimal solution is not reached So we revised the table. Minimu](//img.homeworklib.com/questions/6126f5a0-11ee-11ec-befe-4bbc0f633da1.png?x-oss-process=image/resize,w_560)
Let Xi = 1 if Swimmer j is assigned to stroke i, = 0 if Swimmer j is not assigned to stroke i, Since there are only four stroke styles we use a dummy stroke style with time 0. Where i = backstroke, breaststroke, butterfly, freestyle , dummy and and j = Carl, Chris, David, Tony, Ken (For ease we take backstroke =1, breaststroke=2, butterfly=3, freestyle=4, dummy =5 and and j = Carl=1, Chris=2, David=3, Tony=4, Ken=5. Thus i =1, 2, 3, 4, 5 and j = 1, 2, 3, 4, 5 The objective function is Maximize Z = EE CiXj That is Minimize Z= 37.7X11 + 32.9X12 + 33.8X13+ 37.0X14+ 35.4X15 + 43.4X21+ 33.1X22+ 42.2X23 + 34.7X24 + 41.8X25 + 33.3X31 + 28.5X32 + 38.9X33 + 30.4X34 + 33.6 X35 + 29.2X41 + 26.4X42 + 29.6X43 + 28.5X44 +31.1X45 + 0(X51+X52 + X53 + X 54 +X55) Subject to X11 + X12 + X13 +X14 + X15 51 X21 + X22 + X23 +X24 + X25 51
X31 + X32 + X33 +X34 + X35 <1 X41 + X42 + X43 +X44 + X45 S1 X51 + X52 + X53 +X54 + X55 <1 X11 + X21 + X31 +X41 + X51 S1 X12 + X22 + X32 +X42 + X52 51 X13 + X23 + X33 +X43 + X53 S1 X14 + X24 + X34 +X44 + X54 51 X15 + X25 + X35 +X45 + X55 S1 And Xij 0 or 1, i = 1,2,3,4,5; and j = 1,2,3,4,5 Optimal Solution: Carl Chris David Tony Ken Backstroke 37.7 32.9 33.8 37 35.4 Breaststroke 43.4 33.1 42.2 34.7 41.8 Butterfly 33.3 28.5 38.9 30.4 33.6 Freestyle | 29.2 | 26.4 29.6 28.5 31.1 Dummy 0 0 0 0 0 Subtracting smallest element of a row from every element of that row we get following table: 2.5 Carl Chris David Tony Ken Backstroke 4.8 0.9 4.1 Breaststroke 10.3 9.1 1.6 8.7 Butterfly 10.4 1.9 5.1 Freestyle 2. 8 0 3.2 2.1 4.7 Dummy 0 0 0 Subtracting smallest element of a column from every element of that column we get following table: 4.8 0 O 5.1 Carl | Chris David Tony Ken Backstroke 4.8 0.9 4.1 2.5 Breaststroke 10.3 9.1 1.6 8.7 Butterfly 4.8 0 10.4 1.9 Freestyle 2.8 0 3.2 2.1 2.1 / 4.7 Dummy 0 O 0 Minimum no. of lines covering maximum zeros =2, so optimal solution is not reached So we revised the table. Minimum from uncovered cells =0.9 Chris David Tony Ken Backstroke 3.9 3.2 1.6 Breaststroke 9.4 0.7 / 7.8 Butterfly 3.9 9.5 1 1 4.2 Freestyle 3.8 Dummy 0.9 0 0 0 Carl 0 0 8.2 1.9 2.3 1.2
0 7.1 8.8 Minimum no. of lines covering maximum zeros = 3, so optimal solution is not reached So we revised the table. Minimum from uncovered cells =0.7 Carl Chris David Tony ken Backstroke 3.9 0.7 3.2 1.6 Breaststroke 0 7. 5 0 Butterfly 3.2 0 0.3 3.5 Freestyle 1.2 O 1.6 0.5 3.1 Dummy 0 1.6 0 Minimum no. of lines covering maximum zeros =4, so optimal solution is not reached So we revised the table. Minimum from uncovered cells = 1.2 Carl Chris David Tony Ken Backstroke 3.9 1.9 4.4 1.6 Breaststroke 7.5 0 6.3 / 0 / 5.9 Butterfly 2 0 7.6 0.3 2.3 Freestyle 0 0 0.4 0.5 1.9 Dummy O 2.8 1.2 Minimum no. of lines covering maximum zeros =5, so optimal solution is reached Carl Chris | David Tony Ken Backstroke 3.9 1.9 4.4 1.6 Breaststroke | 7.5 6.3 0 5.9 Butterfly 2 7.6 2.3 Freestyle O A 0.4 0.5 1.9 Dummy @ 2.8 1.2 O The optimal Assignments : Backstroke -> David, Breaststroke -> Tony, Burretfly -> Chris, Freestyle -> Carl, Dummy->ken 0 0 0 0.3 The minimum sum of best times = 33.8 + 34.7+ 28.5+ 29.2 + 0 = 126.2