Question

5. The coach of an age group swim team needs to assign swimmers to a 200-yard medley relay team to send to the Junior Olympics. Since most of his best swimmers are very fast in more than one stroke, it is not clear which swimmer should be assigned to each of the four strokes. The five fastest swimmers and the best times in seconds) they have achieved in each of the strokes (for 50 yards) are Stroke C arl Backstroke 3 7.7 Breaststroke 43.4 Butterfly 3 3.3 Freestyle 29.2 Chris 329 33.1 28.5 26.4 David 33.8 42.2 38.9 29.6 Tony 37 34.7 30.4 28.5 Ken 35.4 41.8 33.6 31.1 The coach wishes to determine how to assign four swimmers to the four different strokes to minimize the sum of the corresponding best times. (a) Formulate this problem as an assignment problem. (b) Obtain an optimal solution

Answer 1

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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 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