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# sol8 - job(nothing Person 4 does Job 4 and Person 5 does...

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AMS 341 Spring, 2011 Homework Set # 8 — Solution Notes #1, 7.5: The assignment problem is given by the table (all “supplies” and “demands” are 1). Since there are 5 workers and only 4 jobs, we add a dummy job with times 0. We convert the dashes to some large cost M : 22 18 30 18 0 18 M 27 22 0 26 20 28 28 0 16 22 M 14 0 21 M 25 28 0 Subtracting the smallest element of each row does not change the table. Subtracting the smallest element of each column, we get: (I assumed here that M is so large that M minus any constant is still M). 6 0 5 4 0 2 M 2 8 0 10 2 3 14 0 0 4 M 0 0 5 M 0 14 0 After applying the first two steps of the Hungarian method, we find only 4 lines which cover all zeros (row 4, and column 2,3,5). After applying step 3, ( k = 2) we get the following table: (again, I am assuming that M s stay M s) 4 0 5 2 0 0 M 2 6 0 8 2 3 12 0 0 6 M 0 2 3 M 0 12 0 The underlined “zeros” give an optimal solution: Person 1 does job 2, Person 2 does job 1, Person 3 does the dummy
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Unformatted text preview: job (nothing), Person 4 does Job 4 and Person 5 does job 3. #6, 8.4 (a) 1 A:2 2 B:4 3 C:3 D:2 4 5 E:10 ±:4 6 ET (1) = 0; ET (2) = 2; ET (3) = 6; ET (4) = 8; ET (5) = 9; ET (6) = 19 LT (1) = 0; LT (2) = 2; LT (3) = 6; LT (4) = 9; LT (5) = 9; LT (6) = 19 TF (1 , 2) = 0; TF (2 , 3) = 0; TF (3 , 4) = 1; TF (4 , 5) = 1; TF (5 , 6) = 0; TF (3 , 6) = 9; TF (3 , 5) = 0 The critical path is 1-2-3-5-6, and the length is 19 weeks. (b) min z = x 6-x 1 s . t . x 2 ≥ x 1 + 2 x 3 ≥ x 2 + 4 x 4 ≥ x 3 + 2 x 5 ≥ x 3 + 3 x 6 ≥ x 5 + 10 x 6 ≥ x 3 + 4 x i unrestricted...
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