Allocation problems 2E 1 The initial cost matrix is shown below (with the numbers representing minutes): Task C Task D Task E Worker L 37 15 12 Worker M 25 13 16 Worker N 32 41 35 The following linear programming problem can be formulated to minimise the total time taken. 1 if worker i does task j 0 otherwise andj C, D, E where i L, M, N To minimise the total time (in minutes) taken P. Minimise P37 x 15 x 12x LC LD LE 25 x 13 x 16x MC MD ME 32 x 41 x 35x NC ND NE Each worker can be assigned to at most one task: x x x 1 or x 1 LC LD LE xmc xmd xme 1 or x 1 xnc xnd xne 1 or x 1 Each task must be done by just one worker: x x x 1 or x 1 LC MC NC xld xmd xnd 1 or x 1 xle xme xne 1 or x 1 Lj Mj Nj ic id ie Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 1
2 The initial cost matrix is shown below (with the numbers representing hundreds of pounds): Task S Task T Task U Task V Worker C 36 34 32 35 Worker D 37 32 34 33 Worker E 42 35 37 36 Worker F 39 34 35 35 The following linear programming problem can be formulated to minimise the total training cost. 1 if worker i is trained in task j 0 otherwise and j S, T, U, V where i C, D, E, F To minimise the total training cost (in hundreds of pounds) P. Minimise P36 x 34 x 32 x 35x CS CT CU CV 37 x 32 x 34 x 33x DS DT DU DV 42 x 35 x 37 x 36x ES ET EU EV 39 x 34 x 35 x 35x FS FT FU FV Each worker is to be trained in exactly one task: x x x x 1 or x 1 CS CT CU CV xds xdt xdu xdv 1 or x 1 xes xet xeu xev 1 or x 1 xfs xft xfu xfv 1 or x 1 Each task must have one worker trained to carry it out: x x x x 1 or x 1 CS DS ES FS xct xdt xet xft 1 or x 1 xcu xdu xeu xfu 1 or x 1 xcv xdv xev xfv 1 or x 1 Ej Fj is it iu iv Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 2
3 The initial cost matrix is shown below (with the figures representing the number of names and addresses): 1 2 3 A 11 15 B 14 18 17 C 16 13 23 D 15 14 22 Replacing the entry with a large value of 1000 gives: 1 2 3 A 11 15 1000 B 14 18 17 C 16 13 23 D 15 14 22 The following linear programming problem can be formulated to maximise the number of names and addresses collected. 1 if worker i is assigned to site j 0 otherwise where i A, B, C, D and j 1, 2, 3 To maximise the number of names and addresses collected P. Maximise P11x A1 15x A2 1000x A3 14 x 18 x 17x B1 B2 B3 16 x 13 x 23x C1 C2 C3 15 x 14 x 22x D1 D2 D3 Each worker must be assigned to just one site: x x x or x 1 A1 A2 A3 1 B1 B2 B3 1 C1 C2 C3 1 D1 D2 D3 1 x x x or x 1 x x x or x 1 x x x or x 1 Each site must be assigned to just one worker: x x x x or x 1 1 A1 B1 C1 D1 1 A2 B2 C2 D2 1 A3 B3 C3 D3 1 Aj Bj x x x x or x 2 1 x x x x or x 3 1 i i i Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 3
4 a The initial cost matrix is shown below (with the numbers representing 100s): A 12 8 11 9 B 14 10 9 13 C 11 9 12 10 D 13 11 10 12 As the table above is an n by n matrix (with n = 4), we do not need to add any additional dummy rows or columns. To maximise the profit, subtract every number from the largest value in the table. b The largest value in the table is 14. Subtracting every value from 14 the table becomes: A 2 6 3 5 B 0 4 5 1 C 3 5 2 4 D 1 3 4 2 The smallest numbers in rows 1, 2, 3 and 4 are 2, 0, 2 and 1. We reduce rows first by subtracting these numbers from each element in the row. The table becomes: A 0 4 1 3 B 0 4 5 1 C 1 3 0 2 D 0 2 3 1 The smallest numbers in columns 1, 2, 3 and 4 are 0, 2, 0 and 1. We reduce columns by subtracting these numbers from each element in the column. Therefore, the reduced cost matrix is: A 0 2 1 2 B 0 2 5 0 C 1 1 0 1 D 0 0 3 0 We can cover all the zeros with four lines, so we have found our optimal solution of: A W (12) B Z (13) C Y (12) D X (11) Maximum profit = 100 12 13 12 11 100 48 4800 Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 4
4 c Using the initial cost matrix below the following linear programming problem can be formulated to maximise the amount of profit made. A 12 8 11 9 1 if worker i does task j 0 otherwise and j W, X, Y, Z where i A, B, C, D To maximise the total profit (in 100s) P. Maximise P12 x 8 x 11 x 9x 14 x 10 x 9 x 13x BW BX BY BZ 11 x 9 x 12 x 10x CW CX CY CZ 13 x 11 x 10 x 12x Each worker can be assigned to at most one task: x x x x 1 or x 1 xbw xbx xby xbz 1 or x 1 xcw xcx xcy xcz 1 or x 1 x B 14 10 9 13 C 11 9 12 10 D 13 11 10 12 x x x 1 or x 1 Each task must be done by just one worker: x x x x 1 or x 1 AW BW CW DW xax xbx xcx xdx 1 or x 1 xay xby xcy xdy 1 or x 1 xaz xbz xcz xdz 1 or x 1 Aj Bj iw ix iy iz Alternatively, the linear programming problem on the next page shows the problem formulated as a minimisation problem using the modified cost matrix found in part b. Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 5
4 c (continued) Modified cost matrix from part b: A 2 6 3 5 B 0 4 5 1 C 3 5 2 4 D 1 3 4 2 1 if worker i does task j 0 otherwise and j W, X, Y, Z where i A, B, C, D Objective: To minimise the objective function P. Minimise P2 x 6 x 3 x 5x 4 x 5 x x BX BY BZ 3 x 5 x 2 x 4x CW CX CY CZ x 3 x 4 x 2x Each worker can be assigned to at most one task: x x x x 1 or x 1 xbw xbx xby xbz 1 or x 1 xcw xcx xcy xcz 1 or x 1 x x x x 1 or x 1 Each task must be done by just one worker: x x x x 1 or x 1 AW BW CW DW xax xbx xcx xdx 1 or x 1 xay xby xcy xdy 1 or x 1 xaz xbz xcz xdz 1 or x 1 Aj Bj iw ix iy iz Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 6