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200 | bed earlier, we specify constraints (16.69) to (16.105) in our model and turn it loose. The result is the spreadsheet in Figure 16.12. Based on the costs we chose, it turns out to be optimal not to use any overtime. (Overtime costs $52.5 - 35 = 15.50 per hour each month, while hiring a new worker costs only $15 per Chapter 16 583 Aggregate and Workforce Planning Figure 16.11 Infeasible “chase” solution. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 3 |
201 | se” solution. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 Parameters: r h l l' e e' b l_0 W_0 t d_t Decision Variables: t Xt Wt Ht Ft It Ot Objective: Profit: Constraints: I1-I0-X1 I2-I1-X2 I3-I2-X3 I4-I3-X4 I5-I4-X5 I6-I5-X6 I7-I6-X7 I8-I7-X8 I9-I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 W6-W5-H6+F6 |
202 | -X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 W6-W5-H6+F6 W7-W6-H7+F7 W8-W7-H8+F8 W9-W8-H9+F9 W10-W9-H10+F10 W11-W10-H11+F11 W12-W11-H12+F12 bX1-W1-O1 bX2-W2-O2 bX3-W3-O3 bX4-W4-O4 bX5-W5-O5 bX6-W6-O6 bX7-W7-O7 bX8-W8-O8 bX9-W9-O9 bX10-W10-O10 bX11-W11-O11 bX12-W12-O12 B C D E F G H I J K L M 1000 10 35 52.5 15 9 12 0 2520 1 200 2 220 3 230 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 1 200.00 2520.00 0.00 0.00 0.00 |
203 | 30 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 1 200.00 2520.00 0.00 0.00 0.00 10.00 2 220.00 2520.00 0.00 0.00 0.00 0.00 3 230.00 2520.00 0.00 0.00 0.00 0.00 4 300.00 2520.00 0.00 0.00 0.00 0.00 5 400.00 2520.00 0.00 0.00 0.00 0.00 6 450.00 2520.00 0.00 0.00 0.00 0.00 7 320.00 2520.00 0.00 0.00 0.00 0.00 8 180.00 2520.00 0.00 0.00 0.00 0.00 9 170.00 2520.00 0.00 0.00 0.00 0.00 10 170.00 2520.00 0.00 0.00 0.00 0.00 11 160.00 2520.00 0.00 0.00 0.00 0.00 12 180.0 |
204 | .00 0.00 0.00 10 170.00 2520.00 0.00 0.00 0.00 0.00 11 160.00 2520.00 0.00 0.00 0.00 0.00 12 180.00 2520.00 0.00 0.00 0.00 0.00 $1,921,600.00 = -d_1 -200.00 -200 = -d_2 -220.00 -220 = -d_3 -230.00 -230 = -d_4 -300.00 -300 = -d_5 -400.00 -400 = -d_6 -450.00 -450 = -d_7 -320.00 -320 = -d_8 -180.00 -180 = -d_9 -170.00 -170 = -d_10 -170.00 -170 = -d_11 -160.00 -160 = -d_12 -180.00 -180 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0 0.00 = 0 0.00 = 0 0.00 = 0 0.00 < |
205 | = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0 0.00 = 0 0.00 = 0 0.00 = 0 0.00 <= 0 -120.00 <= 0 120.00 <= 0 240.00 <= 0 1080.00 <= 0 2280.00 <= 0 2880.00 <= 0 1320.00 <= 0 -360.00 <= 0 -480.00 <= 0 -480.00 <= 0 -600.00 <= 0 -360.00 Note: All decision variables must be >= 0 584 Part III Principles in Practice Figure 16.12 LP optimal solution. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 |
206 | 2 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 Parameters: r h l l' e e' b l_0 W_0 t d_t Decision Variables: t Xt Wt Ht Ft It Ot Objective: Profit: Constraints: I1-I0-X1 I2-I1-X2 I3-I2-X3 I4-I3-X4 I5-I4-X5 I6-I5-X6 I7-I6-X7 I8-I7-X8 I9-I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 W6-W5-H6+F6 W7-W6-H7+F7 W8-W7-H8+F8 W9-W8-H9+F9 W10-W |
207 | W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 W6-W5-H6+F6 W7-W6-H7+F7 W8-W7-H8+F8 W9-W8-H9+F9 W10-W9-H10+F10 W11-W10-H11+F11 W12-W11-H12+F12 bX1-W1-O1 bX2-W2-O2 bX3-W3-O3 bX4-W4-O4 bX5-W5-O5 bX6-W6-O6 bX7-W7-O7 bX8-W8-O8 bX9-W9-O9 bX10-W10-O10 bX11-W11-O11 bX12-W12-O12 B C D E F G H I J K L M 1000 10 35 52.5 15 9 12 0 2520 1 200 2 220 3 230 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 1 302.86 3634.29 1114.29 0.00 102.86 0.00 2 302.86 3634.29 0.00 0.00 18 |
208 | 170 10 170 11 160 12 180 1 302.86 3634.29 1114.29 0.00 102.86 0.00 2 302.86 3634.29 0.00 0.00 185.71 0.00 3 302.86 3634.29 0.00 0.00 258.57 0.00 4 302.86 3634.29 0.00 0.00 261.43 0.00 5 302.86 3634.29 0.00 0.00 164.29 0.00 6 302.86 3634.29 0.00 0.00 17.14 0.00 7 302.86 3634.29 0.00 0.00 0.00 0.00 8 180.00 2160.00 0.00 1474.29 0.00 0.00 9 170.00 2040.00 0.00 120.00 0.00 0.00 10 170.00 2040.00 0.00 0.00 0.00 0.00 11 170.00 2040.00 0.00 0.00 10.00 0.00 12 170.00 2040.00 0.00 0.00 0. |
209 | 0 2040.00 0.00 0.00 0.00 0.00 11 170.00 2040.00 0.00 0.00 10.00 0.00 12 170.00 2040.00 0.00 0.00 0.00 0.00 $1,687,337.14 = -d_1 -200.00 -200 = -d_2 -220.00 -220 = -d_3 -230.00 -230 = -d_4 -300.00 -300 = -d_5 -400.00 -400 = -d_6 -450.00 -450 = -d_7 -320.00 -320 = -d_8 -180.00 -180 = -d_9 -170.00 -170 = -d_10 -170.00 -170 = -d_11 -160.00 -160 = -d_12 -180.00 -180 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 <= 0.00 0 <= 0 0.00 <= |
210 | .00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 <= 0.00 0 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 Note: All decision variables must be >= 0 Chapter 16 585 Aggregate and Workforce Planning hour as a one-time cost.) Instead, the model adds 1,114.29 hours to the workforce, which represents 1,114.29 = 6.6 168 new workers. After the peak season of months 4 to 7, the solution calls for a reduction |
211 | .29 = 6.6 168 new workers. After the peak season of months 4 to 7, the solution calls for a reduction of 1,474.29 + 120 = 1,594.29 hours, which implies laying off 1,594.29 = 9.5 168 workers. Additionally, the solution involves building in excess of demand in months 1 to 4 and using this inventory to meet peak demand in months 5 to 7. The net pro?t resulting from this solution is $1,687,337.14. From a management standpoint, the planned layoffs in months 8 and 9 might be a problem. Although we ha |
212 | rom a management standpoint, the planned layoffs in months 8 and 9 might be a problem. Although we have speci?ed penalties for these layoffs, these penalties are highly speculative and may not accurately consider the long-term effects of hiring and ?ring on worker morale, productivity, and the ?rm’s ability to recruit good people. Thus, it is worthwhile to carry our analysis further. One approach we might consider would be to allow the model to hire but not ?re workers. We can easily do this by |
213 | h we might consider would be to allow the model to hire but not ?re workers. We can easily do this by eliminating the Ft variables or, since this requires fairly extensive changes in the spreadsheet, specifying additional constraints of the form Ft = 0 t = 1, . . . , 12 Rerunning the model with these additional constraints produces the spreadsheet in Figure 16.13. As we expect, this solution does not include any layoffs. Somewhat surprising, however, is the fact that it does not involve any n |
214 | oes not include any layoffs. Somewhat surprising, however, is the fact that it does not involve any new hires either (that is, Ht = 0 for every period). Instead of increasing the workforce size, the model has chosen to use overtime in months 3 to 7. Evidently, if we cannot ?re workers, it is uneconomical to hire additional people. However, when one looks more closely at the solution in Figure 16.13, a problem becomes evident. Overtime is too high. For instance, month 6 has more hours of overtim |
215 | .13, a problem becomes evident. Overtime is too high. For instance, month 6 has more hours of overtime than hours of regular time! This means that our workforce of 15 people has 2,880/15 = 192 hours of overtime in the month, or about 48 hours per week per worker. This is obviously excessive. One way to eliminate this overtime problem is to add some more constraints. For instance, we might specify that overtime is not to exceed 20 percent of regular time. This would correspond to the entire work |
216 | y that overtime is not to exceed 20 percent of regular time. This would correspond to the entire workforce working an average of one full day of overtime per week in addition to the normal 5-day workweek. We could do this by adding constraints of the form Ot = 0.2Wt t = 1, . . . , 12 (16.106) doing this to the spreadsheet of Figure 16.13 and resolving results in the spreadsheet shown in Figure 16.14. The overtime limits have forced the model to resort to hiring. Since layoffs are still not a |
217 | e 16.14. The overtime limits have forced the model to resort to hiring. Since layoffs are still not allowed, the model hires only 508.57 hours worth of workers, or 508.57 =3 168 new workers, as opposed to the 6.6 workers hired in the original solution in Figure 16.12. To attain the necessary production, the solution uses overtime in months 1 to 7. 586 Part III Principles in Practice Figure 16.13 Optimal solution when Ft = 0. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 |
218 | 16.13 Optimal solution when Ft = 0. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 Parameters: r h l l' e e' b l_0 W_0 t d_t Decision Variables: t Xt Wt Ht Ft It Ot Objective: Profit: Constraints: I1-I0-X1 I2-I1-X2 I3-I2-X3 I4-I3-X4 I5-I4-X5 I6-I5-X6 I7-I6-X7 I8-I7-X8 I9-I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5 |
219 | -I7-X8 I9-I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 W6-W5-H6+F6 W7-W6-H7+F7 W8-W7-H8+F8 W9-W8-H9+F9 W10-W9-H10+F10 W11-W10-H11+F11 W12-W11-H12+F12 bX1-W1-O1 bX2-W2-O2 bX3-W3-O3 bX4-W4-O4 bX5-W5-O5 bX6-W6-O6 bX7-W7-O7 bX8-W8-O8 bX9-W9-O9 bX10-W10-O10 bX11-W11-O11 bX12-W12-O12 B C D E F G H I J K L M 1000 10 35 52.5 15 9 12 0 2520 1 200 2 220 3 230 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 1 210.00 |
220 | 2520 1 200 2 220 3 230 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 1 210.00 2520.00 0.00 0.00 10.00 0.00 2 210.00 2520.00 0.00 0.00 0.00 0.00 3 230.00 2520.00 0.00 0.00 0.00 240.00 4 300.00 2520.00 0.00 0.00 0.00 1080.00 5 400.00 2520.00 0.00 0.00 0.00 2280.00 6 450.00 2520.00 0.00 0.00 0.00 2880.00 7 320.00 2520.00 0.00 0.00 0.00 1320.00 8 180.00 2520.00 0.00 0.00 0.00 0.00 9 170.00 2520.00 0.00 0.00 0.00 0.00 10 170.00 2520.00 0.00 0.00 0.00 0.00 11 160.00 2 |
221 | 0 0.00 0.00 9 170.00 2520.00 0.00 0.00 0.00 0.00 10 170.00 2520.00 0.00 0.00 0.00 0.00 11 160.00 2520.00 0.00 0.00 0.00 0.00 12 180.00 2520.00 0.00 0.00 0.00 0.00 $1,512,000.00 = -d_1 -200.00 -200 = -d_2 -220.00 -220 = -d_3 -230.00 -230 = -d_4 -300.00 -300 = -d_5 -400.00 -400 = -d_6 -450.00 -450 = -d_7 -320.00 -320 = -d_8 -180.00 -180 = -d_9 -170.00 -170 = -d_10 -170.00 -170 = -d_11 -160.00 -160 = -d_12 -180.00 -180 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = |
222 | 0 -160 = -d_12 -180.00 -180 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0 0.00 = 0 0.00 = 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 -360.00 <= 0 -480.00 <= 0 -480.00 <= 0 -600.00 <= 0 -360.00 Note: All decision variables must be >= 0 Chapter 16 587 Aggregate and Workforce Planning Figure 16.14 Optimal solution when Ft = 0 and Ot = 0.2Wt . A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 |
223 | en Ft = 0 and Ot = 0.2Wt . A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 Parameters: r h l l' e e' b l_0 W_0 t d_t Decision Variables: t Xt Wt Ht Ft It Ot Objective: Profit: Constraints: I1-I0-X1 I2-I1-X2 I3-I2-X3 I4-I3-X4 I5-I4-X5 I6-I5-X6 I7-I6-X7 I8-I7-X8 I9-I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 |
224 | -I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 W6-W5-H6+F6 W7-W6-H7+F7 W8-W7-H8+F8 W9-W8-H9+F9 W10-W9-H10+F10 W11-W10-H11+F11 W12-W11-H12+F12 bX1-W1-O1 bX2-W2-O2 bX3-W3-O3 bX4-W4-O4 bX5-W5-O5 bX6-W6-O6 bX7-W7-O7 bX8-W8-O8 bX9-W9-O9 bX10-W10-O10 bX11-W11-O11 bX12-W12-O12 B C D E F G H I J K L M 1000 10 35 52.5 15 9 12 0 2520 1 200 2 220 3 230 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 1 302.86 3028.57 5 |
225 | 0 2 220 3 230 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 1 302.86 3028.57 508.57 0.00 102.86 605.71 2 302.86 3028.57 0.00 0.00 185.71 605.71 3 302.86 3028.57 0.00 0.00 258.57 605.71 4 302.86 3028.57 0.00 0.00 261.43 605.71 5 302.86 3028.57 0.00 0.00 164.29 605.71 6 302.86 3028.57 0.00 0.00 17.14 605.71 7 302.86 3028.57 0.00 0.00 0.00 605.71 8 180.00 3028.57 0.00 0.00 0.00 0.00 9 170.00 3028.57 0.00 0.00 0.00 0.00 10 170.00 3028.57 0.00 0.00 0.00 0.00 11 160.0 |
226 | 0.00 0.00 0.00 9 170.00 3028.57 0.00 0.00 0.00 0.00 10 170.00 3028.57 0.00 0.00 0.00 0.00 11 160.00 3028.57 0.00 0.00 0.00 0.00 12 180.00 3028.57 0.00 0.00 0.00 0.00 $1,467,871.43 = -d_1 -200.00 -200 = -d_2 -220.00 -220 = -d_3 -230.00 -230 = -d_4 -300.00 -300 = -d_5 -400.00 -400 = -d_6 -450.00 -450 = -d_7 -320.00 -320 = -d_8 -180.00 -180 = -d_9 -170.00 -170 = -d_10 -170.00 -170 = -d_11 -160.00 -160 = -d_12 -180.00 -180 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 |
227 | 0.00 -160 = -d_12 -180.00 -180 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0 0.00 = 0 0.00 = 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 0.00 <= 0 -868.57 <= 0 -988.57 <= 0 -988.57 <= 0 -1108.57 <= 0 -868.57 Note: All decision variables must be >= 0 588 Part III Principles in Practice Notice that the amount of overtime used in these months is exactly 20 percent of regular time work hours, that is, 3,028.57 × 0.2 = 605.71 Wh |
228 | in these months is exactly 20 percent of regular time work hours, that is, 3,028.57 × 0.2 = 605.71 What this means is that new constraints (16.106) are binding for periods 1 to 7, which we would be told explicitly if we printed out the sensitivity analysis reports generated by the LP solver. This implies that if it is possible to work more overtime in any of these months, we can improve the solution. Notice that the net pro?t in the model of the spreadsheet shown in Figure 16.14 is $1,467,871.4 |
229 | tion. Notice that the net pro?t in the model of the spreadsheet shown in Figure 16.14 is $1,467,871.43, which is a 13 percent decrease over the original optimal solution of $1,687,337.14 in Figure 16.12. At ?rst glance, it may appear that the policies of no layoffs and limits on overtime are expensive. On the other hand, it may really be telling us that our original estimates of the costs of hiring and ?ring were too low. If we were to increase these costs to represent, for example, long-term d |
230 | ing and ?ring were too low. If we were to increase these costs to represent, for example, long-term disruptions caused by labor changes, the optimal solution might be very much like the one arrived at in Figure 16.14. 16.4.3 Modeling Insights In addition to providing a detailed example of a workforce formulation in LP (16.61)– (16.67), we hope that our discussion has helped the reader appreciate the following aspects of using an optimization model as the basis for an AP or WP module. 1. Multi |
231 | te the following aspects of using an optimization model as the basis for an AP or WP module. 1. Multiple modeling approaches. There are often many ways to model a given problem, none of which is “correct” in any absolute sense. The key is to use cost coef?cients and constraints to represent the main issues in a sensible way. In this example, we could have generated solutions without layoffs by either increasing the layoff penalty or placing constraints on the layoffs. Both approaches would achi |
232 | ither increasing the layoff penalty or placing constraints on the layoffs. Both approaches would achieve the same qualitative conclusions. 2. Iterative model development. Modeling and analysis almost never proceed in an ideal fashion in which the model is formulated, solved, and interpreted in a single pass. Often the solution from one version of the model suggests an alternate model. For instance, we had no way of knowing that eliminating layoffs would cause excessive overtime in the solution. |
233 | ce, we had no way of knowing that eliminating layoffs would cause excessive overtime in the solution. We didn’t know we would need constraints on the level of overtime until we saw the spreadsheet output of Figure 16.13. 16.5 Conclusions In this chapter, we have given an overview of the issues involved in aggregate and workforce planning. A key observation behind our approach is that, because the aggregate planning and workforce planning modules use long time horizons, precise data and intric |
234 | the aggregate planning and workforce planning modules use long time horizons, precise data and intricate modeling detail are impractical or impossible. We must recognize that the production or workforce plans that these modules generate will be adjusted as time evolves. The lower levels in the PPC hierarchy must handle the nuts-and-bolts challenge of converting the plans to action. The keys to a good AP module are to keep the focus on long-term planning (i.e., avoiding putting too many short-te |
235 | good AP module are to keep the focus on long-term planning (i.e., avoiding putting too many short-term control details in the model) and to provide links for consistency with other levels in the hierarchy. Some of the issues Chapter 16 Aggregate and Workforce Planning 589 related to consistency were discussed in Chapter 13. Here, we close with some general observations about the aggregate and workforce planning functions. 1. No single AP or WP module is right for every situation. As the e |
236 | and workforce planning functions. 1. No single AP or WP module is right for every situation. As the examples in this chapter show, aggregate and workforce planning can incorporate many different decision problems. A good AP or WP module is one that is tailored to address the speci?c issues faced by the ?rm. 2. Simplicity promotes understanding. Although it is desirable to address different issues in the AP/WP module, it is even more important to keep the model understandable. In general, these |
237 | s in the AP/WP module, it is even more important to keep the model understandable. In general, these modules are used to generate candidate production and workforce plans, which will be examined, combined, and altered manually before being published as “The Plan.” To generate a spectrum of plans (and explain them to others), the user must be able to trace changes in the model to changes in the plan. Because of this, it makes sense to start with as simple a formulation as possible. Additional de |
238 | lan. Because of this, it makes sense to start with as simple a formulation as possible. Additional detail (e.g., constraints) can be added later. 3. Linear programming is a useful AP/WP tool. The long planning horizon used for aggregate and workforce planning justi?es ignoring many production details; therefore, capacity checks, sales restrictions, and inventory balances can be expressed as linear constraints. As long as we are willing to approximate actual costs with linear functions, an LP so |
239 | ar constraints. As long as we are willing to approximate actual costs with linear functions, an LP solver is a very ef?cient method for solving many problems related to the AP and WP modules. Because we are working with speculative long-range data, it generally does not make sense to use anything more sophisticated than LP (e.g., nonlinear or integer programming) in most aggregate and workforce planning situations. 4. Robustness matters more than precision. No matter how accurate the data and h |
240 | planning situations. 4. Robustness matters more than precision. No matter how accurate the data and how sophisticated the model, the plan generated by the AP or WP module will never be followed exactly. The actual production sequence will be affected by unforeseen events that could not possibly have been factored into the module. This means that the mark of a good long-range production plan is that it enables us to do a reasonably good job even in the face of such contingencies. To ?nd such a p |
241 | hat it enables us to do a reasonably good job even in the face of such contingencies. To ?nd such a plan, the user of the AP module must be able to examine the consequences of various scenarios. This is another reason to keep the model reasonably simple. Appendix 16A Linear Programming Linear programming is a powerful mathematical tool for solving constrained optimization problems. The name derives from the fact that LP was ?rst applied to ?nd optimal schedules or “programs” of resource alloc |
242 | rives from the fact that LP was ?rst applied to ?nd optimal schedules or “programs” of resource allocation. Hence, although LP generally does involve using a computer program, it does not entail programming on the part of the user in the sense of writing code. In this appendix, we provide enough background to give the user of an LP package a basic idea of what the software is doing. Readers interested in more details should consult one of the many good texts on the subject (e.g., Eppen, Gould, |
243 | rested in more details should consult one of the many good texts on the subject (e.g., Eppen, Gould, and Schmidt 1988 for an application-oriented overview, Murty 1983 for more technical coverage). Formulation The ?rst step in using linear programming is to formulate a practical problem in mathematical terms. There are three basic choices we must make to do this: 1. Decision variables are quantities under our control. Typical examples for aggregate planning and workforce planning applications o |
244 | ties under our control. Typical examples for aggregate planning and workforce planning applications of LP are production quantities, number of workers to hire, and levels of inventory to hold. 2. Objective function is what we want to maximize or minimize. In most AP/WP applications, this is typically either to maximize pro?t or minimize cost. Beyond simply stating the objective, however, we must specify it in terms of the decision variables we have de?ned. 3. Constraints are restrictions on our |
245 | specify it in terms of the decision variables we have de?ned. 3. Constraints are restrictions on our choices of the decision variables. Typical examples for AP/WP applications include capacity constraints, raw materials limitations, restrictions on how fast we can add workers due to limitations on training capacity, and restrictions on physical ?ow (e.g., inventory levels as a direct result of how much we produce/procure and how much we sell). When one is formulating an LP, it is often useful |
246 | how much we produce/procure and how much we sell). When one is formulating an LP, it is often useful to try to specify the necessary inputs in the order in which they are listed. However, in realistic problems, one virtually never gets the “right” formulation in a single pass. The example in Section 16.4.2 illustrates some of the changes that may be required as a model evolves. To describe the process of formulating an LP, let us consider the problem presented in Table 16.2. We begin by selecti |
247 | rocess of formulating an LP, let us consider the problem presented in Table 16.2. We begin by selecting decision variables. Since there are only two products and because demand and capacity are assumed stationary over time, the only decisions to make concern how much of each product to produce per week. Thus, we let X 1 and X 2 represent the weekly production quantities of products 1 and 2, respectively. Next, we choose to maximize pro?t as our objective function. Since product 1 sells for $90 |
248 | ectively. Next, we choose to maximize pro?t as our objective function. Since product 1 sells for $90 but costs $45 in raw material, its net pro?t is $45 per unit.9 Similarly, product 2 sells for $100 but costs $40 in raw material, so its net unit pro?t is $60. Thus, weekly pro?t will be 45X 1 + 60X 2 - weekly labor costs - weekly overhead costs But since we assume that labor and overhead costs are not affected by the choice of X 1 and X 2 , we can use the following as our objective function for |
249 | re not affected by the choice of X 1 and X 2 , we can use the following as our objective function for the LP model: Maximize 45X 1 + 60X 2 Finally, we need to specify constraints. If we could produce as much of products 1 and 2 as we wanted, we could drive the above objective function, and hence weekly pro?t, to in?nity. This is not possible because of limitations on demand and capacity. 9 Note that we are neglecting labor and overhead costs in our estimates of unit pro?t. This is reasonable |
250 | e that we are neglecting labor and overhead costs in our estimates of unit pro?t. This is reasonable if these costs are not affected by the choice of production quantities, that is, if we won’t change the size of the workforce or the number of machines in the shop. 590 Chapter 16 591 Aggregate and Workforce Planning The demand constraints are easy. Since we can sell at most 100 units per week of product 1 and 50 units per week of product 2, our decision variables X 1 and X 2 must satisfy |
251 | week of product 1 and 50 units per week of product 2, our decision variables X 1 and X 2 must satisfy X 1 = 100 X 2 = 50 The capacity constraints are a little more work. Since there are four machines, which run at most 2,400 minutes per week, we must ensure that our production quantities do not violate this constraint on each machine. Consider workstation A. Each unit of product 1 we produce requires 15 minutes on this workstation, while each unit of product 2 we produce requires 10 minutes. He |
252 | uires 15 minutes on this workstation, while each unit of product 2 we produce requires 10 minutes. Hence, the total number of minutes of time required on workstation A to produce X 1 units of product 1 and X 2 units of product 2 is10 15X 1 + 10X 2 so the capacity constraint for workstation A is 15X 1 + 10X 2 = 2,400 Proceeding analogously for workstations B, C, and D, we can write the other capacity constraints as follows: 15X 1 + 35X 2 = 2,400 workstation B 15X 1 + 5X 2 = 2,400 workstation |
253 | city constraints as follows: 15X 1 + 35X 2 = 2,400 workstation B 15X 1 + 5X 2 = 2,400 workstation C 25X 1 + 14X 2 = 2,400 workstation D We have now completely de?ned the following LP model of our optimization problem: Maximize 45X 1 + 60X 2 (16.107) X 1 = 100 (16.108) X 2 = 50 (16.109) 15X 1 + 10X 2 = 2,400 (16.110) 15X 1 + 35X 2 = 2,400 (16.111) 15X 1 + 5X 2 = 2,400 (16.112) 25X 1 + 14X 2 = 2,400 (16.113) Subject to: Some LP packages allow the user to enter the problem in |
254 | 25X 1 + 14X 2 = 2,400 (16.113) Subject to: Some LP packages allow the user to enter the problem in a form almost identical to that shown in formulation (16.107)–(16.113). Spreadsheet programs generally require the decision variables to be entered into cells and the constraints speci?ed in terms of these cells. More sophisticated LP solvers allow the user to specify blocks of similar constraints in a concise form, which can substantially reduce modeling time for large problems. Finally, with |
255 | ts in a concise form, which can substantially reduce modeling time for large problems. Finally, with regard to formulation, we point out that we have not stated explicitly the constraints that X 1 and X 2 be non-negative. Of course, they must be, since negative production quantities make no sense. In many LP packages, decision variables are assumed to be non-negative unless 10 Note that this constraint does not address such detailed considerations as setup times that depend on the sequence of p |
256 | straint does not address such detailed considerations as setup times that depend on the sequence of products run on workstation A or whether full utilization of workstation A is possible given the WIP in the system. But as we discussed in Chapter 13, these issues are addressed at a lower level in the production planning and control hierarchy (e.g., in the sequencing and scheduling module). 592 Part III Principles in Practice the user speci?es otherwise. In other packages, the user must inc |
257 | Part III Principles in Practice the user speci?es otherwise. In other packages, the user must include the non-negativity constraints explicitly. This is something to beware of when using LP software. Solution To get a general idea of how an LP package works, let us consider the above formulation from a mathematical perspective. First, note that any pair of X 1 and X 2 that satis?es 15X 1 + 35X 2 = 2,400 workstation B 15X 1 + 10X 2 = 2,400 workstation A 15X 1 + 5X 2 = 2,400 workstation |
258 | 35X 2 = 2,400 workstation B 15X 1 + 10X 2 = 2,400 workstation A 15X 1 + 5X 2 = 2,400 workstation C will also satisfy because these differ only by having smaller coef?cients for X 2 . This means that the constraints for workstations A and C are redundant. Leaving them out will not affect the solution. In general, it does not hurt anything to have redundant constraints in an LP formulation. But to make our graphical illustration of how LP works as clear as possible, we will omit constraint |
259 | But to make our graphical illustration of how LP works as clear as possible, we will omit constraints (16.110) and (16.112) from here on. Figure 16.15 illustrates problem (16.107)–(16.113) in graphical form, where X 1 is plotted on the horizontal axis and X 2 is plotted on the vertical axis. The shaded area is the feasible region, consisting of all the pairs of X 1 and X 2 that satisfy the constraints. For instance, the demand constraints (16.108) and (16.109) simply state that X 1 cannot be l |
260 | traints. For instance, the demand constraints (16.108) and (16.109) simply state that X 1 cannot be larger than 100, and X 2 cannot be larger than 50. The capacity constraints are graphed by noting that, with a bit of algebra, we can write constraints (16.111) and (16.113) as X2 = - X2 = - X1 + 2,400 = -0.429X 1 + 68.57 35 (16.114) X1 + 2,400 = -1.786X 1 + 171.43 14 (16.115) 15 35 25 14 If we replace the inequalities with equality signs in equations (16.114) and (16.115), then |
261 | 25 14 If we replace the inequalities with equality signs in equations (16.114) and (16.115), then these are simply equations of straight lines. Figure 16.15 plots these lines. The set of X 1 and X 2 points that satisfy these constraints includes all the points lying below both of these lines. The points marked by the shaded area are those satisfying all the demand, capacity, and non-negativity constraints. This type of feasible region de?ned by linear constraints is known as a polyhedron. N |
262 | ty constraints. This type of feasible region de?ned by linear constraints is known as a polyhedron. Now that we have characterized the feasible region, we turn to the objective. Let Z represent the value of the objective (i.e., net pro?t achieved by producing quantities X 1 and X 2 ). From objective Figure 16.15 25X1 + 14X2 = 2,400 140.00 Feasible region for LP example. X1 = 100 120.00 15X1 + 35X2 = 2,400 X2 100.00 80.00 60.00 X2 = 50 40.00 Feasible region 20.00 0.00 0 50 100 X1 |
263 | X1 + 35X2 = 2,400 X2 100.00 80.00 60.00 X2 = 50 40.00 Feasible region 20.00 0.00 0 50 100 X1 150 Chapter 16 Figure 16.16 593 Aggregate and Workforce Planning 140.00 Solution to LP example. 120.00 Z = 5,557.94 X2 100.00 Z = 7,000 80.00 Optimal solution (75.79, 36.09) 60.00 40.00 Feasible region 20.00 0.00 0 Z = 3,000 50 100 150 X1 (16.107), X 1 and X 2 are related to Z by 45X 1 + 60X 2 = Z We can write this in the usual form for a straight line as Z Z -45 X1 + = |
264 | Z by 45X 1 + 60X 2 = Z We can write this in the usual form for a straight line as Z Z -45 X1 + = -0.75X 1 + X2 = 60 60 60 (16.116) (16.117) Figure 16.16 illustrates equation (16.117) for Z = 3,000, 5,557.94, and 7,000. Notice that for Z = 3,000, the line passes through the feasible region, leaving some points above it. Hence, we can feasibly increase pro?t (that is, Z ). For Z = 7,000 the line lies entirely above the feasible region. Hence, Z = 7,000 is not feasible. For Z = 5,557.94, t |
265 | e line lies entirely above the feasible region. Hence, Z = 7,000 is not feasible. For Z = 5,557.94, the objective function just touches the feasible region at a single point, the point (X 1 = 75.79, X 2 = 36.09). This is the optimal solution. Values of Z above 5,557.74 are infeasible, values below it are suboptimal. The optimal product mix, therefore, is to produce 75.79 (or 75, rounded to an integer value) units of product 1 and 36.09 (rounded to 36) units of product 2. We can think of ?nding |
266 | teger value) units of product 1 and 36.09 (rounded to 36) units of product 2. We can think of ?nding the solution to an LP by steadily increasing the objective value (Z ), moving the objective function up and to the right, until it is just about to leave the feasible region. Because the feasible region is a polyhedron whose sides are made up of linear constraints, the last point of contact between the objective function and the feasible region will be a corner, or extreme point, of the feasible |
267 | en the objective function and the feasible region will be a corner, or extreme point, of the feasible region.11 This observation allows the optimization algorithm to ignore the in?nitely many points inside the feasible region and search for a solution among the ?nite set of extreme points. The simplex algorithm, developed in the 1940s and still widely used, works in just this way, proceeding around the outside of the polyhedron, trying extreme points until an optimal one is found. Other, more m |
268 | und the outside of the polyhedron, trying extreme points until an optimal one is found. Other, more modern algorithms use different schemes to ?nd the optimal point, but will still converge to an extreme-point solution. Sensitivity Analysis The fact that the optimal solution to an LP lies at an extreme point enables us to perform useful sensitivity analysis on the optimal solution. The principal sensitivity information available to us falls into the following three categories. 1. Coef?cients i |
269 | l sensitivity information available to us falls into the following three categories. 1. Coef?cients in the objective function. For instance, if we were to change the unit pro?t for product 1 from $45 to $60, then the equation for the objective function would change from 11 Actually, it is possible that the optimal objective function lies right along a ?at spot connecting two extreme points of the polyhedron. When this occurs, there are many pairs of X 1 and X 2 that attain the optimal value of |
270 | e polyhedron. When this occurs, there are many pairs of X 1 and X 2 that attain the optimal value of Z , and the solution is called degenerate. Even in this case, however, an extreme point (actually, at least two extreme points) will be among the optimal solutions. 594 Part III Figure 16.17 Principles in Practice 140.00 Effect of changing objective coef?cients in LP example. 120.00 45X1 + 60X2 = 5,557.94 X2 100.00 60X1 + 60X2 = 6,712.80 80.00 60.00 40.00 Feasible region 20.00 0.00 |
271 | + 60X2 = 5,557.94 X2 100.00 60X1 + 60X2 = 6,712.80 80.00 60.00 40.00 Feasible region 20.00 0.00 0 50 100 150 X1 equation (16.117) to 60 Z Z X2 = - X1 + = -X 1 + 60 60 60 (16.118) so the slope changes from -0.75 to -1; that is, it gets steeper. Figure 16.17 illustrates the effect. Under this change, the optimal solution remains (X 1 = 75.79, X 2 = 36.09). Note, however that while the decision variables remain the same, the objective function does not. When the unit pro?t for pro |
272 | the decision variables remain the same, the objective function does not. When the unit pro?t for product 1 increases to $60, the pro?t becomes 60(75.79) + 60(36.09) = $6,712.80 The optimal decision variables remain unchanged until the coef?cient of X 1 in the objective function reaches 107.14. When this happens, the slope becomes so steep that the point where the objective function just touches the feasible region moves to the extreme point (X 1 = 96, X 2 = 0). Geometrically, the objective fun |
273 | the feasible region moves to the extreme point (X 1 = 96, X 2 = 0). Geometrically, the objective function “rocked around” to a new extreme point. Economically, the pro?t from product 1 reached a point where it became optimal to produce all product 1 and no product 2. In general, LP packages will report a range for each coef?cient in the objective function for which the optimal solution (in terms of the decision variables) remains unchanged. Note that these ranges are valid only for one-at-a-ti |
274 | s of the decision variables) remains unchanged. Note that these ranges are valid only for one-at-a-time changes. If two or more coef?cients are changed, the effect is more dif?cult to characterize. One has to rerun the model with multiple coef?cient changes to get a feel for their effect. 2. Coef?cients in the constraints. If the number of minutes required on workstation B by product 1 is changed from 15 to 20, then the equation de?ned by the capacity constraint for workstation B changes from e |
275 | d from 15 to 20, then the equation de?ned by the capacity constraint for workstation B changes from equation (16.114) to X2 = - 20 35 X1 + 2,400 = -0.571X 1 + 68.57 35 (16.119) so the slope changes from -0.429 to -0.571; again, it becomes steeper. In a manner analogous to that described above for coef?cients in the objective function, LP packages can determine how much a given coef?cient can change before it ceases to de?ne the optimal extreme point. However, because changing the coe |
276 | ent can change before it ceases to de?ne the optimal extreme point. However, because changing the coef?cients in the constraints moves the extreme points themselves, the optimal decision variables will also change. For this reason, most LP packages do not report this sensitivity data, but rather make use of this product as part of a parametric programming option to quickly generate new solutions for speci?ed changes in the constraint coef?cients. 3. Right-hand side coef?cients. Probably the mos |
277 | for speci?ed changes in the constraint coef?cients. 3. Right-hand side coef?cients. Probably the most useful sensitivity information provided by LP models is for the right-hand side variables in the constraints. For instance, in formulation (16.107)–(16.113), if we run 100 minutes of overtime per week on machine B, then its right-hand Chapter 16 595 Aggregate and Workforce Planning side will increase from 2,400 to 2,500. Since this is something we might want to consider, we would like to |
278 | ll increase from 2,400 to 2,500. Since this is something we might want to consider, we would like to be able to determine its effect. We do this differently for two types of constraints: a. Slack constraints are constraints that do not de?ne the optimal extreme point. The capacity constraints for workstations A and C are slack, since we determined right at the outset that they could not affect the solution. The constraint X 2 = 50 is also slack, as can be seen in Figures 16.15 and 16.16, althou |
279 | he solution. The constraint X 2 = 50 is also slack, as can be seen in Figures 16.15 and 16.16, although we did not know this until we solved the problem. Small changes in slack constraints do not change the optimal decision variables or objective value at all. If we change the demand constraint on product 2 to X 2 = 49, it still won’t affect the optimal solution. Indeed, not until we reduce the constraint to X 2 = 36.09 will it have any effect. Likewise, increasing the right-hand side of this c |
280 | constraint to X 2 = 36.09 will it have any effect. Likewise, increasing the right-hand side of this constraint (above 50) will not affect the solution. Thus, for a slack constraint, the LP package tells us how far we can vary the right-hand side without changing the solution. These are referred to as the allowable increase and allowable decrease of the right-hand side coef?cients. b. Tight constraints are constraints that de?ne the optimal extreme point. Changing them changes the extreme point, |
281 | raints are constraints that de?ne the optimal extreme point. Changing them changes the extreme point, and hence the optimal solution. For instance, the constraint that the number of hours per week on workstation B not exceed 2,400, that is, 15X 1 + 35X 2 = 2,400 is a tight constraint in Figures 16.15 and 16.16. If we increase or decrease the right-hand side, the optimal solution will change. However, if the changes are small enough, then the optimal extreme point will still be de?ned by the sam |
282 | ever, if the changes are small enough, then the optimal extreme point will still be de?ned by the same constraints (i.e., the time on workstations B and D). Because of this, we are able to compute the following: Shadow prices are the amount by which the objective increases per unit increase in the right-hand side of a constraint. Since slack constraints do not affect the optimal solution, changing their right-hand sides has no effect, and hence their shadow prices are always zero. Tight constra |
283 | ng their right-hand sides has no effect, and hence their shadow prices are always zero. Tight constraints, however, generally have nonzero shadow prices. For instance, the shadow price for the constraint on workstation B is 1.31. (Any LP solver will automatically compute this value.) This means that the objective will increase by $1.31 for every extra minute per week on the workstation. So if we can work 2,500 minutes per week on workstation B, instead of 2,400, the objective will increase by 1 |
284 | can work 2,500 minutes per week on workstation B, instead of 2,400, the objective will increase by 100 × 1.31 = $131. Maximum allowable increase/decrease gives the range over which the shadow prices are valid. If we change a right-hand side by more than the maximum allowable increase or decrease, then the set of constraints that de?ne the optimal extreme point may change, and hence the shadow price may also change. For example, as Figure 16.18 shows, if we increase the right-hand side of the c |
285 | rice may also change. For example, as Figure 16.18 shows, if we increase the right-hand side of the constraint on workstation B from 2,400 to 2,770, the constraint moves to the very edge of the feasible region de?ned by 25X 1 + 14X 2 = 2,400 (machine D) and X 2 = 50. Any further increases in the right-hand side will cause this constraint to become slack. Hence, the shadow price is $1.31 up to a maximum allowable Figure 16.18 25X1 + 14X2 = 2,400 140.00 X1 = 100 15X1 + 35X2 = 2,770 120.00 |
286 | maximum allowable Figure 16.18 25X1 + 14X2 = 2,400 140.00 X1 = 100 15X1 + 35X2 = 2,770 120.00 15X1 + 35X2 = 2,400 100.00 X2 Feasible region when RHS of constraint of workstation B is increased to 2,770. 80.00 60.00 X2 = 50 40.00 Feasible region 20.00 0.00 0 50 100 X1 150 596 Part III Principles in Practice increase of 370 (that is, 2,770 - 2,400). In this example, the shadow price is zero for changes above the maximum allowable increase. This is not always the case, howeve |
287 | w price is zero for changes above the maximum allowable increase. This is not always the case, however, so in general we must resolve the LP to determine the shadow prices beyond the maximum allowable increase or decrease. Study Questions 1. Although the technology for solving aggregate planning models (linear programming) is well established and AP modules are widely available in commercial systems (e.g., MRP II systems), aggregate planning does not occupy a central place in the planning func |
288 | stems (e.g., MRP II systems), aggregate planning does not occupy a central place in the planning function of many firms. Why do you think this is true? What difficulties in modeling, interpreting, and implementing AP models might be contributing to this? 2. Why does it make sense to consider workforce planning and aggregate planning simultaneously in many situations? 3. What is the difference between a chase production plan and a level production plan, with respect to the amount of inventory |
289 | between a chase production plan and a level production plan, with respect to the amount of inventory carried and the fluctuation in output quantity over time? How do the production plans generated by an LP model relate to these two types of plan? 4. In a basic LP formulation of the product mix aggregate planning problem, what information is provided by the following? (a) The optimal decision variables. (b) The optimal objective function. (c) Identification of which constraints are tight and wh |
290 | riables. (b) The optimal objective function. (c) Identification of which constraints are tight and which are slack. (d) Shadow prices for the right-hand sides of the constraints. Problems 1. Suppose a plant can supplement its capacity by subcontracting part of or all the production of certain parts. (a) Show how to modify LP (16.28)-(16.32) to include this option, where we define Vit = kit = units of product i received from a subcontractor in period t premium paid for subcontracting produ |
291 | = units of product i received from a subcontractor in period t premium paid for subcontracting product i in period t (i.e., cost above variable cost of making it in-house) ?it = minimum amount of product i that must be purchased in period t = maximum amount of product i that can be purchased in period t (e.g., specified as part of long-term contract with supplier) Vit (e.g., due to capacity constraints on supplier, as specified in long-term contract) (b) How would you modify the formul |
292 | city constraints on supplier, as specified in long-term contract) (b) How would you modify the formulation in part (a) if the contract with a supplier stipulated only that total purchases of product i over the time horizon must be at least ? ? (c) How would you modify the formulation in part (a) if the supplier contract, instead of specifying ? and ii, stipulated that the firm specify a base amount of product i, to be purchased every month, and that the maximum purchase in a given month can exc |
293 | unt of product i, to be purchased every month, and that the maximum purchase in a given month can exceed the base amount by no more than 20 percent? (d) What role might models like those in parts (a) to (c) play in the process of negotiating contracts with suppliers? 2. Show how to modify LP (16.49)-(16.54) to represent the case where overtime on all the workstations must be scheduled simultaneously (i.e., if one resource runs overtime, all resources run overtime). Describe how you would handl |
294 | ously (i.e., if one resource runs overtime, all resources run overtime). Describe how you would handle the case where, in general, different workstations can have different amounts of overtime, but two workstations, say A and B, must always be scheduled for overtime together. Chapter 16 597 Aggregate and Workforce Planning 3. Show how to modify LP (16.61)–(16.67) of the workforce planning problem to accommodate multiple products. 4. You have just been made corporate vice president in charg |
295 | roblem to accommodate multiple products. 4. You have just been made corporate vice president in charge of manufacturing for an automotive components company and are directly in charge of assigning products to plants. Among many other products, the ?rm makes automotive batteries in three grades: heavyduty, standard, and economy. The unit net pro?ts and maximum daily demand for these products are given in the ?rst table below. The ?rm has three locations where the batteries can be produced. The m |
296 | given in the ?rst table below. The ?rm has three locations where the batteries can be produced. The maximum assembly capacities, for any mix of battery grades, are given in the second table below. The number of batteries that can be produced at a location is limited by the amount of suitably formulated lead the location can produce. The lead requirements for each grade of battery and the maximum lead production for each location are also given in the following tables. Product Unit Pro?t ($/ba |
297 | lead production for each location are also given in the following tables. Product Unit Pro?t ($/battery) Maximum Demand (batteries/day) Lead Requirements (lbs/battery) Heavy-duty Standard Economy 12 10 7 700 900 450 21 17 14 Plant Location Assembly Capacity (batteries/day) Maximum Lead Production (lbs/day) 1 2 3 550 750 225 10,000 7,000 4,200 (a) Formulate a linear program that allocates production of the three grades among the three locations in a manner that maximizes pro?t. ( |
298 | llocates production of the three grades among the three locations in a manner that maximizes pro?t. (b) Suppose company policy requires that the fraction of capacity (units scheduled/ assembly capacity) be the same at all locations. Show how to modify your LP to incorporate this constraint. (c) Suppose company policy dictates that at least 50 percent of the batteries produced must be heavy-duty. Show how to modify your LP to incorporate this constraint. 5. Youohimga, Inc., makes a variety of co |
299 | Show how to modify your LP to incorporate this constraint. 5. Youohimga, Inc., makes a variety of computer storage devices, which can be divided into two main families that we call A and B. All devices in family A have the same routing and similar processing requirements at each workstation; similarly for family B. There are a total of 10 machines used to produce the two families, where the routings for A and B have some workstations in common (i.e., shared) but also contain unique (unshared) |