Qdm Assignment
Quantitative Decision Making Models
Assignment # 2
QUESTION ONE
Decision Variables Let, * X1 = number of full-time tellers * Y1 = number of part-time tellers starting at 9 a.m. (leaving at 1 p.m.) * Y2 = number of part-time tellers starting at 10 a.m. (leaving at 2 p.m.) * Y3 = numbers of part-time tellers starting at 11 a.m. (leaving at 3 p.m.) * Y4 = number of part-time tellers starting at noon ( leaving at 4 p.m.) * Y5 = number of part-time tellers starting at 1 p.m. (leaving at 5 p.m.)
Objective Function * MIN 90X1 + 28(Y1 + Y2 + Y3 + Y4 + Y5)
Contraints * X1 + Y1 ≥ 10 (9am-10am) * 0.5X1 + Y1 + Y2 ≥ 12 (10am – 11am) * 0.5X1 + Y1 + Y2 + Y3 ≥ 14 (11am – noon) * X1 + Y1 + Y2 +Y3 + Y4 ≥ 16 (noon – 1pm) * X1 + Y1 + Y2 + Y3 + Y4 + Y5 ≥ 18 (1pm - 2pm) * X1 + Y3 + Y4 + Y5 ≥ 17 (2pm – 3pm) * X1 + Y4 + Y5 ≥ 15 (3pm – 4pm) * X1 + Y5 ≥ 10 (4pm – 5pm) * X < 12 * X1, Y1, Y2, Y3, Y4, Y5 ≥ 0
Part-time workers cannot work more than 50% of the total hours required each day. Therefore, 4(Y1+Y2+Y3+Y4+Y5) ≤ 0.50(10+12+14+16+18+17+15+10)
Optimal Solution * X1 = 10 * Y1 = 0 * Y2 = 7 * Y3 = 2 * Y4 = 5 * Y5 = 0
Optimal Value
$ 1,292 is the optimal value to minimize the total cost of employees working.
QUESTION TWO
A)
Decision Variables
Let,
* O1 = percentage of Oak cabinets assigned to cabinetmaker 1 * O2 = percentage of Oak cabinets assigned to cabinetmaker 2 * O3 = percentage of Oak cabinets assigned to cabinetmaker 3 * C1 = percentage of Cherry cabinets assigned to cabinetmaker 1 * C2 = percentage of Cherry cabinets assigned to cabinetmaker 2 * C3 = percentage of Cherry cabinets assigned to cabinetmaker 3
Objective Function * Min 1800O1 + 1764O2 + 1650O3 + 2160C1 + 2016C2 + 1925C3
Contraints
* 50O1 + 60C1 ≤ 40 *
Assignment # 2
QUESTION ONE
Decision Variables Let, * X1 = number of full-time tellers * Y1 = number of part-time tellers starting at 9 a.m. (leaving at 1 p.m.) * Y2 = number of part-time tellers starting at 10 a.m. (leaving at 2 p.m.) * Y3 = numbers of part-time tellers starting at 11 a.m. (leaving at 3 p.m.) * Y4 = number of part-time tellers starting at noon ( leaving at 4 p.m.) * Y5 = number of part-time tellers starting at 1 p.m. (leaving at 5 p.m.)
Objective Function * MIN 90X1 + 28(Y1 + Y2 + Y3 + Y4 + Y5)
Contraints * X1 + Y1 ≥ 10 (9am-10am) * 0.5X1 + Y1 + Y2 ≥ 12 (10am – 11am) * 0.5X1 + Y1 + Y2 + Y3 ≥ 14 (11am – noon) * X1 + Y1 + Y2 +Y3 + Y4 ≥ 16 (noon – 1pm) * X1 + Y1 + Y2 + Y3 + Y4 + Y5 ≥ 18 (1pm - 2pm) * X1 + Y3 + Y4 + Y5 ≥ 17 (2pm – 3pm) * X1 + Y4 + Y5 ≥ 15 (3pm – 4pm) * X1 + Y5 ≥ 10 (4pm – 5pm) * X < 12 * X1, Y1, Y2, Y3, Y4, Y5 ≥ 0
Part-time workers cannot work more than 50% of the total hours required each day. Therefore, 4(Y1+Y2+Y3+Y4+Y5) ≤ 0.50(10+12+14+16+18+17+15+10)
Optimal Solution * X1 = 10 * Y1 = 0 * Y2 = 7 * Y3 = 2 * Y4 = 5 * Y5 = 0
Optimal Value
$ 1,292 is the optimal value to minimize the total cost of employees working.
QUESTION TWO
A)
Decision Variables
Let,
* O1 = percentage of Oak cabinets assigned to cabinetmaker 1 * O2 = percentage of Oak cabinets assigned to cabinetmaker 2 * O3 = percentage of Oak cabinets assigned to cabinetmaker 3 * C1 = percentage of Cherry cabinets assigned to cabinetmaker 1 * C2 = percentage of Cherry cabinets assigned to cabinetmaker 2 * C3 = percentage of Cherry cabinets assigned to cabinetmaker 3
Objective Function * Min 1800O1 + 1764O2 + 1650O3 + 2160C1 + 2016C2 + 1925C3
Contraints
* 50O1 + 60C1 ≤ 40 *