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 ? 0 (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 = 0Optimal 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 * 42O2 + 48C2? 30 * 30)3 + 35C3? 35 * O1 + O2 + O3=1 * C1 + C2 + C3=1 * O1, O2, O3, C1, C2, C3? 0 B) | Cabinet Maker 1| Cabinet Maker 2| Cabinet Maker 3| Oak| 0. 271| 0| 0. 729| Cherry| 0| 0.

625| 0. 375| Therefore, to complete both projects is assigning 27. 1% of the oak cabinets to Cabinet Maker 1 and 72. 9% to Cabinet Maker 2. For the Cherry, 62.

5% should be assigned to Cabinet Maker 2 and 37. 5% to Cabinet Maker 3. Total cost of completing both projects = \$ 3672. 500 C) Cabinetmaker 1 has a slack of 26. 458 hours and a dual price of 0.

This increases the right hand side of constraint 1 will not change the value of the optimal solution. D) The dual price for constraint 2 is 1. 75. The current Right-Hand-Side is 30 and allowable increase is 11. 143. The upper limit on Right-Hand-Side range is 41.

143. This means for each extra hour of time for Cabinet Maker 2, the total cost will decrease by \$1. 75/hour, to a maximum of 41. 143 hours. E) If cabinetmaker 2 reduces his cost to \$38 per hour, the new objective function coefficients for O2 and C2 would be: 42(38) = 1596 and 48(38) = 1824. The optimal solution does not change but the total cost decreases to \$3552.

0. QUESTION THREE Decision Variables Let, * S =t the amount invested in stocks * B = the amount invested in bonds * M = the amount invested in mutual funds * C = the amount invested in Cash * R = the amount of Risk * AR= annual return Objective Functions * R = 0. 8S + 0. 2B +0. 3M + 0.

0C * AR = 0. 1S + 0. 03B + 0. 04M + 0. 01C Constraints * 10 ? C ? 30 * M ? B * S ? 75 * S, B, M, C ? 0 A) R = 0. 8S + 0.

2B + 0. 3M + 0C R = 0. 8(0. 409) + 0.

2(0. 145) + 0. 3(0. 145) + 0(0. 3) R = 0. 3997 Therefore the optimal allocation for this scenario is as follows; S = 40.

%, B = 14. 5%, M = 14. 5%, C = 30% which will create the optimal amount of risk, 0. 3997, or 0. 4 rounded. AR = 0.

1S + 0. 03B + 0. 04M + 0. 01C AR = 0. 1(0.

409) + 0. 03(0. 145) + 0. 04(0.

145) + 0. 01(0. 3) AR = 0. 05405 Therefore, the annual rate of return from the optimal solution is 0.

05405 or 5. 4 % and total risk is 0. 3997. B) R = 0.

8S + 0. 2B + 0. 3M +0C R = 0. 8(0) + 0.

2(0. 36) + 0. 3(0. 36) + 0(0.

28) R = 0. 18 Therefore the optimal allocation for this scenario is as follows; S = 0% B = 36%, M = 36%, C = 28% which will create the optimal amount of risk, 0. 18.AR = 0.

1S + 0. 03B + 0. 04M + 0. 01C AR = 0. 1(0) + 0.

03(0. 36) + 0. 04(0. 36) + 0. 01(0.

28) AR = 0. 0028 Therefore the annual rate of return for the optimal solution is 0. 0028 or 2. 8 % and total risk is 0. 18 C) R = 0. 8S + 0.

2B + 0. 3M +0C R = 0. 8(0. 75) + 0. 2(0) + 0.

3(0. 15) + 0(0. 10) R = 0. 645 Therefore the optimal allocation for this scenario is as follows; S =75% B = 0%, M = 15%, C = 10% which will create a risk value just under the 0.

7 maximum risk value. AR = 0. 1S + 0. 03B + 0. 04M + 0. 01C AR = 0.

1(0. 75) + 0. 03(0) + 0. 04(0.

15) + 0. 01(0. 10) AR = 0. 82 Therefore the annual rate of return for the optimal solution is 0. 082 or 8.

2 % and total risk is 0. 645. QUESTION FOUR Decision Variables * Xbr – pounds of Brazilian Natural beans used to produce Regular blend * Xgr – pounds of Gourmet beans used to produce Regular blend * Xcr – pounds of Colombian Mild beans used to produce Regular blend * Xbd – pounds of Brazilian Natural beans used to produce DeCaf blend * Xdg – pounds of Gourmet beans used to produce DeCaf blend * Xcd – pounds of Colombian Mild beans used to produce DeCaf blendObjective Function * MAX 2. 75Xbr + 2.

15Xgr + 1. 35Xcr + 3. 3Xbd + 2. 7Xgd + 1. 9Xcd Contraints * Xbr + Xgr + Xcr =1000 * Xbd + Xgd + Xcd =500 * 0.

6Xbr – 0. 4Xgr – 0. 4Xcr<0 * -0. 25Xbr–0. 25Xgr + 0.

75Xcr>0 * -0. 2Xbd + 0. 8Xgd – 0. 2Xcd>0 * -0. 6Xbd – 0.

6Xgd + 0. 4Xcd>0 Optimal Solution * Xbr = 400 * Xgr = 350 * Xcr = 250 * Xbd = 100 * Xgd = 100 * Xcd = 300 This is how many pounds of each type that needs to be produced to optimal profit contribution. Optimal Value * MAX 2. 75Xbr + 2. 15Xgr + 1. 35Xcr + 3.

3Xbd + 2. 7Xgd + 1. 9Xcd = 2. 75(400) + 2. 15(350) + 1. 5(250) + 3.

3(100) + 2. 7(100) + 1. 9(300) =3360 Therefore, the optimal solution creates a profit contribution of \$3360. QUESTION FIVE Decision Variables Let, * X1 = number of units produced in quarter 1 * X2 = number of units produced in quarter 2 * X3 = number of units produced in quarter 3 * X4 = number of units produced in quarter 4 * Y1 = number of units in ending inventory for quarter 1 * Y2 = number of units in ending inventory for quarter 2 * Y3 = number of units in ending inventory for quarter 3 * Y4 = number of units in ending inventory for quarter 4Objective Function * MIN 10000X1 + 11000X2 + 12100X3 + 13310X4 + 250Y1 + 250Y2 + 300Y3 + 300Y3 Constraints * X1-Y1=1900 * X2+Y1-Y2=4000 * X3+Y2-Y3=3000 * X4+Y3-Y4=1500 * X1<4000 * X2<3000 * X3<2000 * X4<4000 * Y4>500 * X1, X2, X3, X4, Y1, Y2, Y3, Y4 ? 0 A) * MIN=10000X1 + 11000X2 + 12100X3 + 13310X4 + 250Y1 + 250Y2 + 300Y3 + 300Y3 * MIN=10000(4000) + 11000(3000) + 12100(2000) + 13310(1900) + 250(2100) + 250(1100) + 300(100) + 300(500) * MIN=40000000 + 33000000 + 24200000 + 25289000 + 525000 + 275000 + 30000 + 150000 * MIN=123469000Therefore, the optimal solution for production to minimize costs will cost \$12369000. The production in units for each variable are as follows; X1 = 4000, X2 = 3000, X3 = 2000, X4 = 1900, Y1 = 2100, Y2 = 1100, Y3 = 100, Y4 = 500.

Quarter| Production| Ending Inventory| 1| 4000| 2100| 2| 3000| 1100| 3| 2000| 100| 4| 1900| 500| B) Quarter| Dual Price| Cost| 1| 12510| 10000| 2| 12760| 11000| 3| 13010| 12100| 4| No Upper Limit| No Upper Limit| C)