1. Analyzing Cost-Time Trade-Offs There are always cost-time trade-offs in project management. – You can completing a project early by hiring more workers or running extra shifts. – There are often penalties if projects extend beyond some specific date, and a bonus may be provided for early completion. Crashing a project means expediting some activities to reduce overall project completion time and total project costs.

2. Cost to Crash To assess the benefit of crashing certain activities, either from a cost or a schedule perspective, the project manager needs to know the following times and costs. Normal time (NT) is the time necessary to complete and activity under normal conditions. Normal cost (NC) is the activity cost associated with the normal time. Crash time (CT) is the shortest possible time to complete an activity. Crash cost (CC) is the activity cost associated with the crash time.

3. Cost to Crash per Period The Cost to Crash per Time Period = CC − NC NT − CT Crash Cost − Normal Cost Normal Time − Crash Time

4. Linear cost assumption 8000 — 7000 — 6000 — 5000 — 4000 — 3000 — 0 — Direct cost (dollars) |||||| 567891011 Time (weeks) Crash cost (CC) Normal cost (NC) (Crash time)(Normal time) Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks 5200 Cost-Time Relationships in Cost Analysis

5. The objective of cost analysis is to determine the project schedule that minimizes total project costs. A minimum-cost schedule is determined by starting with the normal time schedule and crashing activities along the critical path in such a way that the costs of crashing do not exceed the savings in indirect and penalty costs. Minimizing Costs

6. Use these steps to determine the minimum cost schedule: 1.Determine the project’s critical path(s). 2.Find the activity or activities on the critical path(s) with the lowest cost of crashing per time unit. 3.Reduce the time for this activity until… a.It cannot be further reduced or b.Until another path becomes critical, or c.The increase in direct costs exceeds the savings that result from shortening the project (which lowers indirect costs & penalties). 4.Repeat this procedure until the increase in direct costs is larger than the savings generated by shortening the project. Minimum Cost Schedule

7. Regular & Crash Times & Cost to Crash ActivityRegular TimeCrash Time Total Cost to Crash Cost Per Unit of Crash Time $ A*21$1,000 $ B*42$ 500 $ C*22 C $ D*11 C $ E31$ 200 $ F*22 C $ G*11 C $ H* 128$ 800 $ I*85$1,200 $ J66 C $ K53$ 600 $ L* TOTAL * Activities on critical path 33 C $4,300

8. START A2A2 B4B4 C2C2 D1D1 F2F2 G1G1 H 12 J6J6 I8I8 K5K5 L3L3 END E3E3 Network for a Software Purchasing Project using Normal Times

9. IF ALL ACTIVITIES ARE CRASHED START A1A1 B2B2 C2C2 D1D1 F2F2 G1G1 H8H8 J6J6 I5I5 K3K3 L3L3 END E1E1 (1,5,4) (8,8,0)(6,6,0)(5,5,0)(0,0,0)(1,1,0)(3,3,0) (26) (23,23,0) (17,18,1) (17,17,0) (9,20,11) (9,9,0) *What is the minimum amount of time to complete the project? *What is the additional cost involved in achieving this reduction in time?

10. START A2A2 B4B4 C2C2 D1D1 F2F2 G1G1 H 12 J6J6 I8I8 K5K5 L3L3 END E3E3 What is the minimum cost to achieve minimum time schedule for project ActivityRegular TimeCrash Time Total Cost to Crash Cost Per Unit of Crash Time $ A21$1,000 $ B42$ 500$ 250 $ C22 CC $ D11 CC $ E31$ 200$ 100 $ F22 CC $ G11 CC $ H 128$ 800$ 200 $ I85$1,200$ 400 $ J66 CC $ K53$ 600$ 300 $ L33 C $4,300 C

11. START A2A2 B4B4 C2C2 D1D1 F2F2 G1G1 H 12 J6J6 I8I8 K5K5 L3L3 END E3E3 What is the minimum cost to achieve minimum time schedule for project ActivityRegular TimeCrash Time Total Cost to Crash Cost Per Unit of Crash Time $ A21$1,000 $ B42$ 500$ 250 $ C22 CC 11 CC $ E31$ 200$ 100 $ F22 CC $ G11 CC $ H 128$ 800$ 200 $ I85$1,200$ 400 $ J66 CC $ K53$ 600$ 300 $ L33 C $4,300 C

12. START A2A2 B4B4 C2C2 D1D1 F2F2 G1G1 H 12 J6J6 I8I8 K5K5 L3L3 END E3E3 What is the minimum cost to reduce project time to 30 months ActivityRegular TimeCrash Time Total Cost to Crash Cost Per Unit of Crash Time $ A21$1,000 $ B42$ 500$ 250 $ C22 CC 11 CC $ E31$ 200$ 100 $ F22 CC $ G11 CC $ H 128$ 800$ 200 $ I85$1,200$ 400 $ J66 CC $ K53$ 600$ 300 $ L33 C $4,300 C

13. START A2A2 B4B4 C2C2 D1D1 F2F2 G1G1 H 12 J6J6 I8I8 K5K5 L3L3 END E3E3 What is the optimal schedule if I incur a penalty of $280 for each month above the absolute minimum project time of 26 months? ActivityRegular TimeCrash Time Total Cost to Crash Cost Per Unit of Crash Time $ A21$1,000 $ B42$ 500$ 250 $ C22 CC 11 CC $ E31$ 200$ 100 $ F22 CC $ G11 CC $ H 128$ 800$ 200 $ I85$1,200$ 400 $ J66 CC $ K53$ 600$ 300 $ L33 C $4,300 C

14. © 2007 Pearson Education Calculating total Cost Adding Direct, Penalty & Overhead Costs Adding Direct, Penalty & Overhead Costs The overhead (indirect) costs are $200 per day There is a penalty of $100/day for completing project in more than 12 days

15. Network Diagram

16. © 2007 Pearson Education Total Cost Problem

17. Assessing Risks Risk is a measure of the probability and consequence of not reaching a defined project goal. A major responsibility of the project manager at the start of a project is to develop a risk- management plan. A Risk-Management Plan identifies the key risks to a project’s success and prescribes ways to circumvent them.

18. Probabilistic Time Estimates Mean ma b Time Probability Beta Distribution Pessimistic Optimistic

19. Time Probability Normal Distribution Mean abm 3333 3333 Area under curve between a and b is 99.74% Probabilistic Time Estimates

20. t e = a + 4m + b 6 Mean 2 =2 =2 =2 = ( ) b – ab – a66b – ab – a666 2Variance Probabilistic Time Estimates Calculating Means and Variances for a Beta Distribution Where: a is the Optimistic estimate b is the pessimistic estimate m is the most likely estimate

21. OptimisticLikelyPessimistic Activity(a)(m)(b) Time Estimates (wk) A111213 B7815 C51015 D8916 E142530 F6918 G253641 H354045 I101328 J1215 K567 St. John’s Hospital Probabilistic Time Estimates AF I CG Finish D E HBJ K Start

22. Activity B Most OptimisticLikelyPessimistic (a)(m)(b) 7815 AF I CG Finish D E HBJ K Start St. John’s Hospital Probabilistic Time Estimates t e = = 9 weeks 7 + 4(8) + 15 6  2 = = 1.78 ( ) 15 - 7 6 2 Calculating Means and Variances

23. OptimisticLikelyPessimisticExpectedVariance Activity(a)(m)(b)Time (t e )(  2 ) Time Estimates (wk)Activity Statistics A111213120.11 B781591.78 C51015102.78 D8916101.78 E142530247.11 F6918104.00 G253641357.11 H354045402.78 I101328159.00 J121545.44 K56760.11 St. John’s Hospital Probabilistic Time Estimates

24. © 2007 Pearson Education

25.  2 =  (variances of activities) z = T – T E  2  2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11 = 11.89 z = 72 – 69 11.89 What is the Probability of finishing the project in 72 weeks? Given that: Critical Path = B - D - H - J - K T = 72 Weeks T E = 69 Weeks St. John’s Hospital Analyzing Probabilities From Normal Distribution appendix P z =.8078 .81 =.87

26. .00.01.02.03.04.05.06.07.08.09.0.5000.5040.5080.5120.5160.5199.5239.5279.5319.5359.1.5398.5438.5478.5517.5557.5596.5636.5675.5714.5753.2.5793.5832.5871.5910.5948.5987.6026.6064.6103.6141.3.6179.6217.6255.6293.6331.6368.6406.6443.6480.6517.4.6554.6591.6628.6664.6700.6736.6772.6808.6844.6879.5.6915.6950.6985.7019.7054.7088.7123.7157.7190.7224.6.7257.7291.7324.7357.7389.7422.7454.7486.7517.7549.7.7580.7611.7642.7673.7704.7734.7764.7794.7823.7852.8.7881.7910.7939.7967.7995.8023.8051.8078.8106.8133.9.8159.8186.8212.8238.8264.8289.8315.8340.8365.8389 1.0.8413.8438.8461.8485.8508.8531.8554.8577.8599.8621 1.1.8643.8665.8686.8708.8729.8749.8770.8790.8810.8830 1.2.8849.8869.8888.8907.8925.8944.8962.8980.8997.9015 1.3.9032.9049.9066.9082.9099.9115.9131.9147.9162.9177 1.4.9192.9207.9222.9236.9251.9265.9279.9292.9306.9319 1.5.9332.9345.9357.9370.9382.9394.9406.9418.9429.9441 1.6.9452.9463.9474.9484.9495.9505.9515.9525.9535.9545 1.7.9554.9564.9573.9582.9591.9599.9608.9616.9625.9633 1.8.9641.9649.9656.9664.9671.9678.9686.9693.9699.9706 1.9.9713.9719.9726.9732.9738.9744.9750.9756.9761.9767 2.0.9772.9778.9783.9788.9793.9798.9803.9808.9812.9817 2.1.9821.9826.9830.9834.9838.9842.9846.9850.9854.9857 2.2.9861.9864.9868.9871.9875.9878.9881.9884.9887.9890 2.3.9893.9896.9898.9901.9904.9906.9909.9911.9913.9916 2.4.9918.9920.9922.9925.9927.9929.9931.9932.9934.9936 2.5.9938.9940.9941.9943.9945.9946.9948.9949.9951.9952 2.6.9953.9955.9956.9957.9959.9960.9961.9962.9963.9964 2.7.9965.9966.9967.9968.9969.9970.9971.9972.9973.9974 2.8.9974.9975.9976.9977.9978.9979.9980.9981 2.9.9981.9982.9983.9984.9985.9986 3.0.9987.9988.9989.9990 3.1.9990.9991.9992.9993 3.2.9993.9994.9995 3.3.9995.9996.9997 3.4.9997.9998 NORMAL DISTRIBUTION TABLE

27. Project duration (weeks) 6972 Normal distribution: Mean = 69 weeks;  = 3.45 weeks Probability of exceeding 72 weeks is 0.1922 St. John’s Hospital Probability of Completing Project On Time Probability of meeting the schedule is 0.8078 Length of critical path

28. OptimisticLikelyPessimisticExpectedVariance Activity(a)(m)(b)Time (t e )(  2 ) Time Estimates (Months)Activity Statistics A123 B346 C222 D111 E135 F123 G111 H81214 I7810 J567 K456 L236 Software Purchasing Project Probabilistic Time Estimates

29. START A2A2 B 4.17 C2C2 D1D1 F2F2 G1G1 H 11.67 J6J6 I 8.17 K5K5 L 3.33 END E3E3 Network for a Software Purchasing Project using the Mean Times Critical Path: A—B—C—D—F—G—H—I—L Expected Time to complete the project: 2+4.17+2+1+2+1+11.67+8.17+3.33 = 35.33

30. What is the probability in finishing in less than 37 months?

31. What is the probability in finishing in less than 31 months?