The following represents a project that should be scheduled using CPM: Use Appen
ID: 452178 • Letter: T
Question
The following represents a project that should be scheduled using CPM: Use Appendix G. What is the critical path? What is the expected project completion time? (Do not round intermediate calculations. Round your answer to 3 decimal places.) What is the probability of completing this project within 20 days? (Use Excel's NORMSDIST() function to find the correct probability for your computed Z-value. Do not round intermediate calculations. Round "z" value to 2 decimal places and final answer to 4 decimal places.) There is an 80 percent chance the project below can be completed in X weeks or less. What is X? (Round your answer to the nearest whole number.) Given the following project network: Identify the critical path. A-B-D-F A-C-E-F A-C-D-F Identify the earliest start and finish for each task. (Leave no cells blank - be certain to enter "0" wherever required.) Calculate slack for every project activity. (Leave no cells blank - be certain to enter "0" wherever required.)Explanation / Answer
11 Activity Immediate Predecessor Optimistic Most Likely Pessimistic Expected Time Variance A — 1 3 8 3.50 1.36 B — 1 2 9 3.00 1.78 C A 2 3 7 3.50 0.69 D A 1 2 12 3.50 3.36 E B 1 2 3 2.00 0.11 F C,D 1 3 5 3.00 0.44 G D,E 1 7 13 7.00 4.00 H F,G 2 5 13 5.83 3.36 31.33 15.11 *Expected = (Optimistic + 4 * Probable + Pessimistic) / 6 Variance = (Pessimistic - Optimistic / 6)^2 Standard Deviation = Variance Paths Duration Weeks B-E-G-H 3+2+7+5.83 17.833 A-C-F-H 3.5+3.5+3+5.83 15.833 A-D-F-H 3.5+3.5+3+5.83 15.833 A-D-G-H 3.5+3.5+7+5.83 19.833 Critical path Critical path = A-D-G-H Expected Time = 19.833 weeks Variance = 12.0833 weeks Standard Deviation = 3.4761 weeks Probability of completing project within 20 days P (x 20) Z = X - Mean /Sd = 20 - 19.8333 / 0.3461 = 0.04795 Using Normsdist function, by rounding z value to 0.05 Required Probability = 0.5199 Activity Immediate Predecessor Optimistic Most Likely Pessimistic Expected Time Variance A — 4 4 13 5.50 2.25 B A 5 5 5 5.00 0.00 C A 1 5 6 4.50 0.69 D B 4 11 15 10.50 3.36 E C 5 8 11 8.00 1.00 33.50 7.31 *Expected = (Optimistic + 4 * Probable + Pessimistic) / 6 Variance = (Pessimistic - Optimistic / 6)^2 Standard Deviation = Variance Paths Duration Weeks A-B-D 5.5+5+10.5 21.00 Critical path A-C-E 5.5+4.5+8 18.00 Critical path = A-B-D Expected Time = 21.00 weeks Variance = 5.6111 weeks Standard Deviation = 2.3688 weeks There is an 80 percent chance the project below can be completed in X weeks or less Z Value at 80% = 0.841621 Z = X - Mean /Sd 0.841621 = X - 21.00 / 2.3688 or, 0.841621*2.3688 = X - 21.00 or, 1.993614 = X - 21.00 or, X = 21.00 + 1.99 or, X = 22.99 days or 23 days Activity Predecessor Activity Time ES EF LS LF Slack A None 1 0 1 0 1 0 Critical Path Activity B A 2 1 3 5 7 4 C A 6 1 7 1 7 0 Critical Path Activity D B,C 3 7 10 13 16 6 E C 9 7 16 7 16 0 Critical Path Activity F D,E 8 16 24 16 24 0 Critical Path Activity ES = E of Tail Event EF = ES + Activity Duration LF = L of Head Event LS = LF - Activity Duration Slack - LS - ES E = Earliest occurrence time for the event L = Latest allowable time for the occurrence of event Paths Weeks Duration A-B-D-F 1+2+3+8 14 A-C-D-F 1+6+3+8 18 A-C-E-F 1+6+9+8 24 Critical Path The Critical path is A-C-E-F with estimated duration of 24 weeks