Question

Coleman Rich Control Devices, Inc., produces custom-built relay devices for auto makers. The most recent project undertaken by Rich requires 14 different activities. Rich's managers would like to determine the total project completion time (in days) and those activities that lie along the critical path. The appropriate data are shown in the following table: பா் Immediate Immediate Time (days) Time (days) Predecessor(s) Activity Predecessor(s) b b Activity a 6 E, F G, H A 4 5 8 3 3 B 5 3 7 10 6 C 6 6 A J 3 6 8 96 D 4 13 A K 6 8 11 E 3 9 19 B, C L 8 J F 3 5 8 1 3 3 K 5 G 8 D N 7 7 12 L M a) The expected completion time of the project two decimal places). days (round your response to The activities that represent the project's critical path are b) If the time to complete the activities on the critical path is normally distributed, then the probability that the oritical path will be finished in 53 days or less (round your response to two decimal places). c) Using the standard normal table, the number of days that would result in 99 % probability of completion days (round your response to two decimal places). ation rof Enter your answer in each of the answer boxes. Save for Later M Book Pro & ( ) 7 9 0 = delete { } U 0 P K

Answer 1

Procedure Total effective time is calculated using the formula: (a+4m+b)/6 Early Start and Early Finish 1. The activities having no predecessor start at time 0. For them, ES = 0 2. EF = ES + Time 3. If activity A precedes activity B, then ES of B = EF of A + Time of B 4. If 2 activities precede an activity, then the largest value of EF of the 2 activities is considered For example: If activity A and B precede activity C, EF of A =2 and EF of B = 3, then ES of C = EF of B + Time of C Late Start (LS) and Late Finish (LF) 1. We backtrace in this case, so we start from the last activity. 2. LS = LF - Time 3. If activity A precedes activity B, then LF of A = LS of B - Time of A 4. If 2 activities follow an activity, then the smallest value of LS of the 2 activities is considered For example: If activity E and F follow activity B, LS of E=11 and LS of F = 10, then LF of B = LS of F-Time of B Slack = LF - EF or LS - ES Critical Activities: Activities with Slack = 0: Activities which can be crashed

Te = (a+4m+b)/6 EF = ES + Te LS = LF - Te Slack = LF - EF Activity Predecessors Std Deviation =(b a)/6 Variation = std deviation 2 Calculation Value Calculation Value Calculation Value Calculation Value А | NA NA 8 5 0.44 0.25 0.00 2.25 7.11 0.69 B,C 13 19 8 8 6 10 D 1.36 (4+4*5+8)/6 5.33 (2+4*3+5)/6 3.17 (6+4*6+6)/6 6.00 (4+4*9+13)/6 8.83 (3+4*9+19)/6 9.67 (3+4*5+8)/6 5.17 (1+4*5+8)/6 4.83 (3+4*3+6)/6 3.50 (3+4*7+10/6 6.83 (3+4*6+8)/6 5.83 (6+4*8+11/6 8.17 (3+4*6+8/6 5.83 (1+4*3+3)/6 2.67 (7+4*7+12)/6 7.83 Standard deviation (s) 0.00 0.00 5.33 5.33 11.33 14.16 14.16 21.00 24.50 31.33 31.33 37.16 39.50 42.99 0.67 0.50 0.00 1.50 2.67 0.83 1.17 0.50 1.17 0.83 0.83 0.83 0.33 0.83 12.67 0.00+5.33 0.00+3.17 5.33+6.00 5.33+8.83 11.33+9.67 14.16+5.17 14.16+4.83 21.00+3.50 24.50+6.83 31.33+5.83 31.33+8.17 37.16+5.83 39.50+2.67 42.99+7.83 5.33 3.17 11.33 14.16 21.00 19.33 18.99 24.50 31.33 37.16 39.50 42.99 42.17 50.82 5.33-5.33 11.33-3.17 11.33-6.00 19.67-8.83 21.00-9.67 21.00-5.17 24.50-4.83 24.50-3.50 31.33-6.83 37.16-5.83 40.32-8.17 42.99-5.83 42.99-2.67 50.82-7.83 0.00 8.16 5.33 10.84 11.33 15.83 19.67 21.00 24.50 31.33 32.15 37.16 40.32 42.99 H 5.33 5.33-5.33 0.00 11.33 11.33-3.17 8.16 11.33 11.33-11.33 0.00 19.67 19.67-14.16 5.51 21.00 21.00-21.00 0.00 21.00 21.00-19.33 1.67 24.50 24.50-18.99 5 .51 24.50 24.50-24.50 0.00 31.33 31.33-31.33 0.00 37.16 37.16-37.16 0.00 40.32 40.32-39.50 0.82 42.99 42.99-42.99 0.00 42.99 42.99-42.17 0.82 5 0.82 50.82-50.820.00 E,F G,H 7 6 к 0.25 1.36 0.69 0.69 0.69 0.11 0.69 18 J K LM 6 3 L 3 1 7 M N L 12

a) Project completion time (m): 50.82 days Critical Path = A-C-E-H-I-J-L-N b) Let x be the expected project completion time = 53 days s = standard deviation = 12.67 Actual completion time (m) = 50.82 days As we know, z= (x-m) / s Putting values in the formula z = (53 - 50.82) / 12.67 = 2.18 / 12.67 = 0.17206 z score = 0.17206 Probability for z-score (0.167) = 43.17% So, probability that the project will finish in 40 days or less = 43.37% c) Given probability = 99% Probability for 2-score (z) = 99% From z table, value for z = 2.3263 Let x be the expected project completion time s = standard deviation = 12.67 Actual completion time (m) = 50.82 days As we know, z=(x-m) / s Putting values in the formula z = (x - 50.82) / 12.67 We got z= (X-50.82) / 12.67 = 2.3263 x=(2.3263*12.67)+50.82 = 80.29 days So, the expected completion time with 99% probability is 80.29 days