Question

Can someone walk me through manual steps to solve e, f, and g?

The Farmer's American Bank of Leesburg is planning to install a new computerized accounts system. Bank management has determined the activities required to complete the project, the precedence relationships of the activities, and activity time estimates, as shown in the following table: Time Estimates (weeks) Activity Activity Activity Description Position recruiting System development Predecessor a C A 5 17 ? -- 7 3 12 15 System training Equipment training Manual system test Preliminary system changeover Computer-personnel interface Equipment modification Equipment testing System debugging and installation Equipment changeover A 4 7 10 7 1.00 D A 5 23 10 3.00 E B, C B, C 1 1 0.00 F 1 4 13 2.00 G D, E 3 6 6 1.00 H D, E 2.5 7 3 1.00 H 1 1 0.00 J F, G G, I 2 2 2 2 0.00 K 5 11 8 1.00 a. Using the PERT approach, use the three estimates of duration to calculate a mean duration and standard deviation for activities A & B. (2 pts) b. Construct a network for this problem using the mean durations, calculate ES(Foll.), LS(Prec.), and the total float for each task, and hence identify the critical path. What is the standard deviation of the critical path? (24 pts) c. What is the 94% confidence interval for the length of the critical path? (4 pts) d. What is the probability of completing the project within 40 weeks? (4 pts) e. Answer the project manager's question: want to tell the client a project length which I am 89.44% sure that we can meet what figure should I give them?" (4 pts) f. What is the probability of completing the project between 35 and 39 weeks? (4 pts) g. The following table provides the crash data for the network project described above Activity Time (weeks) Activity Cost ($) Crash cost Activity Normal time Crash time Normal cost 7 4,800 6,300 1* 9,100 15,500 7 5 3,000 4,000 d 10 8 3,600 5,000 0 3 1,500 2,000 5 1,800 2,000 3 3 C 1 0 C 2 2 C 6 5,000 7,000 Crash the network to 26 weeks. Indicate how much it would cost the bank and then indicate the critical path. (8 pts) (Total: 50 pts) F

Answer 1

Answer:

89.44% confidence interval for length of project: Z value corresponding to 94% probability= 1.250 NORMSINV(0.8944) B17 c D E G 1.250 17 Given time-Expected time Standard deviation 1.250 Given time-33 4+9+1+ Given time-33 = 1.250x 15 -1.250x3.87 Given time 4.837+33 37.837 days Length of project corresponding to 89.44% 37.837 days.

f)

39 weeks Given time-Expected time Standard deviation 39-33 4+9+1 6 6 VIS 3.87 1.55 Note: Calculate the probability for z value, using probability chart for standard normal distribution Probability corresponding to z value 1.55 is 0.9394 or 93.94% 35 weeks: Given time-Expected time Standard deviation 35-33 4+9+1+ 2 2 I5 3.87 =0.52 Note: Calculate the probability for z value, using probability chart for standard normal distribution Probability corresponding to z value 0.52 is 0.6984 or 69.84%

Probability of completing the project between 35 and 39 weeks = 93.94% - 69.84% = 24.1%

g. Normal cost of project 4800910030003600 01500 1800 +0+0+0+5000 $ 28800 Crash as shown below Process: Step 1: Determine the cost slope (Crash cost - Normal cost)/ (Normal time Crash time) Step 2: Crash that activity on the critical path which has minimum cost slope. Step 3: Repeat step 2 until 26 weeks durations is achieved. A C D E F Activity Time (weeks) Activity Cost (S) Activity Normal time Crash time Normal cost Crash cost 1 Slope |=(E3-D3)/(B3-C3) |(E4-D4)/ (B4-C4) |=(E5-D5)/(B5-C5) |(E6-D6) / (B6-C6) 6300 15500 7 4800 9 a b 9100 4 11 5 4000 5 7 3000 C 5000 10 3600 6 0 7 1 e |=(E8-D8)/(B8-C8) |=(E9-D9)/(B9-C9) 1500 3 2000 f 5 8 2000 5 1800 3 0 10 0 11 0 2 0 0 2 12 =(E13-D13) / (B13-C13) 8 k 6 5000 7000 13 oo o

A B C D F Activity Cost (S) Activity Time (weeks) Normal time Crash time Normal cost 1 Activity Crash cost Slope 2 7 4,800 6,300 3 750 a b 9,100 15.500 4 11 3200 5 7 5 3,000 4,000 500 C d 10 3,600 5,000 6 700 0 0 7 e f 8 5 3 1,500 2,000 250 6 5 2.000 1,800 200 10 3 0 i 0 11 1 0 12 2 2 0 5.000 13 k 6 7,000 1000 Note: Activity e, h, 1,j cannot be crashed. Crash in the following order: n 0 n 1 n 2 n = 3 n 4 n= 5 n = 6 n 7 n= 8 9+10+6+2 A-D-G-J 22 27 26 25 24 23 22 22 22 9+10+6+8 A-D-G-k 33 32 31 30 29 28 27 26 26 A-D-H-I-K 9+10+3+1+8 31 31 30 29 28 27 26 25 25 A-C-E-G- 9+7+1+6+2 22 25 24 24 24 23 22 22 22 28 A-C-E-G-K 9+7+1+6+8 31 30 30 30 29 27 26 26 A-C-E-H-I-K A-C-F- 9+7+1+3+ 1+8 29 29 29 29 28 27 26 25 25 9+7+5+8 29 29 29 29 28 27 27 27 26 B-E-G- 11+1+6+2 20 19 19 19 19 19 19 19 19 B-E-G-K 11+1+6+8 26 25 25 25 25 25 25 25 19 11+1+3+1+8 B-E-H-I-K 24 24 24 24 24 24 23 22 19 18 11+5+2 B-F-J 18 18 18 18 18 18 18 17

Activity crashed k Total crash cost k f g a 200 700 700 750 750 1000 1000 250 Cost 5350 Note: Though activity f has a lower cost slope it is crashed in last because, the main aim is to reduce the critical path to shorten the project duration. Total cost to make the project 26 weeks $ 28800 + $5350 $34150 New critical paths are shown below: Project duration (weeks) A-D-G-k 9+10+6+8 26 A-C-E-G-K 9+7+1+6+8 26 A-C-F-J 9+7+5+8 26