# The Following Values For The Project

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QUESTION 1 a) The following values for the project: Activity time Early Start Early Finish Late Start Late Finish Slack Standard Deviation Project 36.33 1.56 A 10 0 10 0 10 0 0.67 B 9 0 9 9 18 9 2 C 8 0 8 2 10 2 0.33 D 2 9 11 18 20 9 0.33 E 10 10 20 10 20 0 0.67 F 6 20 26 20 26 0 0.33 G 3 20 23 23 26 3 0.67 H 5 26 31 26 31 0 1 I 4 23 27 32.33 36.33 9.33 0.67 J 5.33 31 36.33 31 36.33 0 0.67 K 2 31 33 34.33 36.33 3.33 0 We see from the output that the activities which have zero slack are activities A, E, F, H, J. Therefore, these activities are the critical path and thus the total duration is: 10 + 10 + 6 + 5 + 5.33 = 36.33 The project is expected to take 36.33 weeks b) We solve this using the following formula to get the…show more content…
(0.6)2 = 0.144 Probability (3) = 0.4*(0.6)3 = 0.0864 Probability of more than 3 = 1 – (0.4 + 0.24 + 0.144 + 0.0864) Probability of more than 3 = 1 – 0.8704 Probability of more than 3 = 0.1296 g) 15 minutes = 0.25 hours W=1/(μ-λ) 0.25=1/(μ-3) 0.25μ – 0.75 = 1 0.25μ = 1.75 μ = 7 The required service rate should be 7 customers per hour h) p = (0.05) 1/4 = 0.473 3 / μ = 0.473 μ = 6.34 The required service rate should be 6.34 customers per hour QUESTION 3 a) Observation Sample 1 2 3 4 Mean R 1 12 11.97 12.1 12.08 12.0375 0.13 2 11.91 11.94 12.1 11.96 11.9775 0.19 3 11.89 12.02 11.97 11.99 11.9675 0.13 4 12.1 12.09 12.05 11.95 12.0475 0.15 5 12.08 11.92 12.12 12.05 12.0425 0.2 6 11.94 11.98 12.06 12.08 12.015 0.14 7 12.09 12 12 12.03 12.03 0.09 8 12.01 12.04 11.99 11.95 11.9975 0.09 9 12 11.96 11.97 12.03 11.99 0.07 10 11.92 11.94 12.09 12 11.9875 0.17 11 11.91 11.99 12.05 12.1 12.0125 0.19 12 12.01 12 12.06 11.97 12.01 0.09 13 11.98 11.99 12.06 12.03 12.015 0.08 14 12.02 12 12.05 11.95 12.005 0.1 15 12 12.05 12.01 11.97 12.0075 0.08 12.0095 0.1267 From the table which has factors, we see that with sample size of 4 the values are: D3 = 0 D4 = 2.282 A2 = 0.729 LCL of range = D3R LCL of range = 0 * 0.1267 = 0 UCL of range = D4R UCL of range = 2.282 * 0.1267 = 0.28905 The range is within statistical control as the value of R is between the LCL and UCL of range LCL of process average = Mean of all averages + A2R LCL of process average = 12.0095 -