Question

Problem #3 (25 points) For the construction project whose network diagram is shown below, a time-cost optimization procedure is required to determine the shortest duration and its corresponding lowest cost. Activity ID ES Darwin EF LS UF 1. Fill in values for the normal cost and crashed cost in the table below. Activities A and B already have costs associated with them. For activities through I do not use the same costs for any activity 2. Calculate the slope and determine normal duration and normal cost for the project. Fill the answers in the table below. B Slope (S/day) Activity Normal Duration Crashed Duration Normal Cost (5) Crashed Cost (5) days) (days) A 3 3 600 600 B 5 4 250 340 с 6 3 D 6 3 E 8 4 F 2. 1 G 3 2 н 3 2 5 5 J

Answer 1

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Normal Crashed Acuity- Duration_duration Normal ast CRashel Cost. Crash slupe preference NC-CC orcher Spay hash A 3 3 600 600 B 4 250 340 490/y 6 3 300 450 50$/42y D 3 240 300 20$/day ☺ € 8 4 400 SDO 25+ Iday @ 2 1 200 225 G 3 2 210 250 40fday H 3 2 270 340 70$444 S 5 700 4 700 - D 6 B 5 E 11 3 F J 即 2 2 6 3

with @ The Critical Path is the longest path of the network no slack or float along it path Duration Normal cost along that the of of path A-B-D-H-] 2060 $ 3170 Total wst 22 Days A-B-E-H-J 24 Days 2220$ 3170 A-B-E-H-I 18 days 2020$ 3170 A-C-F-H-) 19 Days 20708 3170 Atc-a-J 17 Days 1810 3170 Hence, Project Duration along the critical path. A-B-E-H- J is A 24 Days. & Normal cost associated is a $with the 3170 Best fausachvoly are D, E, F 4 GCcrash Slope wise). 26 Total Crash Duration of frugect = 4 days (since, only Elles glong critical Path). _CRashed cost 3395$ © Best two achiv Hy to crash 1 day (Best_on_Cost Slope 4 offect on op?misat Llony. Two Activity are D& E. Total Crashed Duration = 23 Days( since a day crashing of Df E actrites). CRashed cost 3170+20+&5 = 3215$.

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