M5 IND6. Bowman Builders manufactures steel storage sheds for commercial use. Jo

Question

M5 IND6. Bowman Builders manufactures steel storage sheds for commercial use. Joe Bowman, president of Bowman Builders, Inc. is contemplating producing sheds for home use. The activities necessary to build an experimental model and related data are given in the table below STANDARD STANDARD CRASH ACTIVITY IMMEDIATE TIME weeks TIME PREDECESSORS COST COST weeks) $1,000 $2,000 $300 $1,300 $850 4,000 $1,600 $2,700 1,600 $5,000 D, E 2,000 Crash the project from 14 weeks to 10 weeks by 1 week increments and answer the following questions a) b) c) To crash the project from 14 weeks to 13 weeks, which activity would you crash by one week? To crash the project from 13 weeks to 12 weeks, which activity would you crash by one week? To crash the project from 12 weeks to 11 weeks, which activity(s) would you crash by one week? (Select all that apply) To crash the project from 11 weeks to 10 weeks, which activity(s) would you crash by one week? (Select all that apply) What is the total crashing cost to crash the project from 14 weeks to 10 weeks? d) e)

Explanation / Answer

We need to find the critical path So that we can crash the activities on the critical path first

Path A-D-G = 14 days

Path B-E-G = 2+6+4= 12 days

path C-F= 1+2=3 days

The critical path is A-D-G so we need to crash one activity which is having least crash cost/period

D can be crashed because the least cost/period = 75

a)So to make the project to 13 weeks we need to crash D activity by 1 day

b) Again to make project completion to 12 days we need to crash activity D by 1 day as it is having the least cost

c) Now there are two critical paths having equal time 12 days i.e A-D-G and B-E-G

so we need to crash one activity each from both the paths by 1 days to make the project completion to 11 days

In path A-D-G least crash cost activity = D and in path B-E-G least crash cost activity = E

so we need to crash D and E

d)To crash the project to 10 days again the same procedure has to be taken i.e. D and E should be crashed simultaneously

e)Total Crashing cost = 75+75+(75+50)+(75+50) = $400

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Normal time Crash time Normal Cost Crash Cost Crash cost/pd Project 14 A 3 2 1000 1600 600 B 2 1 2000 2700 700 C 1 1 300 300 0 D 7 3 1300 1600 75 E 6 3 850 1000 50 F 2 1 4000 5000 1000 G 4 2 1500 2000 250

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