Question

rash C Duration (day) Cost (SR) Cost Slope (Skany) 300 500 Normal Activity by Duration .com (day) on A - 7 3600 BA55500 CB96350 DB 19 4700 Е в 10 | 2050 FC 8 1200 G F 5 7200 Γ Η Τ Ε Ι 9450 I D, G, H 4500 Cost (SK) 3900 6500 7200 4900 2200 1450 7200 10000 4600 ABD OTO D ESTU It is required to crash the project duration from its original duration to a final duration of 38 days. Assume an indirect cost of SR 200/day. Use Project information to perform all needed cycles and use them to answer the following questions: a) What is the cost slope for activity C and G? b) In order to reduce the project duration which activity activities would be crashed in cyclel? and by how many days the project can be shortened? c) In order to reduce the project duration which activity activities would be crashed in cycle 2? and by how many days the project can be shortened? d) In order to reduce the project duration further die less than 38 days) which activity activities would be crashed next? and by how many days the project can be shortened? e) At what duration the total cost direct cost indirect cost) is the smallest (optimal duration)

Engineering Management

Answer 1

So set up the table with Duration and Crash Durations and calculate the Cost Slope

Cost Slope = (Crash Cost - Normal Cost)/(Normal Duration - Crash Duration)

So we get Cost Slope for Activity C = 425, Cost Slope for Activity G = 0

Cost Slope (SR/day) Normal Activity Predecessor Duration Cost (SR) (day) 3600 А 5500 B 9 6350 4700 B 10 2050 C,D 8 1200 7200 Crash Duration Cost (SR) (day) 3900 3 6500 7 7200 4900 2200 7 1450 7200 10 10000 5 4600 300 500 425 200 150 250 0 9450 550 50 D,G,H 7 4500L

Set up the precedence network. For each activity set up a box like and get the precdence network

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Early Start for Acitivity A = 0, Early finish for Activity A = 0 + duration = 0+7 =7

so that way we get EF for Activity I = 45. Assume this to be the LF for Activity I

ES for Activity I = maximum of EF of D,G,H

29 33 25 8 в 11 5 12 19 D 1 31 1938 17 | 45 45 38 | | 7 | н 33 27 11 38

So now we know the ES, EF, LS, LF of all activities, so we have

Normal Activity Predecessor Duration Cost (SR) Cost Slope (SR/day) EF Crash Duration Cost (SR) (day) 3900 6500 7200 4900 300 500 425 200 7 12 7 12 21 31 3600 5500 6350 4700 2050 1200 7200 9450 4500 2200 150 250 21 1450 7200 33 38 10000 550 33 38 TD,G,H 1 22 38 7 5 4600 45

so Critical path = A-B-C-F-G-I as shown below

12 с 16 9 25 в 33 | G 1 5 1 38 38 38 38 145 7 1 45

Critical path = A-B-C-F-G-I

Time to complete project if you go in critical path = Sum of durations of A-B-C-F-G-I= 7+5+10+11+5+7 = 41

Normal project completion time = 41 days, Total direct normal cost = 44550

Indirect cost = 200/day. So multipy indirect cost by duration = 200*45 = 9000

So total cost = 44550+9000= 53550 - this is the total cost for normal completion time

For crashing the project, we need to find the crash limit and slope for the critical path

So we find out the crash limit and slope for activities in the critical path. These are

Critical Path Critical Activity Cost Scope Crash Limit = Normal time- Crash time | А в A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 300 500 150 550 0 50 Least

b) So Activity I has the least slope i.e 50

So in CYCLE 1, Crash Activity by 1 day, this wil cost 50. So the project can be shortened by 1 day i.e total duration of project reduces to 40

c) So for cycle 1 we have to re-calculate critical path, we have this as follows

So now new critical path is still the same

Critical path = A-B-C-F-G-I

Time to complete project if you go in critical path = Sum of durations of A-B-C-F-G-I= 7+5+10+11+5+6 = 40days

First iteration total cost = 53550

So add 50 to this and subtract 200/day (i.e indirect cost)

So total cost after cycle 1 = 53550+50-200 = 53550-150 = 53400

c) So now cyle 2, Total cost = 53400

So we find out the crash limit and slope for activities in the critical path. These are

Cycle 2 Critical Path Critical Activity Crash Limit = Normal 'Cost Scope time- Crash time 300 500 425 250 A B A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G- | A-B-C-F-G- I G 50 Least

c) so in cycle 2 - again Activity I can be shortened by 1 day. So project duration reduces from 40 days to 39 days

Previous iteration Total cost = 53400

So add 50 to this and subtract 200/day (i.e indirect cost)

So total cost after Cycle 2 = 53400+50-200 = 53250

So now our new critical path will stil be same

d) Cycle 3 we want to reduce project duration to 37 days. Sofirst reduce from 39 to 38 as follows

Cycle 3 Critical Path Critical Activity Crash Limit | = Normal time- Crash time Cost Scope A 300 500 425 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G-1 A-B-C-F-G- I 250 crash this F G 0 50 cannot crash this

So now you cant crash activity I, so you have to crash the next activity with least slope i.e Activity F

so reduce activity F by 1 day total duration is now 38days

Previous iteration Total cost = 53400

So add 250 to this and subtract 200/day (i.e indirect cost)

So total cost after Cycle 3 = 53400+250-200 = 53450

Now we still need to reduce by one more week, so look at critical path now when Activity F has reduced by 1 day

Critical path is still same so we have

so we havef for Cycle 4 - now we reduce from 38 days to 37 days

Cycle 4 Critical Path Critical Activity Crash Limit = Normal Cost Scope 1 time- Crash time 300 A crash this B 500 A-B-C-F-G-1 A-B-C-F-G- I A-B-C-F-G-1 A-B-C-F-G-I A-B-C-F-G-1 A-B-C-F-G-1 cannot crash this L cannot crash this

so now you cannot crash activity F or Activity I, so you have to crash the next least slope, so crash Activity A by 1 day

so project duration is shortened from 38 days to 37 days.

Previous iteration Total cost = 53400

So add 300 to this and subtract 200/day (i.e indirect cost)

So total cost after Cycle 3 = 53400+300-200 = 53500