Question

Given the following data and network, find answer the questions 18-20 if you're asked to finish this project 2 days earlier than its normal duration. Indirect cost (overhead) is $200 per day. 9.12 1.8 E 3 12. 19 B 10.13 H 7 1.8 13.20 0.1 1.7 8.15 15.20 20.25 A F I с 6 u K 5 20.25 0.1 8.15 15.20 4. 12 2.8 1.4 D 3 6,9 15.18 G 8 9.17 3 17.20 Duration Normal Crash Days Shortened 7 6 A B с D E F G H Cost Duration Cost Crash Cost Normal Crash Difference Difference per Day 1 5800 $800 4 51 000 51600 4 $300 $500 21 $4001 $800 11 5100 5200 61 5500 5800 7 7 61 5600 5050 51.000 5800 31 4 5450 Orect cos tv BB

Answer 1

Answer :

Path duration:

A-B-F-J-K =25 days (critical path)

A-B-E-H-K = 23 days

A-C-F-H-I-K = 24 days(sub critical path)

A-DG-J-K = 20 days

A-B-F-J-K = 23 days

18) Normal cost of the project = direct cost of all activities + overall cost of critical path

= 800+1000+300+400+100+500+1200+350+700+500+450+(200*25)

=6300+5000

So, Normal cost of the project = $11300

19)After calculating path slope ,we found slope I has least cost slope of 75 in path A-B-F-J-K.So,we will crash cost slope I by one day,

New Total cost = 11300+75-200 = $11175.

We have two critical bath(A-B-F-I-K and A-C-F-I-K).I has found common in both the paths

Boe,crash cost slope of I by one more day.

Now,Total cost = 11175+75-200 = $11050.

Hence,Cost of the project after 2 days shortening = $11050.

20) Crashing cost per day for activity G = Cost Slope of G = 1400-1200/8-4 = 50.

Crashing cost per day for activity G = $50.