18 Office Automation Inc Developed A Proposal For

Question

18. Office Automation, Inc., developed a proposal for introducing a new computerized office system that will standardize the electronic archiving of invoices for a particular company. Contained in th…

18. Office Automation, Inc., developed a proposal for introducing a new computerized office system that will standardize the electronic archiving of invoices for a particular company. Contained in the proposal is a list of activities that must be accomplished to complete the new office system project. Use the following relevant information about the activities Crash Cost ($1000s) Activity Description Immediate NormalCrash Normal Predecessor Time TimeCost (weeks) (weeks ($1000s) 10 70 150 160 50 75 Plan needs Order equipment Install equipment Set up training lab Conduct training course Test system 30 120 100 40 50 6 10 6 10 60 a) b) Identify the critical path, and what is the expected project completion time? Assume that the company wants to complete the project in six months or 26 weeks, use the trial and error method to recommend crashing decisions that will meet the desired completion time at the least possible cost. What will be the total crash cost? And what will be the total cost for the project? Formulate the crash problem as a Linear Programming problem. Do not solve the LP problem c)

Solutions

Expert Solution

(a)

The critical path is the longest path i.e. A-B-C-F and the
expected duration of the critical path (and that of the project) is
31 weeks.

(b)

Activity Normal Time Crash Time Normal Cost Crash Cost Cost slope
A 10 8 30 70 20
B 8 6 120 150 15
C 10 7 100 160 20
D 7 6 40 50 10
E 10 8 50 75 12.5
F 3 3 60

Cost Slope = (Crash cost – Normal cost) / (Normal time – Crash
time)

(b)

Let Tj be the start time of the activity-j and Cj be the crash
weeks fro activity-j for j=A, B,…, F

Minimize Z = total cost = 400,000 + 20,000 CA + 15,000 CB +
20,000 CC + 10,000 CD + 12,500 CE

Subject to,

Precedence constraints:

TB – TA + CA >= 10
TD – TA + CA >= 10
TC – TB + CB >= 8
TE – TD + CD >= 7
TF – TC + CC >= 10
TF – TE + CE >= 10

Crash limit constraints:

CA <= 2
CB <= 2
CC <= 3
CD <= 1
CE <= 2
CF <= 0

Project completion constraint:

TF + 3 – CF <= 26 i.e. TF – CF <= 23

Tj, Cj >= 0 for j=A, B, …, F


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