Question

In: Operations Management

Consider the project described in the below Excel a. Use Excel to find the critical path...

Consider the project described in the below Excel

a. Use Excel to find the critical path and the project duration using normal activity times.

b. What is the shortest duration the project can be crashed down to?

c. If you were to crash the project manually and your goal is to minimize crashing cost, which activity would you start with?

d. Find the cheapest way to crash the project to 33 weeks. What are the critical activities in the crashed network, and the cost of crashing?

Activity Predecessor Normal time (months) Crashed time (months) Normal Cost ($) Crashed Cost ($)
1 -- 8 5 700 1,200
2 -- 10 9 1,600 2,000
3 -- 9 7 900 1,500
4 1 4 2 500 700
5 1 6 3 500 900
6 2 5 4 500 800
7 3 7 5 700 1,000
8 3,5,6 15 12 1,400 2,000
9 3,5,6 12 10 1,800 2,300
10 4 18 14 1,400 3,200
11 7,9 4 3 500 800
12 12 7 6 800 1,400

Solutions

Expert Solution

Network diagram is following:

(a) Project duration is determined using Excel Solver as follows

Formulas:

E3 =C3

E4 =C4

E5 =C5

E6 =C6+D3

E7 =C7+D3

E8 =C8+D4

E9 =C9+D5

E10 =C10+D5

F10 =C10+D7

G10 =C10+D8

E11 =C11+D5

F11 =C11+D7

G11 =C11+D8

E12 =C12+D6

E13 =C13+D9

F13 =C13+D11

E14 =C14+D13

D19 =D17

Project duration = 38 weeks

Critical path is the longest path along the network

Critical path is: 2-6-9-11-12

(b) Minimum project duration is determined by crashing the project activities by maximum time.

K3 =C3-I3

K4 =C4-I4

K5 =C5-I5

K6 =C6-I6+K3

K7 =C7-I7+J3

K8 =C8-I8+J4

K9 =C9-I9+J5

K10 =C10-I10+J5

L10 =C10-I10+J7

M10 =C10-I10+J8

K11 =C11-I11+J5

L11 =C11-I11+J7

M11 =C11-I11+J8

K12 =C12-I12+J6

K13 =C13-I13+J9

L13 =C13-I13+J11

K14 =C14-I14+J13

J19 =J17

(c) The activity on critical path having the minimum crash cost slope is activity 9 with a crash cost slope of 250

(d) Cheapest way to crash the project is determined using Excel Solver as follows:

Formulas:

J19 =SUMPRODUCT(I3:I14,H3:H14) All other formulas are same as in part b

Minimum crashing cost = $ 1633.33


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