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emey.b
mgt_604_assignment_000_0-1-1.docx
MGT 604
Assignment #2
(To be uploaded to Blackboard by midnight 7/23/2016)
Consider a construction project with the following (8) activities:
1) Determine Activities Schedule for this project using a Gantt chart on Microsoft Project.
2) Identify the Critical Path of activities for this project.
Paths and durations are:
A, C, E, G, H
tA + tC + tE + tG + tH = 2+2+4+5+2 = 15
A, C, F, H
tA + tC + tF + tH = 2+2+3+2 = 9
A, D, G, H
tA + tD + tG + tH = 2+4+5+2 = 13
B, D, G, H
tB + tD + tG + tH = 3+4+5+2 = 14
A, C, E, G, H gives the longest duration thereby being the Critical Path
3) Determine those activities that have Slack / Float, and their amount of Slack/ Float.
|
Activity |
ES |
EF |
LS |
LF |
Slack (LS-ES) |
Critical |
|
A |
0 |
2 |
0 |
2 |
0 |
Yes |
|
B |
0 |
3 |
1 |
4 |
1 |
No |
|
C |
2 |
4 |
2 |
4 |
0 |
Yes |
|
D |
3 |
7 |
4 |
8 |
1 |
No |
|
E |
4 |
8 |
4 |
8 |
0 |
Yes |
|
F |
4 |
7 |
10 |
13 |
6 |
No |
|
G |
8 |
13 |
8 |
13 |
0 |
Yes |
|
H |
13 |
15 |
13 |
15 |
0 |
Yes |
4) Given the Optimistic and Pessimistic time estimate for each activity in the table below, a) What is the probability of project completion in 16 weeks or less? b) With 99% confidence, when would this project be completed?
Part (a)
Expected time: t = (a + 4m + b)/6 where a = optimistic, b = pessimistic, m = most likely time
Variance: v = [(b-a)/6]2
|
Activity |
a |
m |
b |
t |
v |
|
A |
1 |
2 |
3 |
(1 + 4*2 +3)/6 = 2 |
[(3 - 1)/6]2 = 0.11 |
|
B |
2 |
3 |
4 |
(2 + 4*3 + 4)/6 = 3 |
[(4 - 2)/6]2 = 0.11 |
|
C |
1 |
2 |
3 |
(1 + 4*2 +3)/6 = 2 |
[(3 - 1)/6]2 = 0.11 |
|
D |
2 |
4 |
6 |
(2 + 4*4 + 6)/6 = 4 |
[(6 - 2)/6]2 = 0.44 |
|
E |
1 |
4 |
7 |
(1 + 4*4 + 7)/6 = 4 |
[(7 - 1)/6]2 = 1.00 |
|
F |
1 |
2 |
9 |
(1 + 4*2 + 9)/6 = 3 |
[(9 - 1)/6]2 = 1.78 |
|
G |
3 |
4 |
11 |
(3 + 4*4 + 11)/6 = 5 |
[(11 - 3)/6]2 = 1.78 |
|
H |
1 |
2 |
3 |
(1 + 4*2 +3)/6 = 2 |
[(3 - 1)/6]2 = 0.11 |
Total variance = 0.11 + 0.11 + 1.00 + 1.78 + 0.11 = 3.11
Completion on or before 16 weeks
Z = 0.5681818181818182 ≈ 0.57
Using normal distribution tables;
i.e. P (T ≤ 16 weeks) = 0.7157 = 71.57%
Part (b)
5) Given activity Crash Costs in the table below:
a) What is the Total Cost for completion of this construction project in normal time?
Solution
Total Cost in normal time = Cost of Critical tasks
Total Cost = CA, CC, CE, CG, CH
Total Cost = $22000 + $26000 + $56000 + $80000 + $16000
Total Cost = $200000
b) What is the additional cost for completing the project two (2) weeks earlier than expected?
Solution
Crash cost per period = Crash cost – Normal cost/Normal time – Crash time
|
Activity |
N. time |
C. time |
N. cost |
C. cost |
CCpW |
Critical |
|
A |
2 |
1 |
22000 |
22750 |
750 |
Yes |
|
B |
3 |
1 |
30000 |
34000 |
2000 |
No |
|
C |
2 |
1 |
26000 |
27000 |
1000 |
Yes |
|
D |
4 |
2 |
48000 |
49000 |
1000 |
No |
|
E |
4 |
2 |
56000 |
58000 |
1000 |
Yes |
|
F |
3 |
2 |
30000 |
30000 |
500 |
No |
|
G |
5 |
2 |
80000 |
84000 |
1500 |
Yes |
|
H |
2 |
1 |
16000 |
19000 |
3000 |
Yes |
Activities on critical path are A, C, E, G, H
Crash A by 1 week with Crash Cost per Week of $750
Crash E by 1 week, at $1000
That creates a new critical path, B, D, G, H with completion period of 14 weeks
Crash D with Crash Cost per Week, $1000
Therefore, total additional cost = $750 + $1000 + $1000 = $2750
[The Normal Distribution Cumulative Probability Table is attached]
MGT-604_Week- 2_ Assignment _RS
And3.111.76
Variance
s
===
1615
Z
1.76
X
m
s
--
==
(X16)P(Z0.57)0.7157
P
£=£=
99%Z2.33 (-)/15/1.76
cmsc
³===-
2.33*1.761519.1008weeks
c
=+=
