Review problems and solutions

Review problems and solutions
1.
For the Project network shown answer the following questions based on the data
provided. You can use Excel to assist you in solving the problem, but you still
need to show your work pretending you solve by hand.
A
B
E
G
J
D
C
Activity
A
B
C
D
E
F
G
H
I
J
K
(a)
(b)
(c)
Expected time(weeks)
3
2
8
1
6
4
5
1
3
5
6
I
F
H
K
Standard deviation
.3
.5
2.0
.2
1.0
.2
.4
.1
3
1.0
.6
What is the critical path?
What is the probability of meeting a desired 26 weeks completion time?
Several activities can be expedited for a cost. Specifically, activity D expected
time could be reduced by 10% for a cost of $500, or by 20% for a cost of $2000.
If the project is not completed in 26 weeks a penalty of $100,000 will be paid.
What is your recommendation regarding expediting activity D? Assume the
standard deviations don’t change.
Solution
To solve with Excel enter the data to the PERT-CPM. Since the standard deviation
and the expected time for each activity are given, the three time estimates are not
needed.
Run the CPM-PERT template and obtain the following result:
(a) The critical path is CDEGIK.
(b) P(X<26)= [Use PERT-CPM]=P(Z<(26-29)/3.8158)=.2159. You need to
explain this result. X is normally distributed with mean = 29 and standard
deviation = [22+.22+12+.42+32+.62]1/2=3.8158
(c)
We apply the expected value criterion:
(i)
Not reducing the completion time of activity D:
E(Cost)=0P(X<26)+100,000P(X>26)=100,000(.7841)=78410
(ii)
Reducing the expected completion time of activity D by 10%. New
time=.9. From the template we have P(X<26)=.2236 so
P(X>26)=.7764.
E(Cost)=0P(X<26)+100,000P(X>26)+500=100,000(.7764)+500=78140
(iii)
Reducing the expected completion time of activity D by 20%. New
time=.8. From Excel we have P(X<26)=.2315 so P(X>26)=.7685.
E(Cost)=0P(X<26)+100,000P(X>26)+2000=100,000(.7685)+2000=78850
Conclusion: Allocate $500 to expedite activity D by 10%.
2.
The following data pertains to a certain project:
Immediate
Normal
Normal
Activity
Predecessor.
Time(weeks) Cost
__
A
10
30
B
A
8
120
C
B
10
100
D
A
7
40
E
D
10
50
F
C, E
3
60
(a)
(b)
(c)
(d)
Crash
Time(weeks)
8
6
7
6
8
1
Crash
Cost
70
150
160
50
75
95
Show the project network
Assume the company wants to complete the project in 28 weeks. Should any
activity be crashed? Explain.
Which activities should be crashed, by how much, and at what total crashing
cost, if the objective is to minimize the total crashing costs, while meeting the
deadline of 28 weeks?
Assuming there is a penalty of 20 paid for each week the project is not
completed beyond the deadline of 28 weeks. What is the optimal crash plan
that minimizes total costs?
Solution:
(b) We run Excel and realize that the project needs to be crash since the normal
project completion time is 31 weeks.
(c) From the Excel printout we have:
Activity B is crashed by 1 week and activity F is crashed by 2 weeks.
(d) We need to formulate this problem as a linear programming model, because the
PERT-CPM template does not provide a solution to this problem.
The linear programming model:
Minimize Crash cost + penalties =
20YA+15YB+20YC+10YD+12.5YE+17.5YF+20(Delta)
S.T.
Activity F
Delta = XF+3-YF-28
28
XF
XF+3-YF
XB>=XA+10-YA
Delta
XCXB+8-YB
XDXA+10-YA
XEXD+7-YD
XFXE +10-YE
XFXC+10-YC
YA2
YB2
………
XF-YF-DELTA=25
-XA+XB+YA 10
-XB+XC+YB 8
-XA+XD+YA 10
-XD+XE+YD 7
-XE+XF+YE 10
-XC+XF+YC 10
Using Solver to solve this linear programming model, the start times of each activity
appears in the list of X values. Also, read the amount of time reduction for each
activity from the list of Y values. For example, activity B starts at time 10 and
takes 8 – 1 = 7 weeks to complete [Normal time – time reduction = 8 –1]. The
project is completed in 30 weeks [XF+3-YF = 27+3-0], therefore, the delay
measured by the variable DELTA is equal to 2 weeks [30-28]. Thus, with the
penalty of 20 per week of tardiness, it pays to be late by two weeks; the project is
crashed by 1 week (from 31 to 30).
Another way to solve the problem is by running the CPM-Deadline multiple times,
changing the deadline, and adding the appropriate penalty manually. For example:
 Let deadline = 28. This yields a certain crash costs, say C1. No need to add any
penalty.
 Let deadline = 29. This yields a crash cost of C2, smaller than C1, but we need to
add a penalty of 20. Continue with this manner.
Note that you do not need to go below a deadline of 28, because the crash cost will
increase compared to C1. Also note, that there is no need to go beyond 31(the completion
normal time), because no crash will be needed but the penalties accumulate.
3.