Critical Path Method (CPM) with Crashing The film The Lord of the Rings: The Ret
ID: 458523 • Letter: C
Question
Critical Path Method (CPM) with Crashing
The film The Lord of the Rings: The Return of the King, the last installment of The Lord of the Rings trilogy, was directed by Peter Jackson and released on December 17 2003 in the U.S. with huge success. Eventually, the film won 11 Oscars for all of its nominations. The film's box office has been more than $1 billion. In the film, there are long sequences in which not all main characters appear at the same time. And there are big scenes where everybody shows up. Thinking about the making of the film, behind the scenes, we can hypothetically break down the entire cinematic project into the following aggregate activities (We assume away all the complexities in shooting the three installments at the same time and other production details):
Plan, screenplay, finance, cast, film crew, location
Construction of movie sets
Shooting sequence 1
Shooting sequence 2
Shooting sequence 3
Big scene sequence
Postproduction & finalization before release
We have the following hypothetical data on time in months and cost in millions of dollars:
ID
IP
T
ES
EF
LS
LF
SLK
NT
NC
CT
CC
CCR
A
6.0
6.0
13.0
6.0
13.0
B
A
4.0
4.0
9.0
3.0
12.0
C
B
4.0
4.0
12.0
4.0
12.0
D
B
5.0
11.0
5.0
14.0
4.0
16.0
E
B
6.0
0.0
6.0
16.0
4.0
19.0
F
C,D,E
4.0
4.0
19.0
4.0
19.0
G
F
6.0
6.0
11.0
5.0
13.5
Note: IP = Immediate Predecessor; T = NT = Normal Time; ES = Earliest Start; EF = Earliest Finish; LS = Latest Start; LF = Latest Finish; SLK = Slack Time; NT = Normal Time; NC = Normal
Cost; CT = Crash Time; CC = Crash Cost; & CCR = Crash Cost/Time Unit (Crash Cost Rate).
Copy the table to your homework and complete the following parts:
By looking at the chart in part (a), find all possible paths through the network with their corresponding path times. Based on these path times, determine the critical path.
Based on part (a) By using the forward and backward passes together with the four time determination rules, fill your calculations for ES,
EF, LS, LF, and SLK in the appropriate cells.
(b)[2] Try to independently fill in the columns ES, EF, LS, LF, and SLK of the table. If you have difficulties, refer to part (a). Determine the activity with the most slack time.
c) [2] Based on part (a), find the critical path by going through the activities with no slack time.
Compare this result with the critical path found in part (b).
d) Use appropriate data to compute crash costs per time unit and fill in the CCR column. Cross out any CCR cells where the corresponding activities cannot be crashed.
(e)[2] Find the total normal cost of the film. Find the normal cost on the original critical path. Compute the cost proportion of the critical path in the film's budget.
ID
IP
T
ES
EF
LS
LF
SLK
NT
NC
CT
CC
CCR
A
6.0
6.0
13.0
6.0
13.0
B
A
4.0
4.0
9.0
3.0
12.0
C
B
4.0
4.0
12.0
4.0
12.0
D
B
5.0
11.0
5.0
14.0
4.0
16.0
E
B
6.0
0.0
6.0
16.0
4.0
19.0
F
C,D,E
4.0
4.0
19.0
4.0
19.0
G
F
6.0
6.0
11.0
5.0
13.5
Explanation / Answer
Part a:
Variable paths to complete project are ABEFG, which is the critical path.
ABDFG and ABCFG are other paths, though non critical.
ES, EF, LS, LF and SLACK are filled in table below.
Part b:
Activity with most slack time is C with slack of 2 units.
Part c:
Critical path, as found through both methods, is ABEFG where no activity has any slack.
Part d:
Crash cost per unit time is calculated as (CC -NC)/(NT-CT).
Numbers are filled in table below.
Part e:
Total normal cost of film is sum of normal cost of all activities. Similarly calculated for critical path also.
Total normal cost = 94 million dollar
Critical path normal cost = 68 million dollar.
Considering total budget is 94 as above, critical path forms 72% of the proportion.
IP
T
ES
EF
LS
LF
SLK
NT
NC
CT
CC
CCR
A
6.0
0
6
0
6
0
6.0
13.0
6.0
13.0
X
B
A
4.0
6
10
6
10
0
4.0
9.0
3.0
12.0
3
C
B
4.0
10
14
12
16
2
4.0
12.0
4.0
12.0
X
D
B
5.0
10
15
11.0
16
1
5.0
14.0
4.0
16.0
2
E
B
6.0
10
16
10
16
0.0
6.0
16.0
4.0
19.0
1.5
F
C,D,E
4.0
16
20
16
20
0
4.0
19.0
4.0
19.0
X
G
F
6.0
20
26
20
26
0
6.0
11.0
5.0
13.5
2.5
IP
T
ES
EF
LS
LF
SLK
NT
NC
CT
CC
CCR
A
6.0
0
6
0
6
0
6.0
13.0
6.0
13.0
X
B
A
4.0
6
10
6
10
0
4.0
9.0
3.0
12.0
3
C
B
4.0
10
14
12
16
2
4.0
12.0
4.0
12.0
X
D
B
5.0
10
15
11.0
16
1
5.0
14.0
4.0
16.0
2
E
B
6.0
10
16
10
16
0.0
6.0
16.0
4.0
19.0
1.5
F
C,D,E
4.0
16
20
16
20
0
4.0
19.0
4.0
19.0
X
G
F
6.0
20
26
20
26
0
6.0
11.0
5.0
13.5
2.5
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.