Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Crashing Example, Study notes of Project Management

Crashing Example: The network and durations given below shows the normal schedule for a project. You can decrease (crash) the durations at an additional ...

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

gerrard
gerrard 🇮🇹

3.9

(7)

213 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Crashing Example: The network and durations given below shows the normal schedule for a
project. You can decrease (crash) the durations at an additional expense. The Table given
below summarizes the time-cost information for the activities. The owner wants you to you
to finish the project in 110 days. Find the minimum possible cost for the project if you want to
finish it on 110 days. (Assume that for each activity there is a single linear, continuous
function between the crash duration and normal duration points).
Activity
Normal
duration
(days)
Crash
Duration
(days) Normal
Cost Crash
Cost
A 120 100 12000 14000
B 20 15 1800 2800
C 40 30 16000 22000
D 30 20 1400 2000
E 50 40 3600 4800
F 60 45 13500 18000
Solution:
Assume that the duration-cost relationship for each activity is a single linear, continuous
function between the crash duration and normal duration points. Using the normal duration
(ND), crash duration (CD), normal cost (NC), and crash cost (CC), the crash cost slope for
each activity can be determined as follows;
CDND
NCCC
SA
=
daySA/100$
100120
1200014000 =
=
SB = $200/day
SC = $600/day
SD = $60/day
SE = $120/day
SF = $300/day
A
120
B
20
C D E
F
40 30 50
60
1 2 3 4 5
pf3

Partial preview of the text

Download Crashing Example and more Study notes Project Management in PDF only on Docsity!

Crashing Example: The network and durations given below shows the normal schedule for a project. You can decrease (crash) the durations at an additional expense. The Table given below summarizes the time-cost information for the activities. The owner wants you to you to finish the project in 110 days. Find the minimum possible cost for the project if you want to finish it on 110 days. (Assume that for each activity there is a single linear, continuous function between the crash duration and normal duration points).

Activity

Normal duration (days)

Crash Duration (days)

Normal Cost

Crash Cost A 120 100 12000 14000 B 20 15 1800 2800 C 40 30 16000 22000 D 30 20 1400 2000 E 50 40 3600 4800 F 60 45 13500 18000

Solution:

Assume that the duration-cost relationship for each activity is a single linear, continuous function between the crash duration and normal duration points. Using the normal duration (ND), crash duration (CD), normal cost (NC), and crash cost (CC), the crash cost slope for each activity can be determined as follows;

ND CD

CC NC

S A

S (^) A $ 100 / day 120 100

S (^) B = $200/day

S (^) C = $600/day

S (^) D = $60/day

S (^) E = $120/day

S (^) F = $300/day

A

B

C D E

F

The normal cost for the project is the sum of a normal cost for each activity. The normal cost for the project is $48300 and the normal duration is 140 days. The activity which should be crashed is the one on the critical path which will add the least amount to the overall project cost. This will be the activity with the flattest or least-cost slope. The duration can be reduced as long as the critical path is not changed or a new critical path is created. In addition, the activity duration cannot be less than the crash duration.

S (^) D = $60/day (least-cost slope) Maximum of 10 days can be cut from this schedule by reducing the duration of activity D to the crash duration of 20 days.

Overall duration is 130 days and there are multiple critical paths (B-F-E and B-C-D-E). Total project cost at this duration is the normal cost of $48300 plus the cost of crashing the activity D by 10 days (60 * 10 = $600) for a total of $48900.

The next activity to be crashed would be the activity E, since it has the least-cost slope ($ per day) of any of the activities on the critical path. Activity E can be crashed by a total of 10 days. Crashing the activity E by 10 days will cost an additional $120 per day or $1200.

The project duration is now 120 days and the total project cost is $50100. There are now three critical paths (A, B-C-D-E, and B-F-E). The next stage of crashing requires a more through analysis since it is impossible to crash one activity alone and achieve a reduction in the overall project duration. Activity A is paired with each of the other activities to determine which has the least overall cost slope for those activities which have remaining days to be crashed. Activity A ($100) + activity B ($200) Activity A ($100) + activity C ($600) + activity F ($300)

A

B

C D E

F

A

B

C D E

F