Here is a list of activity times for a project as well as crashing costs for its activities. Determine which activities should be crashed and total cost of crashing if the goal is to shorten the project by three weeks as cheaply as possible. First construct a network diagram. You can use either an AOA or and AON
Path Activity "Duration
(weeks)" Duration Total (weeks) First Crash "Second
Crash"
TOP A 5 14 $8 $10
B 6 7 9
C 3 14 15
MIDDLE D 3 13 9 11
E 7 8 9
C 3 14 15
F 5 15 10 15
BOTTOM G 5 11 13
H 5 12 14
Critical path: 1-5-6-7
To shorten the project by three weeks as cheaply as possible, we need to find the cheapest combination of crashing activities.
First, let's look at the current critical path:
1-5-6-7 (TOP: A-B-C / 14 weeks)
Now, we need to find activities that can help shorten the project duration:
1. Activity A (First Crash): Reduces duration by 1 week; cost: $8
2. Activity B (First Crash): Reduces duration by 1 week; cost: $7
3. Activity D (First Crash): Reduces duration by 1 week; cost: $9
4. Activity E (First Crash): Reduces duration by 1 week; cost: $8
5. Activity F (First Crash): Reduces duration by 1 week; cost: $10
6. Activity G (First Crash): Reduces duration by 1 week; cost: $11
7. Activity H (First Crash): Reduces duration by 1 week; cost: $12
Since our goal is to shorten the project by three weeks, we will choose the three activities with the lowest crashing costs:
1. Activity B (First Crash): Reduces duration by 1 week; cost: $7
2. Activity A (First Crash): Reduces duration by 1 week; cost: $8
3. Activity E (First Crash): Reduces duration by 1 week; cost: $8
Total cost of crashing: $7 + $8 + $8 = $23
By crashing activities A, B, and E by one week each, we can shorten the project by three weeks, and the total cost of crashing is $23.
To determine which activities should be crashed and the total cost of crashing, let's first construct a network diagram. We can use the Activity on Arrow (AOA) method.
Here is the network diagram for the given project:
A (5) B (6)
/ \ /
/ \ /
START E(7) --------> F(5)
\ / \
\ / \
D (3) C (3)
/ \ /
/ \ /
G (5) H (5)
\ /
\ /
END
In the above diagram:
- The activities are represented by letters (A, B, C, D, E, F, G, H).
- The numbers in parentheses next to the activity letters represent their durations in weeks.
- The arrows indicate the dependencies between activities.
- The complete path from START to END consists of activities A, E, F, G, and H.
To determine the critical path, we need to find the longest path from START to END. In this case, the critical path is A (5) - E (7) - F (5) - G (5) - H (5), with a total duration of 27 weeks.
To shorten the project by three weeks, we need to crash activities on the critical path.
Let's consider crashing each activity on the critical path and calculate the cost of crashing for each:
- Activity A can be crashed by 2 weeks to a duration of 3 weeks. The cost for the first crash is $8, and the cost for the second crash is $10.
- Activity E can be crashed by 1 week to a duration of 6 weeks. The cost for crashing is $8.
- Activity F can be crashed by 5 weeks to a duration of 0 weeks (eliminated). The cost for crashing is $15.
- Activity G can be crashed by 2 weeks to a duration of 3 weeks. The cost for the first crash is $11, and the cost for the second crash is $13.
- Activity H can be crashed by 2 weeks to a duration of 3 weeks. The cost for the first crash is $12, and the cost for the second crash is $14.
To shorten the project by three weeks as cheaply as possible, we should crash activities in the following order:
1. Crash activity F (5 weeks) for a cost of $15.
- The new critical path will be A - E - G - H, with a total duration of 23 weeks.
2. Crash activity G (2 weeks) for a cost of $11.
- The new critical path will be A - E - H, with a total duration of 21 weeks.
3. Crash activity E (1 week) for a cost of $8.
- The new critical path will be A - H, with a total duration of 20 weeks.
Therefore, the activities that should be crashed are F, G, and E. The total cost of crashing is $15 + $11 + $8 = $34.
By crashing these activities, we have shortened the project by three weeks (from 27 weeks to 20 weeks) as cheaply as possible.
To determine which activities should be crashed and the total cost of crashing, we first need to construct a network diagram to identify the critical path.
1) Constructing the network diagram:
The activities and their durations are as follows:
Activity A (Duration: 5 weeks)
Activity B (Duration: 6 weeks)
Activity C (Duration: 3 weeks)
Activity D (Duration: 3 weeks)
Activity E (Duration: 7 weeks)
Activity F (Duration: 5 weeks)
Activity G (Duration: 5 weeks)
Activity H (Duration: 5 weeks)
Based on the provided information, the network diagram can be constructed as follows:
A (5)
/ \
/ \
B (6) D (3)
/ \
/ \
G (5) E (7) F (5) H (5)
\ /
\ /
C (3)
2) Determine the critical path:
The critical path is the longest path through the network diagram and determines the shortest time needed to complete the project. Based on the provided durations, the critical path can be identified as follows:
A (5) -> B (6) -> D (3) -> E (7) -> F (5)
3) Determine activities to be crashed:
To shorten the project by three weeks, we need to identify activities on the critical path that can be crashed. Since the critical path is A -> B -> D -> E -> F, the activities on this path should be considered for crashing.
Crashing the critical path activities:
- Activity A: The first crash option reduces the duration to 7 weeks with a cost of $8. The second crash option further reduces it to 9 weeks with a cost of $10.
- Activity B: The first crash option reduces the duration to 7 weeks with a cost of $9.
- Activity D: The first crash option reduces the duration to 9 weeks with a cost of $9. The second crash option further reduces it to 11 weeks with a cost of $11.
- Activity E: The first crash option reduces the duration to 8 weeks with a cost of $9.
4) Calculate the total cost of crashing:
To calculate the total cost of crashing, we need to sum the costs of crashing each activity.
Total cost of crashing:
= (Cost of crashing Activity A first option) + (Cost of crashing Activity B first option) + (Cost of crashing Activity D first option) + (Cost of crashing Activity E first option)
= $8 + $9 + $9 + $9
= $35
Therefore, if we want to shorten the project by three weeks as cheaply as possible, we should crash Activity A (first option), Activity B (first option), Activity D (first option), and Activity E (first option). The total cost of crashing would be $35.