4. 4
4
Schedule crashing – homework
Act. NT NC CT CC Predecessor
A 7 3000 4 6000 -
B 3 4000 2 5500 A
C 4 15000 2 20000 A
D 8 10000 5 19000 B, C
E 9 7000 6 9100 C
Provide answers at:
Lecture 5 homework
Given data
• Activities
• Normal time
• Normal cost
• Crash time
• Crash cost
• Indirect cost (overhead) = £1400 per time period
Identify
• Cost to crash per period for each activity
• Maximum reduction of project duration
• Cost-optimal project duration
• Nominal, Minimum, Optimal project cost
Draw Activity on Node diagram
5. 5
5
Schedule crashing – Finding the
minimum cost schedule
To shorten a project, crash only activities that are critical
Crash from least expensive to most expensive to crash
Each activity can be crashed until
• It reaches its minimum time duration
• It causes another path to become critical
• It is more expensive to crash than not to crash
Continue until no more activities should be crashed.
6. 6
6
Act. NT NC CT CC Cost to crash per period
A 7 3000 4 6000 (6000-3000)/(7-4) = 1000
B 3 4000 2 5500 (5500-4000)/(3-2) = 1500
C 4 15000 2 20000 (20000-15000)/(4-2) = 2500
D 8 10000 5 19000 (19000-10000)/(8-5) = 3000
E 9 7000 6 9100 (9100-7000)/(9-6)= 700
Example: Finding the cost to crash
per period
A
7
B
3
C
4
D
8
E
9
ABD duration: 18
ACD duration: 19
ACE duration: 20
What is the critical path?
7. 7
7
A
7
B
3
C
4
D
8
E
9
ABD duration: 18
ACD duration: 19
ACE duration: 20
Example: Critical path
Act. NT NC CT CC Cost to crash per period
A 7 3000 4 6000 (6000-3000)/(7-4) = 1000
B 3 4000 2 5500 (5500-4000)/(3-2) = 1500
C 4 15000 2 20000 (20000-15000)/(4-2) = 2500
D 8 10000 5 19000 (19000-10000)/(8-5) = 3000
E 9 7000 6 9100 (9100-7000)/(9-6)= 700
8. 8
8
A
4
B
2
C
2
D
5
E
6
ABD duration: 11
ACD duration: 11
ACE duration: 12
Example: Max reduction of project
duration
Act. NT NC CT CC Cost to crash per period
A 7 3000 4 6000 (6000-3000)/(7-4) = 1000
B 3 4000 2 5500 (5500-4000)/(3-2) = 1500
C 4 15000 2 20000 (20000-15000)/(4-2) = 2500
D 8 10000 5 19000 (19000-10000)/(8-5) = 3000
E 9 7000 6 9100 (9100-7000)/(9-6)= 700
Does it make sense?
9. 9
9
A
4
B
2
C
2
D
5 6
E
6
ABD duration: 12
ACD duration: 12
ACE duration: 12
Example: How much would the
shortest project cost
Act. NT NC CT CC Cost to crash per period
A 7 3000 4 6000 (6000-3000)/(7-4) = 1000
B 3 4000 2 5500 (5500-4000)/(3-2) = 1500
C 4 15000 2 20000 (20000-15000)/(4-2) = 2500
D 8 10000 5 19000 (19000-10000)/(8-5) = 3000
E 9 7000 6 9100 (9100-7000)/(9-6)= 700
Shortest project cost =
= Crash cost + MinTime Indirect cost - 1 day crashing cost for D =
= 6000 + 5500 + 20000 + 19000 + 9100 + (12days x 1400/day) - (1day x 3000/day)
= 73400
10. 10
10
A
7
B
3
C
4
D
8
E
9
ABD duration: 18
ACD duration: 19
ACE duration: 20
Indirect costs
Under normal conditions, this project takes 20 days. Suppose
each day the project incurs an indirect cost of 1400
(overhead).
Example: Which activities to crash?
Act. NT NC CT CC Cost to crash per period
A 7 3000 4 6000 (6000-3000)/(7-4) = 1000
B 3 4000 2 5500 (5500-4000)/(3-2) = 1500
C 4 15000 2 20000 (20000-15000)/(4-2) = 2500
D 8 10000 5 19000 (19000-10000)/(8-5) = 3000
E 9 7000 6 9100 (9100-7000)/(9-6)= 700
11. 11
11
A
7
B
3
C
4
D
8
E
9
ABD duration: 18
ACD duration: 19
ACE duration: 20
Indirect cost of 1400 / period.
Example: Cost optimal project duration
• Act. E is critical and the least expensive to crash
• If it is crashed by 1 day, we would spend 700 and save 1400
• E can be crashed only by 1 day (although its Max reduction is 3days)
because this causes ACD to become critical
Act. NT NC CT CC Cost to crash per period
A 7 3000 4 6000 (6000-3000)/(7-4) = 1000
B 3 4000 2 5500 (5500-4000)/(3-2) = 1500
C 4 15000 2 20000 (20000-15000)/(4-2) = 2500
D 8 10000 5 19000 (19000-10000)/(8-5) = 3000
E 9 7000 6 9100 (9100-7000)/(9-6)= 700
12. 12
12
Act. NT NC CT CC Cost to crash per period
A 7 3000 4 6000 (6000-3000)/(7-4) = 1000
B 3 4000 2 5500 (5500-4000)/(3-2) = 1500
C 4 15000 2 20000 (20000-15000)/(4-2) = 2500
D 8 10000 5 19000 (19000-10000)/(8-5) = 3000
E 9 7000 6 9100 (9100-7000)/(9-6)= 700
A
7
B
3
C
4
D
8
E
9 8
ABD duration: 18
ACD duration: 19*
ACE duration: 19*
• After crashing E by 1 day: What to crash next?
Crash E by 1 day:
1000
2500
3700
Example: Cost optimal project
duration – next step
Cost/period
Indirect cost of £1400/period.
• Option 1: A
• Option 2: C
• Option 3: both D and E
• Option 1: A
• Option 2: C
• Option 3: both D and E
13. 13
13
A
7 4
B
3
C
4
D
8
E
8
ABD duration: 18
ACD duration: 19*
ACE duration: 19*
Crash E by 1 day:
ABD duration: 15
ACD duration: 16*
ACE duration: 16*
Crash A by ? days:
What to
crash next?
• Option 1: C
• Option 2: D and E
2500 (cost to crash) > 1400 (indirect cost saving)
3700 (cost to crash) > 1400 (indirect cost saving)
Stopping condition verified
Example: Cost optimal project
duration – next step
3
Indirect cost of 1400/period.
Cost/period
14. 14
14
A
4
B
3
C
4
D
8
E
8
ABD duration: 18
ACD duration: 19*
ACE duration: 19*
Crash E by 1 day:
ABD duration: 15
ACD duration: 16*
ACE duration: 16*
Crash A by 3 days:
The cost-optimal project duration
is 16 days.
What the project total cost with
crashed duration (16 days)?
Crashed project cost =
= Direct cost + CT Indirect cost +
crashing cost of E and A =
= 39000 + (16days x 1400/day) +
(1day x 700/day + 3days x
1000/day) =
= 65100
Normal project cost = Direct cost + NT Indirect cost =
= 39000 + 20days x 1400 = 67000
Example: Optimal project cost
calculation
18. 2
Earned Value Analysis – Tutorial solution
The total ACWP is the sum of ACWP of Task 1, 2,
3, 4 and 5
19. 3
Earned Value Analysis – Tutorial solution
To calculate the BCWP, each of the percentage
completion must be multiplied by original
budgets:
20. 4
Earned Value Analysis – Tutorial solution
At month 3: BCWS = £23500 BCWP = £17750 ACWP = £24300
The project is over budget
The project is behind schedule
26. 17
17
Bonus homework 2
• Based on the defined and derived EV metrics,
define Time Variance (TV). Derive equation(s)
to express TV as a function of the other EV
metrics. Derive equation to express Projected
Slippage as a function of TV
28. 19
19
• How was fast tracking achieved?
• Why did the costs increase?
• What were the limitations experienced with fast
tracking in this case study?
• What cost saving measures were applied? Were they
effective?
Case study homework
29. 20
20
Another crashing homework (2)
The network and durations given
below show the normal schedule for
a project. You can decrease (crash)
the durations at an additional
expense. The Table summarizes the
time-cost information for the
activities. The owner wants you to
finish the project faster. Find the
minimum possible cost for the
project if you want to finish in (105,
110, 115, 120, 125) days
