1. The document discusses reasons and methods to reduce project durations, including to earn incentive pay, avoid penalties, and fit contractual timelines. It then describes four common methods: using overtime, adding workers, better equipment, and subcontracting labor. 2. The optimal project duration balances minimizing total project costs, including direct costs from activities as well as indirect overhead costs. The network compression algorithm is used to iteratively shorten the critical path by accelerating individual activities until no further reduction is possible without increasing total costs. 3. An example construction project is provided to demonstrate calculating the least cost project duration using crash costs, indirect costs, and shortening cycles to optimize the project schedule.

1. Reasons to Reduce Project Durations:
1. To realize incentive pay
2. To fit within the contractually required time (influences
Bid price).
3. To avoid paying Liquated Damages/Damage for delay
causes (or avoid damaging the company’s
relationship).
4. To complete a project early & move on to another
project (Although the Job-specify profit may reduce,
company profile may increase).
5. To complete a project when weather conditions make
it less expensive (Avoid temporary Heating, avoid
completing site work during raining season).
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King Saud University Dr. Khalid Al-Gahtani

2. Methods to reduce durations:
1. Have the existing crew work overtime. This increase
the labor costs due to increase pay rate and decrease
productivity.
2. Bring in additional workers to enlarge crew size. This
increases labor costs due to overcrowding and poor
learning curve.
3. Use better/more advanced equipment. This will usually
increase costs due to rental and transport fees. If labor
costs (per unit) are reduced, this could reduce costs.
4. Add subcontracted labor to the activity. This almost
always increases the cost of an activity unless the
subcontracted labor is for more efficient.
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3. Project Costs Definitions
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4. Practices to Estimate the Project cost
• Some company assigns Overhead office as % of
direct cost.
• Most companies don’t consider profit as a cost
of the job. Instead, analysis are completed by
using Direct costs + Indirect costs + Company
overhead vs. Budgeted (estimated costs).
• Similar to each activity, the project as a whole
has an ideal (Lease expensive) Duration.
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5. Spring 2008, Time-Cost-Trade-off 5
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6. Accelerating the Critical and
Noncritical path
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7. Network Compression Algorithm
• Determine Normal project duration, and
cost.
• Identify Normal duration Critical Path.
• For large network, using CRITICALITY
THEORM to eliminate the noncritical
paths that do not need to be crashed.
• Compute the cost slop :
co st slo pe = crash co st - no rmal co st
no rmal duratio n- crash duratio n
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8. Network Compression Algorithm
• shortening the CRITICAL ACTIVITIES
beginning with the activity having the
lowest cost-slope
• Determine the compression limit (Nil)
- C rash L im it, or
N il = M in - Free F lo at o f any o f the no n crit ica l a ctivit ie s
in the para lle l paths co m pet ing fo r crit ica l path.*
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9. Network Compression Algorithm
• Organize the data as in the following table:
• yc le toAS ho rten S C artened N il S ho rtened C oDsta y CC yc le r T otal DPuratio n
C
#
ct ivit y
ho
n Be D a ys per o st fo
C o st
ro ject
0
1
2
3
• Update the project network
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10. Network Compression Algorithm
• When a new Critical path is formed:
– Shorten the combination of activity which
Falls on Both Critical Paths, OR
– Shorten one activity from each of the critical
paths. Use the combined cost of shortening
both activities when determining if it is cost
effective to shorten the project.
• At each shortening cycle, compute the
new project duration and project cost
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11. Network Compression Algorithm
• Continue until no further shortening is possible
• Tabulate and Plot the Indirect project Cost on
the same time-cost graph
C yc le P ro ject D irect Ind ire ct T otal
# D uratio n C o st C o st C o st
0
1
2
3
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12. Network Compression Algorithm
• Add direct and indirect cost to find the
project cost at each duration.
• Use the total project cost-time curve to
find the optimum time.
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13. E xa m p le 1:
T he duratio ns a nd d irect co sts fo r each activity in the netw o rk o f a sm a ll co nstructio n
co ntract under bo th no rm a l a nd crash co nd it io ns are g ive n in be lo w T able. E stablish the least
co st fo r exped iting the co ntract . D eterm ine the o ptim u m duratio n o f the co ntract assu m ing the
ind irect co st am o unts to S R 125/w eek.
N o rm a l C rash
A ctivity P receded by D uratio n C o st D uratio n C o st
(w eeks) (S R ) (w eeks) (S R )
A — 12 7000 10 7200
B A 8 5000 6 5300
C A 15 4000 12 4600
D B 23 5000 23 5000
E B 5 1000 4 1050
F C 5 3000 4 3300
G E ,C 20 6000 15 6300
H F 13 2500 11 2580
I D, G, H 12 3000 10 3150
36,500
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14. 20 23 43
D 4
24 4 47
0
0 12 12 12 8 20 20 5 25 27 20 47 47 12 59
A B E G I
0 0 12 14 2 22 22 2 27 27 0 47 47 0 59
2@ 100 2@ 150 1@ 50 5@ 60 2@ 75
2
12 15 27 27 5 32 32 13 45
ES D EF
C F H
A ctivity
12 0 27 29 2 34 34 2 47
LS F LF
3@ 200 1@ 300 2@ 40
C rash lim it
C yc le A ct ivit y Can Be D a ys C o st per C o st fo r T otal P ro ject
N il
# to S ho rten S ho rtened S ho rtened W e ek C yc le C o st D uratio n
0 3 6 ,5 0 0 59
1 G 5 2 2 60 120 3 6 ,6 2 0 57
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15. 20 23 43
D
22 2 45
0
0 12 12 12 8 20 20 5 25 27 18 45 45 12 57
A B E G I
0 0 12 14 2 22 22 2 27 27 0 47 45 0 55
2@ 100 2@ 150 1@ 50 3@ 60 0
12 15 27 27 5 32 32 13 45
ES D EF
C F H
A ctivity
12 0 27 27 0 32 32 0 45
3@ 200 1@ 300 2@ 40 LS F LF
C rash lim it
C yc le A ct ivit y Can Be D a ys C o st per C o st fo r T otal P ro ject
N il
# to S ho rten S ho rtened S ho rtened W e ek C yc le C o st D uratio n
0 36,500 59
1 G 5 2 2 60 120 36,620 57
2 I 2 2 2 70 140 36,760 55
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King Saud University Dr. Khalid Al-Gahtani

16. 20 23 43
D
22 2 45
0
0 12 12 12 8 20 20 5 25 27 18 45 45 10 55
A
0 B E G I
0 0 12 14 2 22 22 2 27 27 0 45 45 0 55
2@ 100 2@ 150 1@ 50 3@ 60 0
12 15 27 27 5 32 32 13 45
ES D EF
C F H
A ctivity
12 0 27 27 0 32 32 0 45
LS F LF
3@ 200 1@ 300 2@ 40
C rash lim it
C yc le A ct ivit y Can Be D a ys C o st per C o st fo r T otal P ro ject
N il
# to S ho rten S ho rtened S ho rtened W e ek C yc le C o st D uratio n
0 36,500 59
1 G 5 2 2 60 120 36,620 57
2 I 2 2 70 140 36,760 55
3 A 2 2 100 200 36,960 53
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17. 18 23 41
D 2
20 2 43
0
0 10 10 10 8 18 20 5 25 25 18 43 43 10 53
A B E G I
0 0 10 12 2 20 22 2 27 25 0 43 43 0 53
0 2@ 150 1@ 50 3@ 60 0
10 15 25 25 5 30 30 13 43
ES D EF
C F H
A ctivity
10 0 25 25 0 30 30 0 43
LS F LF
3@ 200 1@ 300 2@ 40
C rash lim it
C yc le A ct ivit y Can Be D a ys C o st per C o st fo r T otal P ro ject
N il
# to S ho rten S ho rtened S ho rtened W e ek C yc le C o st D uratio n
0 36 ,500 59
1 G 5 2 2 60 1 20 36 ,620 57
2 I 2 2 70 140 36 ,760 55
3 A 2 2 100 200 36 ,960 53
4 H, G 2 2 2 60+ 40 200 37 ,160 51
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18. 18 23 41
D
18 0 41
0
0 10 10 10 8 18 18 5 23 25 16 41 41 10 51
A B 0 E G I
0 0 10 10 0 18 20 2 25 25 0 41 41 0 51
0 2@ 150 1@ 50 1@ 60 0
10 15 25 25 5 30 30 11 41
ES D EF
C F H
A ctivity
10 0 25 25 0 30 30 0 41
LS F LF
3@ 200 1@ 300 0
C rash lim it
C yc le A ct ivit y Can Be D a ys C o st per C o st fo r T otal P ro ject
N il
# to S ho rten S ho rtened S ho rtened W e ek C yc le C o st D uratio n
0 36,500 59
1 G 5 2 2 60 120 36,620 57
2 I 2 2 70 140 36,760 55
3 A 2 2 100 200 36,960 53
4 G, H 2 2 2 60+ 40 200 37,160 51
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5 B, C
King Saud University 2 2 150+ 200
Dr. Khalid Al-Gahtani 700 37,860 49

19. 16 23 39
D
16 0 39
0
0 10 10 10 6 16 16 5 21 23 16 39 39 10 49
A B E G I
0 0 10 10 0 16 18 2 23 23 0 39 39 0 49
0 0 1@ 50 1@ 60 0
10 13 23 23 5 28 28 11 39
ES D EF
C F H
A ctivity
10 0 23 23 0 28 28 0 39
1@ 200 1@ 300 0 LS F LF
C rash lim it
C yc le P ro ject D irect Ind ire ct T otal
# D uratio n C o st C o st C o st
0 59 36500 7375 43875
1 57 36620 7125 43745
2 55 36760 6875 43635
3 53 36960 6625 43585
4 51 37160 6375 43535
5 49 37860 6125 43985
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20. Project Optimal Duration
50000
45000
40000
35000 Direct
Cost
Cost (SR)
30000
Indirect
25000
cost
20000 Total
15000 Cost
10000
5000
0
48 49 50 51 52 53 54 55 56 57 58 59 60
Project Duration (Week)
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21. C la ss W o rk #2
D ata o n sm a ll m a inte na nce pro ject is g ive n a s be lo w :
D epend s N o rm a l C rash
A ct ivit y on T im e C o st T im e C o st
A 6 da ys $700 4 da ys $800
B 4 da ys 400 4 da ys 400
C 5 da ys 650 4 da ys 700
D A 8 da ys 625 5 da ys 700
E B 10 da ys 200 7 da ys 350
F B 7 da ys 500 5 da ys 700
G C 3 da ys 600 3 da ys 600
H D, E 6 da ys 300 5 da ys 400
I F, G 7 da ys 350 4 da ys 425
O n co m p let io n, the pro ject w ill g ive a return o f $110/da y. U sing t im e-co st trade-o ff m etho d,
ho w m u c h w o uld yo u like to co m press the pro ject fo r m a xim iz ing the return? S ho w all
ca lcu lat io ns.
Spring 2008, Time-Cost-Trade-off 21
King Saud University Dr. Khalid Al-Gahtani