Construction Management

Time Cost Trade Off

Construction Management
(Year –IV, Semester- II)

Tutorial on

Time Cost Tradeoff

Asst professor
Er. Sandip Duwadi
Pashchimanchal Campus

Terminologies
Project cost
1) Project Cost • For any project total expenditure incurred in terms of manpower,
➢Direct Cost equipment, machinery, materials and time to achieve a
➢Indirect Cost
particular goal is known as the total cost of the project.
2) Normal Time • The total cost of the project is the sum of the two distinct cost.
3) Crash Time ▪ Direct cost
4) Normal Cost ▪ Indirect cost
5) Crash Cost
6) Cost Slope
7) Minimum Duration

Prepared By: Er. Sandip Duwadi 1

Construction Management
Time Cost Trade Off

➢ Direct cost
• The cost of the materials, equipment and money spent on manpower Normal time
form the direct cost. • The time usually allowed for an activity by the estimator is known as
• The direct costs of the project are of major concern and behavior pattern normal time. It is the standard time for that activity and its denoted by tn
of the direct costs with the time is of importance
• Direct cost of a project depends on the completion time of project but the Crash time
variation is not linear.
• The minimum possible time in which an activity can be completed by
➢Indirect cost deploying the extra resources is known as crash time.
• The expenditures which cannot be allotted clearly to the individual • Beyond crash time the duration of the activity can not be shortened by
activities of the project but are assessed as the whole are called indirect
costs. any amount of increase in the resources mobilization. It is denoted by tc
• It indulges overhead charges, administrative charges, establishment
charges, supervision charges, loss of revenue, etc.

Normal cost Normal and Crash time relationship of an activity

• The direct cost required to complete the activity in the normal duration is
called normal cost. It is denoted by Cn
Crash cost
• The direct cost corresponding to the crash time of completing is known as
crash cost. It denoted by Cc
Cost slope
• The direct cost curve generally is a curve. But this curve can be
approximated by the straight line or more than one straight line depending
upon flatness of the curve. Thus the slope of the straight line is known as
cost slope.
𝐶𝑟𝑎𝑠ℎ 𝐶𝑜𝑠𝑡 −𝑁𝑜𝑟𝑚𝑎𝑙 𝐶𝑜𝑠𝑡 𝐶 −𝐶𝑛
• 𝐶𝑜𝑠𝑡 𝑆𝑙𝑜𝑝𝑒 𝐶𝑆 = 𝑁𝑜𝑟𝑚𝑎𝑙 𝑇𝐼𝑚𝑒 −𝐶𝑟𝑎𝑠ℎ 𝑇𝑖𝑚𝑒 = 𝑇𝑐 −𝑇𝑐
𝑛

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Construction Management
Time Cost Trade Off

Steps in cost optimization Steps in cost optimization

1. First draw the network diagram and find out all the critical path and 6. During the process of crashing there is possible of emerging new critical
critical activities path. Therefore the crashing of project is done simultaneously by re-
crashing the parallel activities
2. Calculate the cost slope of all activities and rank them in ascending 7. The activities for crashing are considered serially in the ascending order
order of cost slope to their cost slope. The activity lying in common path is selected for
3. Calculate direct cost by adding normal cost of all activities and indirect optimal solution.
cost is calculated by multiplying the longest duration times the 8. While crashing activities on the critical path or paths which became
overhead expense critical, the activity on this new critical path are also crashed in the same
4. Crash the project activity as per ranking, crashing is done starting from way.
the activity of critical path having least cost slope to the maximum 9. If there are more than one critical path then one activity from each path
possible extent. is selected at one time for crashing (simultaneous crashing) and so on.
5. Calculate the direct cost by adding extra cost of crashing to the normal This process is continued till the project shortening is possible.
cost and corresponding indirect cost of reduced project duration.

Solution: 6 6
Example • Step 1: First Draw the Network Diagram - 3
A C
Consider the data of a project as shown in table below Activity Time Cost 0 0 5
(Weeks) (in ‘000) 6 11 11
Activity Predecessor Time (Weeks) Cost (in Thousand)
Normal Crash Normal Crash N C N C 1 4
A 6 3 20 41
A - 6 3 20 41 B
B 7 40 58 D
B - 7 4 40 58 4 7 3
C 5 3 60 76
C A 5 3 60 76 2
D 3 1 30 38
D B 3 1 30 38 7 8
Summation = 150 213

If the indirect cost per week is Rs. 6250, find the optimal crashed ➢ Normal Project Cost of all activities =
project completion time and corresponding minimum cost. Without Crashing, We Found that: ∑ normal cost = 150,000.00
- Maximum Project Duration = 11 Weeks ➢ Indirect cost = critical duration * 6250
- Critical path is A-C or 1-3-4 = 11 * 6250 = 68,750.00
- Critical Activities are A and C So, Total Project Cost = Rs. 218,750.00

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Construction Management
Time Cost Trade Off

Step 2 - Now Starting Crashing of Project :

- Calculate Cost Slope of all activities : (Step 3: Iteration 1) (Note: A can be crashed by 3 weeks but if we crash
Now, Crashing Activity A by 1 week: A by 3 weeks, the total duration will be 8 weeks but
Activity Time (Weeks) Cost (in Thousand) Cost Slope, Path Duration (weeks) the next path still will be of 10 weeks and there will
Normal Crash Normal Crash 𝐶𝑐 − 𝐶𝑛 A-C 11 -1 = 10 be change in CP )
𝑇𝑛 − 𝑇𝑐 B-D 10
A 6 3 20 41 (41-20)/(6-3) = 7
B 7 4 40 58 6 ➢ Here, Both paths becomes critical
C 5 3 60 76 8
D 3 1 30 38 4 Now,
➢ Extra Cost of Crashing = crashed duration * Cost slope of crashed activity = 1 * 7000 = 7000
Now, We have to find critical path(CP): ➢ Direct Cost = 150,000 + 7000 = 157000
Path Duration Next, Look for the critical activity
➢ Indirect cost = critical duration * 6250 = 10 * 6250 = 62,500
A-C 11 with Least Cost Slope in the CP.
So, Total Project Cost = Rs. 218,750.00
B-D 10
Here, It is Activity A.

Again,
➢ Extra Cost of Crashing = 2 * (7000+4000)= 22,000
Step 7:
(Step 4: Iteration 2)
➢ Direct Cost = 157000 + 22000 = 179,000
Again, Performing Next Crashing
Path Duration least CS
➢ Indirect cost = critical duration * 6250 = 8 * 6250 Time Cost Trade-off Chart
= 50,000
A-C 10 -2 = 8 A 250
So, Total Project Cost = Rs. 229,000.00
B-D 10-2 = 8 D 244.5
245

Cost in thousands
240
Again,
(Step 5: Iteration 3) ➢ Extra Cost of Crashing = 2 * (6000+8000)= 28,000 235
Again, Performing Next Crashing ➢ Direct Cost = 179000 + 28000 = 207,000 229
230
Path Duration least CS ➢ Indirect cost = critical duration * 6250 = 6 * 6250
A-C 8 -2 = 6 C = 37500
225
B-D 8-2 = 6 B 219.5 218.75
So, Total Project Cost = Rs. 244,500.00 220

Step 6: 215
0 2 4 6 8 10 12
Conclusion: Duration (in Weeks)
✓ Optimal Project Duration is 11 weeks with corresponding cost Rs. 218,750.00
✓ Least(Minimum) Project Duration is 6 weeks with corresponding cost Rs. 244,500.00

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Construction Management
Time Cost Trade Off

Alternatively (Tabular Method) Cost slope Project's Project's

Activity Crash Project's Project's Total Project
(in Rs Critical Crashing Overhead
S.N Under Duration Duration Normal Cost Cost Remarks
thousand/ Path(s) Cost Cost
Crash (week) (weeks) (Rs '000') (Rs '000')
(Iteration 1) (Iteration 2) week) (Rs '000') (Rs '000')
Now, We have to find critical path(CP): Now, Crashing Activity A by 1 week:
Path Duration Path Duration (weeks) 1 - - - A-C 11 150 - 68.75 218.75least cost
A-C 11 (CP) A-C 11 -1 = 10 (CP) 2 A 1 7 A-C, B-D 10 150 7 62.5 219.5
B-D 10 B-D 10 (CP) 3 A,D 2 11 A-C, B-D 8 150 22 50 229

4 B,C 2 14 A-C, B-D 6 150 28 37.5 244.5least duration

(Iteration 3) (Iteration 4) Note:

Again, Performing Next Crashing Again, Performing Next Crashing 1) Project Crashing Cost = Cost slope of Critical Activity * Crash Duration
Path Duration least CS Path Duration least CS 2) Project Overhead Cost (Indirect Cost) = Project Duration * Indirect Cost (Over head Cost )
A-C 10 -2 = 8 A (CP) A-C 8 -2 = 6 C (CP) 3) Total Cost = Project Normal Cost + Project Over head Cost + Sum of all Project Crashing Cost
B-D 10-2 = 8 D (CP) B-D 8-2 = 6 B (CP) E.g : Say for 3rd Iteration: Total Cost = 150 + 50 + (0+7+22) = 229

(8,11)
Example 2 2
• Consider the data of a project as shown in table below:
C
Activity Predecessor Time (Weeks) Cost A D
6
Normal Crash Normal Crash 8 9
A - 8 5 2000 2300 (20,20) (25,25)
(0,0) G
B - 10 8 4000 4300 1 4 5
5
C A 6 5 3000 3125
E
D A 9 6 5000 5225 B F
10 10
E B 10 9 2500 2700 13 Here , Possible Paths are :
F B 13 13 5000 - 3 Path Duration(weeks)
G D,E 5 3 1000 1700 A-C 14
(10,10) A-D-G 22
• If the indirect cost per week is Rs. 300, Find the optimal crashed project B-E-G 25 (C.P)
completion time and corresponding minimum cost. B-F 23

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Construction Management
Time Cost Trade Off

➢ Direct Project Cost of all activities = ∑ normal cost = 22500 (Iteration ): Crashing Activities in CP
➢ Indirect cost = critical duration * 300 = 25 * 300 = 7500 Path Duration (It -1) Iteration 2 Iteration 3
So, Total Project Cost = Rs. 30,000 A-C 14 14 14
Activity Time (Weeks) Cost Cost Slope, A-D-G 22 22 22-1 = 21 (G-1)
Calculate Cost Slope of all
𝐶𝑐 − 𝐶𝑛 B - E- G 25 -2 =23 (B-2) 23-1 =22 (E-1) 22-1 = 21 (G-1)
activities : Normal Crash Normal Crash
𝑇𝑛 − 𝑇𝑐 B-F 21 21 21
A 8 5 2000 2300 100 Project's Project's
Activity Crash Project's Project's Total Project
B 10 8 4000 4300 150 Cost slope Crashing Overhead
S.N Under Duration Critical Path(s) Duration Normal Cost Remarks
(Rs /week) Cost Cost
C 6 5 3000 3125 125 Crash (week) (weeks) Cost (Rs) (Rs)
(Rs ) (Rs)
D 9 6 5000 5225 75
1 - - - B-E-G 25 22500 - 7500 30000
E 10 9 2500 2700 200
2 B 2 150 B-E-G 23 22500 300 6900 29700
F 13 13 5000 - -
3 E 1 200 B-E-G 22 22500 200 6600 29600least Cost
G 5 3 1000 1700 350 4 G 1 350 B-E-G, A-D-G 21 22500 350 6300 29650Least duration
Now, Crashing of project is started from the CP with least Cost Slope • Conclusion:
The CP is B-E-G, the least CS amongst there activities is activity ‘B’ ✓ Optimal Project Duration is 22 weeks with corresponding minimum cost Rs. 29,600.00

Time Cost Trade-off Chart Example

Perform time-cost trade-off based on given information regarding a
30050
30000 project comprising five different activities. Also, draw the time-cost trade-
30000
off chart. Take time dependent cost per day of Rs 1.6 millions.
29950
29900 Time (DAYS) Cost (MILLIONS RS.)
29850 Activity Predecessors
Normal Crash Normal Crash
Cost

29800
A - 26 18 4 20
29750
29700 B - 28 22 5 20
29700
29650
29650 C A 10 5 6 11
29600
29600 D A 4 2 8 11
29550 E B,D 12 6 3 14
20.5 21 21.5 22 22.5 23 23.5 24 24.5 25 25.5
Duration (in Weeks)

Prepared By: Er. Sandip Duwadi 6