Project duration can be reduced by assigning more labor to project activities, often in the form of overtime, and by assigning more resources (material, equipment, etc). However, additional labor and resources cost money and hence increase the overall project cost. Thus, the decision to reduce the project duration must be based on an analysis of the trade-off between time and cost.
Project crashing is a method for shortening the project duration by reducing the time of one or more of the critical project activities to a time that is less than the normal activity time. This reduction in the normal activities times is called ‘crashing’. Crashing is achieved by devoting more resources (in terms of money) to the activities to be crashed.
Many times it is felt that the project duration as estimated from arrow diagram is long and it is desired by management to accomplish the project in a shorter duration of time in order to secure progress payments or to avoid lateness penalties, etc. To do so various possibilities are explored. Work study techniques of systematic questioning (as in critical examination under method study) are employed to every (critical) activity to seek the possibilities of reducing their duration.
The critical path activities as classified as Do and Ancillary activities. Ancillary activities are those which support Do activities. For example, cutting of threads on a bar is Do activity whereas making the set up for cutting threads is an Ancillary activity. After identification, Do activities are subjected to systematic questioning technique embodying series of questions as regards purpose, place, sequence, person and means.
The second method is to trade off or transfer some resources from the activities having float to the critical activities, in order to reduce their duration. Trading-off redistributes the resources and ac- companies changes in duration. The resources maybe workforce, amount of equipment and machinery, money, (better and more suitable) materials, etc.
Even by using work study techniques, trade off and other possible methods, if efficiency does not improve and the project duration is not shortened, then the only alternative left is network contrac- tion or compression.
In other words, when all those techniques, which can reduce project duration at almost zero additional cost, fail then network, contraction, expedition or crashing of activities is resorted to. This system of improvement involves extra cost because extra money is spent on overtime engagement of workforce, purchase of additional machinery, use of better materials, skilled workers, etc. The cost increases, as more and more activities are crashed and in turn, of course the project duration decreases.
One has to strike a balance between the extra money spent and the project time saved.
There are two types of activities- critical and non-critical. Crashing non-critical activities does not serve any purpose as they do not serve any purpose because they do not control the project duration.
Completing the non-critical activities earlier does no benefit rather it increases work-in-progress. There- fore, crashing is centered on critical activities only which can reduce project duration if completed earlier.
Crashing of critical activities is started from that critical activity which has least cost-time slope, i.e., which is cheapest to crash. Activities are crashed one after the other till the activity duration cannot be reduced further or the duration of original critical path gets shortened beyond a certain limit that another path becomes critical. In that case, one activity in each of the critical paths is chosen for crashing by the same amount of time. The crashing continues to the point whereafter further decrease in project duration is not possible, because either the network reaches the compression limit or the cost of crashing is more than the amount of saving in return.
The following terms should be conceptualized before actually crashing the network.
Normal Cost (Nc): It is the lowest cost of completing an activity in the minimum time, employ- ing normal means, i.e., not using overtime or other special resources.
Normal Time (NT): it is the minimum time required to achieve the normal cost.
Crash Cost (Cc): It is the least cost of completing an activity by employing all possible means like overtime, additional machinery, proper materials, etc.
Crash Time (CT): It is the absolute minimum time associated with the crash cost.
Cc
Nc C
N
CT NT
Cost-time slope of activity, s = (Cc – Nc)/(NT – CT)
As seen from the figure above, as the time to complete a project is reduced the cost involved gets increased. There are two types of cost in a project: Direct and Indirect or overheads. When the project time is reduced, the indirect cost decreases while the direct cost increases.
Example 9.3. For the given information in Table 9.3, the critical path of the project is 36 weeks. The builder wants to complete the project of house construction in just 30 weeks. How much extra cost would be incurred to complete the house by this time.
Table 9.3
Activity Description Normal time Crash time Normal cost Crash cost Allowable
(weeks) (weeks) ($) ($) crash time (weeks)
1-2 Design house 12 7 3000 5000 5
& get financing
2-3 Lay foundation 8 5 2000 3500 3
2-4 Order materials 4 3 4000 7000 1
3-4 Dummy 0 0 0 0 0
4-5 Select paint 4 1 500 1100 3
4-6 Build house 12 9 50000 71000 3
5-6 Select carpet 4 1 500 1100 3
6-7 Finish work 4 3 15000 22000 1
Sum 75000 110700
Solution: Let
TN = Normal time (weeks) TC = Crash time (weeks) CN = Normal cost ($) CC = Crash cost ($)
TN – TC = allowable crash time (weeks) – (CN – CC)/(TN – TC) = Crash cost/ week
If we assume that the relationship between crash cost and crash time is linear, then activity 1-2 can be crashed by any amount of time (not exceeding the maximum allowable crash time) at a rate of $400 per week as shown in Table 9.4.
The critical path of the network = 1-2-3-4-6-7 and the project duration is 36 weeks.
Table 9.4
Activity TN TC CN ($) CC ($) (TN – TC) (CN – CC)/(TN – TC)
(week) $/week
1-2 12 7 3,000 5,000 5 400
2-3 8 5 2,000 3,500 3 500
2-4 4 3 4,000 7,000 1 3,000
3-4 0 0 0 0 0 0
4-5 4 1 500 1,100 3 200
4-6 12 9 50,000 71,000 3 7,000
5-6 4 1 500 1,100 3 200
6-7 4 3 15,000 22,000 1 7,000
Σ 75,000 110,700
1 2
3
4 6 7
5 Figure 9.3
• The objective of project crashing is to reduce the project duration while minimizing the cost of crash- ing. Since, the project completion time can be shortened only by crashing activities on the critical path, it may turn out that not all activities have to be crashed. However, as the activities are crashed, the CP may change, requiring crashing of previously non-critical activities to further reduce the project com- pletion time.
• We start the crashing process by looking at the CP and seeing which activity has the minimum crash cost per week. We see from the Table-9.4 that activity 1-2 has the minimum crash cost of $400 (exclud- ing the dummy activity 3-4, which cannot be reduced). Thus activity 1-2 will be reduced as much as possible (5 weeks), but we can reduce activity 1-2 only to the point where another path becomes critical. When two paths simultaneously become critical, activities on both must be reduced by the same amount.
• If we reduce the activity time beyond the point where another path becomes critical, we may be incur- ring an unnecessary cost. This means we must keep a watch on the total network, when we reduce the individual activities. This makes the manual crashing very cumbersome.
• It turns out that activity 1-2 can be crashed by the total amount of 5 weeks without another paths becoming critical, since activity 1-2 is common in all four paths in the network. Crashing this activity gives the new project duration of 31 weeks at a crashing cost of $2,000.
• The process must now be repeated. The CP remains the same, and the minimum CC on the CP is $500 for activity 2-3, which can be crashed a total of 3 weeks, but since the contractor only desires to crash the network to 30 weeks, we only need to crash activity 2-3 by one week. Crashing activity 2-3 to 7 weeks (i.e., a one week reduction) costs $500 and reduces the project duration to 30 weeks.
• The total cost of crashing the project to 30 weeks is $2500.
Example 9.4. Given the following network and PERT activity time estimates, determine the expected project completion time and variance, and the probability that the project will be completed in 28 days or less.
Table 9.5a
Time estimates (weeks)
Activity to tm tp
1-2 5 8 17
1-3 7 10 13
2-3 3 5 7
2-4 1 3 5
3-4 4 6 8
3-5 3 3 3
4-5 3 4 5
Solution:
2
1 4
3
5
Figure 9.4
Compute the expected activity times and variances using the formulae discussed before. These are shown in the Table 9.5b.
Table 9.5b
Activity Te = (to + 4tm + tp)/6 Vt = {(tp – to)/6}2
1-2 9 4
1-3 10 1
2-3 5 4/9
2-4 3 4/9
3-4 6 4/9
3-5 3 0
4-5 4 1/9
Determine the earliest and latest time at each node as shown below in the order of (ES, EF, LS, LF).
2
1 4
3
5 [9, 12, 17, 20]
[14, 17, 21, 24]
[9, 14, 9, 14]
[20, 24, 20, 24]
[0, 9, 0, 9]
Figure 9.5
• Identify the critical path and compute expected project completion time.
Slack = (LS-ES) or (LF-EF). Path joining the activities with slack = 0 is called the critical path. Thus the critical path is 1-2-3-4-5 and the project duration is 24 days
Variance of CP = VCP = V1–2 + V2–3 + V3–4 + V4–5 = 4 + 4/9 + 4/9 + 1/9 = 5 days
• Determine the probability that the project will be completed in 28 days or less.
Compute Z using the formula Z = (x-à)/σ = (28 – 24)/ 5 =4/ 5 = 1.79
The corresponding probability is 0.4633; thus P(x ≤ 28) = 0.50 + 0.4633 = 0.9633
Example 9.5. Three different time estimates for various activities of a project are given in Table 9.6. Construct a PERT network and find out (a) The earliest possible time (TE) to complete the different activities, (b) The latest allowable time (TL) for them, (c) The slack values, (d) The critical paths, and (e) The probability factor for completing the project in 30 weeks.
Table 9.6
Activity i – j Optimistic time Most likely time Pessimistic time
1 – 2 1 2 3
2 – 3 1 2 3
2 – 4 1 3 5
3 – 5 3 4 5
4 – 5 2 3 4
4 – 6 3 5 7
5 – 7 4 5 6
6 – 7 6 7 8
7 – 8 2 4 6
7 – 9 4 6 8
8 – 10 1 2 3
9 – 10 3 5 7
Solution: The calculations are shown in Table 9.7
Table 9.7
i – j to tm tp te σt Vt C.P.
1 – 2 1 2 3 2.00 0.33 0.11 0.11
2 – 3 1 2 3 2.00 0.33 0.11
2 – 4 1 3 5 3.00 0.67 0.44 0.44
3 – 5 3 4 5 4.00 0.33 0.11
4 – 5 2 3 4 3.00 0.33 0.11
4 – 6 3 5 7 5.00 0.67 0.44 0.44
5 – 7 4 5 6 5.00 0.33 0.11
6 – 7 6 7 8 7.00 0.33 0.11 0.11
7 – 8 2 4 6 4.00 0.67 0.44
7 – 9 4 6 8 6.00 0.67 0.44 0.44
8 – 10 1 2 3 2.00 0.33 0.11
9 – 10 3 5 7 5.00 0.67 0.44 0.44
VCP 2.00
σCP 1.41
3 5 7 8
10 2
1
4 6
9
2 2
3 3
4 5 4
5
7 6
5 (0, 0) (2, 2) 2
(4, 8) (8, 12) (17, 17) (21, 26)
(28, 28)
(23, 23) (5, 5) (10, 10)
(Te, TL)
Te : earliest possible time TL : latest allowable time
Figure 9.6 (iv) Critical Path
1–2–4–6–7–9–10 = 28 weeks.
VCP = 2 and σCP = 1.41
(v) The probability factor for completing the project in 30 weeks.
Z = X− à
σ where X = 30 weeks, à = 28 weeks and σCP = 1.41 Hence, Z = 30 28
1 41
−
. = 1.41
Z is the number of standard deviations by which X exceeds à à = Te is the total project duration, and
X = due or scheduled date or time
From the table of Standard Normal Distribution Function, the corresponding percentage prob- abilities are given as follows:
For Z = 1.41 the probability is 92.07 % and for Z = 1.42 the probability is 92.22 %.
Interpolation
1.41 ………..0.9207 1.414 …………. p 1.42 ………..0.9222
1 414 1 41 1 42 1 41
0 9207 0 9222 0 9207
. .
. .
.
. .
−
− = −
− p
p = ( . . )( . . )
( . . )
1 414 1 41 0 9222 0 9207 1 42 1 41
− −
− + 0.9207 = 0.9213
Therefore, there is 92.13% probability that the project will be finished in 30 weeks.
Example 9.6. A reactor and storage tank are interconnected by a 3” insulated process line that needs periodic replacement. There are valves along the lines and at the terminals and these need replacing as well. No pipe and valves are in stock. Accurate, as built, drawings exist and are available. You are the maintenance and construction superintendent responsible for this project. The works engineer has requested your plan and schedule for a review with the operating supervision. The plant methods and standards section has furnished the following data. The precedents for each activity have been determined from a familiarity with similar projects.
Table 9.8
Symbol Activity Description Precedent Time (Hrs.)
A Develop required material list — 8
B Procure Pipe A 200
C Erect Scaffold — 12
D Remove Scaffold I, M 4
E Deactivate Line — 8
F Prefabricate Sections B 40
G Place New Pipes F, L 32
I Fit up Pipe and Valves G, K 8
J Procure Valves A 225
K Place Valves J, L, F 8
L Remove old pipe and valves C, E 35
M Insulate G, K 24
N Pressure Test I 6
O Clean-up and Start-up D, N 4
(i) Sketch the arrow diagram of this project plan.
(ii) Make the forward pass and backward calculations on this network, and indicate the critical path and its length.
(iii) Calculate total float and free float (both early and late) for each of the non-critical activities.
The network (arrow diagram) is given in the figure below.
7
2 5 6 8 9 11 12
1 4
3
10 (0, 0)
(8, 8)
(12, 213)
(8, 213)
(248, 272)
(208, 208)
(248, 248) (280, 280)
(288, 288) (308, 308)
(312, 312)
(304, 304) A
B
C D
F G I N O
J K
0
200 40 32 8 6 4
M 4 12 24
8E 0
L 35
Critical path Figure 9.7
Critical Path: 1 - 2 - 5 - 6 - 8 - 10 - 11 - 12 = 312 Hrs.
Table 9.9
Activity Precedence Earliest Latest Earliest Latest Total Slack Free slack Start ES Start LS Finish EF Finish LF (LF-EF) (FS)
(hrs) (hrs) (hrs) (hrs) (hrs)
1–2 — 0 0 8 0 8 8–8 = 0
2–5 A 8 8 208 208 0 208–208 = 0
1–3 — 0 205 8 213 205 8–8 = 0
3–4 — 8 213 8 213 205 12–8 = 0
10–11 I M 304 304 308 308 0 308–308 = 0
1–4 — 0 201 12 213 201 12–12 = 0
5–6 B 208 208 248 248 0 248–248 = 0
6–8 F L 248 248 280 280 0 280–280 = 0
8–9 G K 280 294 288 302 14 288–288 = 0
2–7 A 8 34 233 272 26 248–233 = 15
6–7 — 248 272 248 272 24 248–248 = 0
7–8 J L F 248 272 256 280 24 280–256 = 24
4–6 C E 12 213 47 248 201 248–47 = 201
8–10 G K 280 280 304 304 0 304–304 = 0
9–10 — 288 304 288 304 16 304–288 = 16
9–11 I 288 302 294 308 14 308–294 =14
11–12 D N 308 308 312 312 0 0
Example 9.7. The following Table-9.10 gives the activities in a construction project with other relevant information.
(a) Draw the network of the project.
(b) Crush the project step-by-step until the shortest duration is reached.
Table 9.10
Activity Preceding activity TN (days) TC (days) CN ($) CC ($)
1–2 — 20 17 600 720
1–3 — 25 25 200 200
2–3 1–2 10 8 300 440
2–4 1–2 12 6 400 700
3–4 1–3, 2–3 5 2 300 420
4–5 2–4, 3–4 10 5 300 600
Solution:
(a) The Project Network is shown in the following figure using Normal Time (TN).
2
1
4
3
5 (20, 20) (35, 35)
(0, 0)
(30, 30)
(45, 45) 25
20 10
12 5
10
Figure 9.8 CP1 = 1 – 2 – 3 – 4 – 5.
CP2 = 1 – 2 – 4 – 5.
Normal duration of the project is 45 days.
(b) Compute the different min. cost schedule that can occur between normal and crash times, which are dependent on the cost time slopes for different activities. These are computed by the formula :
Cost slope = Crash cost Normal cost Normal time Crash time
−
−
The slopes of the activities of the above network are calculated as shown in Table-9.11:
Table 9.11
Activity 1-2 1-3 2-3 2-4 3-4 4-5
Slope (∆C/∆T) 40 0 70 50 40 60
Total Direct Cost (TC1) = ΣCN = $2100.
Step 1. Since the project duration is controlled by the activities on the critical path, the duration of some activities lying on the critical path is reduced. First the duration of that activity is reduced which has the minimum cost slope. Since the activities (1, 2) and (3, 4) both give the minimum cost slope, the duration of activity (1, 2) is compressed from 20 to 17 days with an additional cost of $ 3 × 40 = $120.
Thus the revised schedule corresponds to 42 days with a cost of TC2 = $(2100 + 120) = $2220.
Thus the revised schedule corresponds to 42 days with a cost of TC2 = $(2100 + 120) = $2220.
Step 2. Further if the duration of activity (3, 4) of critical path is reduced from 5 days to 2 days = 3 days, the two critical paths of the project become the same. Therefore, we crash activity (3, 4) by 3 days with an additional cost of $ 3 × 40 = $120.
Thus the revised schedule corresponds to 39 days with a cost of TC3 = $(2220 + 120) = $2340.
It should be noted that I have chosen the activity (3, 4) because in doing so two parallel critical paths 1 → 2 → 4 → 5 and 1 → 2 → 3 → 4 → 5 are obtained.
Step 3. Now both CP1 and CP2 are equal and there are two possibilities of crashing the project duration.
• Crashing activity (4, 5) by 10 days – 5 days = 5 days at a cost of $60 per day, and
• Crashing both activities (2, 3) and (2, 4) by 2 days at a cost of
$70 + $50 = $120.
Therefore, we crash activity (4, 5) by 5 days with an additional cost of $ 5 × 60 = $300.
Thus the revised schedule corresponds to 34 days with a cost of TC4 = $(2340 + 300) = $2640.
Step 4. The remaining projects that can be crashed are both activities (2, 3) and (2, 4) at the same time.
Activity (2,3) can be crushed by TN – TC = 10 – 8 = 2 days Activity (2,4) can be crushed by TN – TC = 12 – 6 = 6 days
Therefore, we can only crush both activities by 2 days at a cost of ($70 + $50) × 2 = $240.
Thus the revised schedule corresponds to 32 days with a cost of TC5 = $(2640 + 240) = $2880.
Conclusion: Just by paying ∆C = TC5 – TC1 = $2880 – $2100 = $780, The project duration can be reduced from 45 days to 32 days or by 13 days.
Example 9.8. In the project network shown in the Figure 9.9 below, the nodes are denoted by numbers and the activities by letters. The normal and crash durations of the various activities along with the associated costs are shown below in Table 9.12.
4
1
2 3
5 6
12
15
10
8 14 A 16
B
C F E
D
Figure 9.9 Table 9.12
Activity Normal Duration Normal cost Crash duration Crash cost
(days) ($) (days) ($)
A 8 1800 6 2200
B 16 1500 11 2200
C 14 1800 9 2400
D 12 2400 9 3000
E 15 800 14 2000
F 10 2000 8 4000
Determine the least cost for 36 days schedule.
Solution : First assume that all activities occur at normal times. Then the following network shows the critical path computations under normal conditions.
4
1
2 3
5 6
12
15
10
8 14 A 16
B
C F E
D
E4 = 12 L4 = 23
E1 = 0 L1 = 0
E5 = 38 L5 = 38
E6 = 48 L6 = 48
E3 = 24 L3 = 24 E2 = 8
L2 = 8
Figure 9.10
The critical path is A → B → C → F. The schedule of the project is 48 days and its associated normal cost becomes:
= $(1800 + 1500 + 1800 + 2400 + 800 + 2000 ) = $ 10,300.
The different minimum cost schedule that can occur between normal and crash times, which are mainly dependent on the cost time slopes for different activities. The cost time slopes can be computed by the formula:
Cost – time slope = Crash cost Normal cost Normal time Crash time
−
− The cost slopes for the activities of the above network are:
Table 9.13
Activity A B C D E F
Slope 200 140 120 200 1200 1000
Now we proceed step-by-step as follows:
Step-1. Since the present schedule consumes more time, the schedule can be reduced by crashing some of the activities. Since the project duration is controlled by the activities lying on the critical path, the duration of some activities on the critical path is reduced. First reduce the duration of that activity which involves minimum cost. Activity C with minimum slope gives the minimum cost. So the dura- tion of activity C is compressed from 14 days to 11 days with an additional cost $ 5 × 120 = $600. There fore, the new schedule corresponds to 43 days with a cost of $(10,300 + 600) = $10,900.
Step-2. Now, it can be observed that the present schedule still consumes more time and also not all the activities on the critical path are at their crash durations. Hence the project duration can be reduced by crashing some other activity. Out of the remaining activities on the critical path, activity B has the least slope. So reduce the duration activity B from 16 days to 11 days at a cost of $ 5 × 140 = $700. Thus the new project duration becomes 38 days with a cost of $(10,900 + 700) = $11,600.
Step 3. This project duration is still more than the required duration of 36 days. So select some other activity lying on the critical path for crashing. Obviously, only activities A and F on the critical path can be considered for crashing. Since activity A has the smaller slope, the duration of A can be compressed.
Compress A only by one day although it can be compressed by two days (from 8 to 6 days). Because, the path 1 → 4 → 5 → 6 becomes a parallel critical path as soon as A is compressed by one day. Thus the new schedule corresponds to 37 days with a cost of $(11,600 + 200) = $11,800.
Step 4. Since only 36 days schedule is required, compress some activity by one day. To do so compress one day in each of the two parallel critical paths. So there are three choices:
• Activity F can be compress by one day at a cost of $1000.
• Activities A and D can be compressed by one day each (since B and C are already at their crash points). This gives the total cost of $(200 + 200) = $400.
• Activities A and E can be compressed by one day each at a total cost of $(200 + 1200) =
$1400.
But, the second choice gives the least cost schedule and hence it should be selected. This involves a 36 days schedule with a cost of $(11,800 + 400) = $ 12,200.