Project Crashing | Sahithyan's Notes

Project crashing reduces a project’s duration by shortening critical activities. It increases direct cost in exchange for time saved.

The project duration is the duration of its longest path. A path with this duration is critical. Crashing an activity on a shorter path does not immediately reduce the project duration. The critical path method identifies which activities can reduce the project duration.

Activity Limits

Each activity has a normal duration and a minimum duration.

To crash an activity is to shorten its duration below its normal time, usually by using additional resources or faster methods. This increases the activity’s direct cost. Its duration cannot be reduced below its crash time.

Normal time
Activity duration under regular execution.
Crash time
Shortest achievable activity duration.

The crash allowance is the maximum time reduction for an activity.

\text{Crash allowance} = \text{Normal time} - \text{Crash time}

For activity $i$ , its reduction $x_i$ must satisfy:

0 \leq x_i \leq t_i^{N} - t_i^{C}

Here:

$x_i$: time reduction for activity $i$
$t_i^{N}$: normal time for activity $i$
$t_i^{C}$: crash time for activity $i$

Crashable Activity

An activity can still be crashed when its current duration is greater than its crash time.

t_i > t_i^{C}

Here:

$t_i$: activity’s current duration
$t_i^{C}$: activity’s crash time

If $t_i = t_i^{C}$ , the activity has no remaining crash allowance.

Cost Slope

The cost slope is the additional direct cost per unit of activity time reduced. A constant slope assumes each reducible time unit has the same cost.

\text{Cost slope}_i = -\frac{C_i^{N} - C_i^{C}} {t_i^{N} - t_i^{C}}

Here:

$C_i^{N}$: direct cost of activity $i$ at normal time
$C_i^{C}$: direct cost of activity $i$ at crash time

An activity can only be crashed when its duration is above its crash time. The direct cost of reducing activity $i$ by $x_i$ time units is:

\text{Added direct cost}_i = x_i \times \text{Cost slope}_i

Crashing Procedure

Use these steps for each reduction:

Find critical paths.
Calculate the duration of every path. Mark every longest path as critical.
Find crashable activities on each critical path.
On each critical path, compare each activity’s current duration with its crash time. Keep activities whose current duration is greater.
Choose activities to crash.
Choose low-cost activities that shortens all the critical paths. One activity may shorten several paths. Otherwise, choose a combination with at least one activity from each critical path and add their cost slopes.
Crash activities.
Reduce the chosen activity durations. Recalculate path durations as soon as an activity reaches its crash time, another path becomes critical, or the target duration is reached.
Repeat from step 1.
Stop when the target or cost limit is reached, or when a critical path has no crashable activity.

Crashing one time unit means reducing an activity’s duration by one unit. The project duration only falls if this reduction shortens every path that currently has the longest duration. A new critical path can appear after a reduction, so recalculate before choosing the next crash action.

Project Cost

Total direct cost equals the normal direct cost plus the added cost of crashing activities.

C_D = \sum_i C_i^{N} + \sum_i x_i s_i

Here:

$C_D$: total direct cost
$x_i$: time reduction for activity $i$
$s_i$: cost slope of activity $i$

Indirect costs are project-wide expenses that continue while the project is in progress, such as site rent, utilities, and administration. They depend on project duration, rather than on the duration of a specific activity.

If indirect costs accrue at a constant rate $r$ per project time unit, total project cost is:

C_T = C_D + rT

Here:

$C_T$: total project cost
$r$: indirect cost rate per time unit
$T$: project duration

Reducing the project by one time unit is economically justified when the added direct cost does not exceed the indirect cost saved. Deadline penalties can be included as additional indirect costs.

Crashing Offer

A crashing offer gives a maximum net additional cost for reducing the project duration. The net additional cost compares added direct cost with indirect cost saved.

C_{\text{net}} = \Delta C_D - r\Delta T

Here:

$C_{\text{net}}$: net additional cost
$\Delta C_D$: added direct cost
$\Delta T$: reduction in project duration

The indirect cost saving depends on project time saved, not on the number of activities crashed.

Worked Example

A project has 3 paths. Its normal duration is $10$ days, the duration of its longest path.

Path	Activities	Normal Duration
$P_1$	$A \to C$	$10$
$P_2$	$B \to D$	$9$
$P_3$	$E$	$8$

The activity data are:

Activity	Normal Time	Crash Time	Normal Cost	Crash Cost	Cost Slope per Day
$A$	$6$	$5$	Rs. $40{,}000$	Rs. $70{,}000$	Rs. $30{,}000$
$B$	$5$	$4$	Rs. $30{,}000$	Rs. $50{,}000$	Rs. $20{,}000$
$C$	$4$	$3$	Rs. $20{,}000$	Rs. $60{,}000$	Rs. $40{,}000$
$D$	$4$	$3$	Rs. $35{,}000$	Rs. $85{,}000$	Rs. $50{,}000$

Indirect cost is Rs. $50{,}000$ per day. Find the maximum possible duration reduction and determine whether it lowers total project cost.

Solution Reveal Hide

The initial path durations are $P_1=10$ , $P_2=9$ , and $P_3=8$ days. Only $P_1$ is critical. Activities $A$ and $C$ can shorten $P_1$ . Activity $A$ costs Rs. $30{,}000$ per day to crash, while $C$ costs Rs. $40{,}000$ per day. Choose $A$ .
Crash $A$ by $1$ day, from $6$ to $5$ days. Then $P_1$ falls from $10$ to $9$ days. The paths are now $P_1=9$ , $P_2=9$ , and $P_3=8$ . Both $P_1$ and $P_2$ are critical. The project duration has fallen by $1$ day, and the added direct cost is Rs. $30{,}000$ .
To reduce the project duration again, shorten both critical paths. Activity $A$ is already at its crash time, so choose $C$ on $P_1$ and $B$ on $P_2$ . Their combined cost per day is Rs. $40{,}000 + \text{Rs. }20{,}000 = \text{Rs. }60{,}000$ .
Crash $C$ and $B$ by $1$ day each. The path durations become $P_1=8$ , $P_2=8$ , and $P_3=8$ . Added direct cost for this step is Rs. $60{,}000$ , bringing the total added direct cost to Rs. $90{,}000$ .
All 3 paths are now critical. Path $P_3$ contains only activity $E$ , which has no crash data and therefore cannot be shortened. The project cannot be reduced below $8$ days.

The project duration falls from $10$ to $8$ days. This saves $2 \times \text{Rs. }50{,}000 = \text{Rs. }100{,}000$ in indirect cost. The added direct cost is Rs. $90{,}000$ .

C_{\text{net}} = \text{Rs. }90{,}000 - \text{Rs. }100{,}000 = -\text{Rs. }10{,}000

The $8$ -day plan reduces total project cost by Rs. $10{,}000$ .