Crashing the project

Covers AS91576

What crashing means

Crashing a project means shortening it by speeding up one or more activities — usually by paying more (extra crew, overtime, express delivery).
Each activity may have a crash time (its shortest possible duration) and a crash cost per day (what one day saved costs).
The goal is to reach a target completion time for the least extra cost, or to find the cheapest way to save time.

Only the critical path controls the time

Shortening a non-critical activity does not shorten the project — the critical path still runs the same length.
To reduce the project duration you must shorten an activity on the critical path.
Speeding up F (float 1) or C (float 5) in the hall build saves no project time; you must crash one of A, B, D, E or G.

Crash the cheapest critical activity first

To save time most cheaply:
- List the critical activities and their crash cost per day.
- Crash the cheapest critical activity first, one day at a time.
- Recheck the network after each day — the critical path can change.

When the critical path changes

As you crash the critical path, a parallel path may catch up and become critical too.
Once two paths are both critical, saving another day means shortening both at once — often both activities in the same time slot — which costs more.
Stop when the target is met, or when no critical activity can be crashed further (all at their crash times).

Worked ExampleCrashing the hall build

The hall build takes 19 days on the critical path $A\to B\to D\to E\to G$ . F (float 1) lies on the parallel path $A\to B\to D\to F\to G$ . Each day saved costs: A $400, B $300, D $500, E $250, G $350 (each can be crashed by at least one day). Find the cheapest way to finish the project one day early, and comment on what happens next.

Step 1 — Only crash the critical path

Reducing the project below 19 days needs a critical activity, so choose from A, B, D, E, G. C and F have float and would save nothing.

Step 2 — Pick the cheapest critical activity

The crash costs per day are A $400, B $300, D $500, E $250, G $350. The cheapest is E at $250 per day.

Step 3 — Crash E by one day and recheck

Cutting E from 3 to 2 days shortens the critical path to $3+4+5+2+4 = 18$ days.

Check the parallel path through F: $A\to B\to D\to F\to G = 3+4+5+2+4 = 18$ days.

Step 4 — Note the changed structure

Both paths are now 18 days, so F's float has fallen to 0 — the network now has two critical paths ( $\ldots\to E\to G$ and $\ldots\to F\to G$ ).

$\boxed{\text{Crash E by 1 day for \$250} \Rightarrow 18 \text{ days}}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A project's critical path is P → Q → R. A manager crashes activity S, which has a float of 3, to save time. Explain why the project completion time does not change.

Merit

In the hall build (critical path A-B-D-E-G, 19 days), the crash costs per day are A $400, B $300, D $500, E $250, G $350. Find the cheapest single day to save and the new duration, and state which activity you crash.

Excellence

Continuing the hall build after E is crashed to 2 days (project 18 days, with A-B-D-E-G and A-B-D-F-G both critical), the crash costs per day are A $400, B $300, D $500, E $250 (no further), F $200, G $350. Devise the cheapest way to save one more day (to 17 days) and justify it against the alternatives.