Crashing (VCE SSCE General Mathematics): Revision Notes

Worked Example: Analyzing Crashing Options

If we want to reduce the overall completion time, we need to consider which activities to crash:

Activities A, B, or C: These are not on the critical path. They already have slack time, so reducing their duration won't shorten the project's total time. Crashing these would be ineffective and wasteful.

Activities D or E: These are on the critical path. Reducing either of these activities will reduce the overall project time.

Example: If activity D is reduced from $6$ hours to $4$ hours, the project completion time decreases from $13$ hours to $11$ hours. This achieves the desired outcome.

Worked Example: Optimal Crashing Strategy

Part a: Identifying the critical path

To find the critical path, list all possible routes from Start to Finish and calculate each path's total duration:

The critical path is C-F-G with a duration of $16$ days (the longest path).

Part b: Minimum completion time

The minimum completion time is $16$ days (the duration of the critical path).

Part c: New completion time after crashing

Activity F can be reduced by a maximum of $3$ days at a cost of $100 per day.

Let's examine what happens if we crash activity F by different amounts:

Analysis:

• If we crash activity F by the maximum $3$ days, the path C-F-G reduces from $16$ to $13$ days
• However, this creates a new critical path: A-D-H with $14$ days
• The overall project time becomes $14$ days (limited by the new critical path)

Strategic decision:

Crashing activity F by only $2$ days is more efficient:

• Path C-F-G becomes $14$ days
• This equals the path A-D-H ($14$ days)
• We now have two equal critical paths
• We achieve the same result ($14$ days total) but save $100 compared to crashing by $3$ days

The new minimum completion time is $14$ days.

Part d: Minimum cost for maximum reduction

To achieve the new minimum completion time of $14$ days, we need to crash activity F by $2$ days only.

Cost calculation:

$\text{Cost} = 2 \text{ days} \times \$100 \text{ per day} = \$200$

The minimum cost is $200.

Worked Example: Complex Multi-Activity Crashing

Part a: New minimum completion time

First, list all possible paths and their durations:

After applying the maximum reduction to all five activities:

• Path A-C-E becomes the new critical path
• The new minimum completion time is $25$ days

Part b: Minimum cost for maximum reduction

Now we need to determine the cheapest combination of crashes that achieves the $25$-day completion time.

Step-by-step approach:

Step 1: Consider the new critical path A-C-E

• Activity E must be crashed by $2$ days to achieve the $25$-day target
• Cost: $2 \times `$2000 = $`4000$

Step 2: Ignore path A-F-H

• This path only takes $19$ days
• It's already well below the critical path duration
• No crashing needed

Step 3: Consider path B-D-F-H ($28$ days)

• This must be reduced to $25$ days (a reduction of $3$ days)
• Options: Crash B by $2$ and H by $1$, or crash B by $1$ and H by $2$
• Activity B costs $1500 per day; activity H costs $900 per day
• Although H is cheaper per day, crashing B is strategic because...

Strategic insight: Crashing activity B affects two paths (B-D-F-H and B-G-I)

• This makes crashing B more cost-effective overall
• Decision: Crash B by $2$ days and H by $1$ day
• Cost: $(2 \times `$1500) + (1 \times $`900) = `$3900$`

Step 4: Consider path B-G-I ($28$ days)

• Already reduced by $2$ days due to crashing B (now $26$ days)
• Needs one more day of reduction to reach $25$ days
• Options: Crash G or I
• Activity G costs $700 per day; activity I costs $800 per day
• Choose G (cheaper)
• Cost: $1 \times `$700 = $`700$

Total cost calculation:

The minimum cost to achieve maximum reduction is $8600.