Topics

Scheduling, resources and shortening the project

Covers AS91576

From analysis to a schedule

The network says what is possible; the schedule says what will happen.
A schedule assigns each activity a start date, choosing within its float where there is any.
Two natural extremes:
- Earliest-start schedule — everything begins as soon as it can. Safest, since all float is kept in reserve.
- Latest-start schedule — everything begins as late as it can. Cheapest for cash flow, since money is spent later, but every activity becomes critical and there is no protection against delay.
Most real schedules sit between the two, using float deliberately to smooth resources.

The Gantt chart

A Gantt chart plots time along the horizontal axis and one row per activity.
Each activity is a bar from its start to its finish, with its float shown as a lighter extension.
Critical activities have no extension, so the critical path is visible as an unbroken run of solid bars from left to right.

What the chart shows that the network does not:
- What is running at the same time — read a vertical line down the chart
- When the workload peaks
- Where float sits, and therefore where activities could be moved

Resource scheduling

The network assumes unlimited resources. In reality, two activities needing the same crew cannot both run on Tuesday.
The method:
- Draw the Gantt chart on an earliest-start basis.
- Add up the resource needed on each day, and plot that as a resource histogram.
- Where a peak exceeds what is available, move non-critical activities within their float to flatten it.
- If flattening is impossible without moving a critical activity, the project must get longer — and that is a genuine finding to report.
Only float can absorb resource clashes. This is what float is for, and it is the main reason for calculating free float separately.

Shortening the project

Only the critical path controls the duration, so shortening anything else achieves nothing.
The procedure:
- Shorten a critical activity by one time unit (more staff, overtime, plant hire).
- Recalculate both passes. The project falls by one unit — unless another path becomes critical.
- Repeat, watching for the moment a second path becomes critical.
Once two paths are critical, both must be shortened together to gain anything further. That is where the cost of acceleration rises sharply, and saying so is a Merit/Excellence-level observation.
There is a floor. Eventually every path is critical, or an activity cannot be shortened further, and no amount of money buys more time.

When a deadline is imposed

If the project duration exceeds a required deadline, the shortfall is the finding, and the report should say:
- by how much the project is late,
- which critical activities could be shortened, and
- what happens to the critical path if they are.
Do not report a negative float. If a deadline earlier than the computed duration is imposed, the correct statement is "the project is 3 days over the deadline", not "float $= -3$ ".

What the model leaves out

Deterministic durations. Every duration is treated as exactly known. Real tasks vary, and a near-critical path with small float can become critical through ordinary variability.
Finish-to-start only. The model assumes an activity cannot begin until its predecessor is entirely complete. Real projects often overlap tasks.
Unlimited resources, unless resource scheduling is done separately.
No cost. The network optimises time alone; the cheapest schedule and the fastest schedule are rarely the same.
Naming one of these limits in the report is a genuine and quick way to show critical evaluation.

Tips and Tricks

Draw the Gantt chart on earliest starts and show float as a lighter bar. It makes the critical path visible as a solid unbroken chain and shows instantly what could be moved.

Only critical activities can shorten the project. Before recommending overtime, check the activity is on the critical path — otherwise the money buys nothing but extra float.

Recalculate after every change. Shortening a critical activity by 3 days rarely saves 3 days, because another path takes over part-way. Re-running both passes is the only reliable way to find out.

A latest-start schedule leaves no protection. It is worth stating as a risk: every activity becomes critical, so any delay anywhere delays the project.

Report a missed deadline as a shortfall, not as negative float. "Four days over" is a finding; "float = −4" is a symptom being reported as a number.

Name one limitation of the model — fixed durations, no resource limits, or no overlap. One sentence is enough, and it is exactly the critical evaluation the standard asks for.

Worked ExampleMeeting a deadline on the deck project

The deck project takes 26 days with critical path $A \to C \to D \to F \to G \to J$ . The client needs it finished in 22 days. The builder can shorten activities by hiring extra labour:

Activity	Duration	Can be shortened by	Cost per day saved
C (clear site)	4	2 days	$400
D (foundations)	6	3 days	$900
F (frame)	7	3 days	$700
G (decking)	5	2 days	$500

Determine the cheapest way to meet the deadline, and state what the critical path becomes.

Step 1 — Establish what must be saved

$26 - 22 = 4 \text{ days}$

Step 2 — Identify the competing paths, before shortening anything

The critical path $A\,C\,D\,F\,G\,J$ : $3 + 4 + 6 + 7 + 5 + 1 = 26$ days.

The parallel route through B and E replaces C and D with B and E:

$A\,B\,E\,F\,G\,J = 3 + 5 + 2 + 7 + 5 + 1 = 23 \text{ days}$

The route through H instead of G:

$A\,C\,D\,F\,H\,(\text{dummy})\,J = 3 + 4 + 6 + 7 + 3 + 0 + 1 = 24 \text{ days}$

$A\,B\,E\,F\,H\,J = 3 + 5 + 2 + 7 + 3 + 1 = 21 \text{ days}$

Step 3 — Rank the shortenable activities by cost

Activity	Cost per day	Max days	On which paths?
C	$400	2	the $C,D$ routes only
G	$500	2	the $G$ routes only
F	$700	3	every path
D	$900	3	the $C,D$ routes only

Step 4 — Take the cheapest saving first: C, 1 day at $400

New path lengths: $A\,C\,D\,F\,G\,J = 25$ ; $A\,C\,D\,F\,H\,J = 23$ ; $A\,B\,E\,F\,G\,J = 23$ .

Project: 25 days. Still 3 to save.

Step 5 — Take C's second day, also $400

New lengths: $A\,C\,D\,F\,G\,J = 24$ ; $A\,B\,E\,F\,G\,J = 23$ ; $A\,C\,D\,F\,H\,J = 22$ .

Project: 24 days. Running total $800. C cannot be shortened further.

Step 6 — Next cheapest is G at $500

Shorten G by 1 day. $G$ appears on both the 24-day and 23-day routes:

$A\,C\,D\,F\,G\,J = 23$
$A\,B\,E\,F\,G\,J = 22$
$A\,C\,D\,F\,H\,J = 22$ (unchanged — H route)

Project: 23 days. Running total $1,300.

Step 7 — One more day needed, and now TWO paths are at 23 and 22

Shorten G again? $A\,C\,D\,F\,G\,J$ falls to 22, but $A\,C\,D\,F\,H\,J$ is already at 22 — so the project stops at 22.

$\text{Project} = \max(22,\ 22,\ 21,\ 21) = 22 \ \checkmark$

Cost of G's second day: $500. Running total $1,800.

Step 8 — Check the alternative: use F instead of G's second day

F is on every path. Shortening F by 1 day at $700 gives:

$A\,C\,D\,F\,G\,J = 22$
$A\,C\,D\,F\,H\,J = 21$
$A\,B\,E\,F\,G\,J = 21$

Project: 22 days, at a cost of $\$800 + \$500 + \$700 = \$2{,}000$ .

Step 9 — State the cheapest solution

Step 10 — Verify by recomputing every path

Path	Original	Shortened	Length
$A\,C\,D\,F\,G\,J$	$3,4,6,7,5,1$	$3,\mathbf{2},6,7,\mathbf{3},1$	22
$A\,C\,D\,F\,H\,J$	$3,4,6,7,3,1$	$3,\mathbf{2},6,7,3,1$	22
$A\,B\,E\,F\,G\,J$	$3,5,2,7,5,1$	$3,5,2,7,\mathbf{3},1$	21
$A\,B\,E\,F\,H\,J$	$3,5,2,7,3,1$	unchanged	21

$\text{Project duration} = 22 \text{ days} \ \checkmark$

Step 11 — State the new critical path(s)

H, which previously had 2 days of float, is now critical. The handrail crew no longer has any flexibility.

Step 12 — Advise the client

Two consequences the client should know:

The project is now far more fragile. With two critical paths and only the B–E chain retaining float, a delay to almost any activity now delays the whole project. At 26 days, three activities had slack; at 22 days, only two do.
Further acceleration is expensive. To reach 21 days, both critical paths must shorten, which means shortening F (on every path) at $700, or shortening G and H together — and H has no stated shortening option at all. The $450-per-day average cost of the first four days rises to at least $700 for the fifth.

A limitation worth stating: these durations are treated as exact. At 22 days the schedule has almost no slack, so ordinary variability — a wet week, a late inspection — is now much more likely to push the finish out. The client is buying speed at the cost of resilience.

Scheduling, resources and shortening the project

Covers AS91576

From analysis to a schedule

The network says what is possible; the schedule says what will happen.
A schedule assigns each activity a start date, choosing within its float where there is any.
Two natural extremes:
- Earliest-start schedule — everything begins as soon as it can. Safest, since all float is kept in reserve.
- Latest-start schedule — everything begins as late as it can. Cheapest for cash flow, since money is spent later, but every activity becomes critical and there is no protection against delay.
Most real schedules sit between the two, using float deliberately to smooth resources.

The Gantt chart

A Gantt chart plots time along the horizontal axis and one row per activity.
Each activity is a bar from its start to its finish, with its float shown as a lighter extension.
Critical activities have no extension, so the critical path is visible as an unbroken run of solid bars from left to right.

What the chart shows that the network does not:
- What is running at the same time — read a vertical line down the chart
- When the workload peaks
- Where float sits, and therefore where activities could be moved

Resource scheduling

The network assumes unlimited resources. In reality, two activities needing the same crew cannot both run on Tuesday.
The method:
- Draw the Gantt chart on an earliest-start basis.
- Add up the resource needed on each day, and plot that as a resource histogram.
- Where a peak exceeds what is available, move non-critical activities within their float to flatten it.
- If flattening is impossible without moving a critical activity, the project must get longer — and that is a genuine finding to report.
Only float can absorb resource clashes. This is what float is for, and it is the main reason for calculating free float separately.

Shortening the project

Only the critical path controls the duration, so shortening anything else achieves nothing.
The procedure:
- Shorten a critical activity by one time unit (more staff, overtime, plant hire).
- Recalculate both passes. The project falls by one unit — unless another path becomes critical.
- Repeat, watching for the moment a second path becomes critical.
Once two paths are critical, both must be shortened together to gain anything further. That is where the cost of acceleration rises sharply, and saying so is a Merit/Excellence-level observation.
There is a floor. Eventually every path is critical, or an activity cannot be shortened further, and no amount of money buys more time.

When a deadline is imposed

If the project duration exceeds a required deadline, the shortfall is the finding, and the report should say:
- by how much the project is late,
- which critical activities could be shortened, and
- what happens to the critical path if they are.
Do not report a negative float. If a deadline earlier than the computed duration is imposed, the correct statement is "the project is 3 days over the deadline", not "float $= -3$ ".

What the model leaves out

Deterministic durations. Every duration is treated as exactly known. Real tasks vary, and a near-critical path with small float can become critical through ordinary variability.
Finish-to-start only. The model assumes an activity cannot begin until its predecessor is entirely complete. Real projects often overlap tasks.
Unlimited resources, unless resource scheduling is done separately.
No cost. The network optimises time alone; the cheapest schedule and the fastest schedule are rarely the same.
Naming one of these limits in the report is a genuine and quick way to show critical evaluation.

Tips and Tricks

Draw the Gantt chart on earliest starts and show float as a lighter bar. It makes the critical path visible as a solid unbroken chain and shows instantly what could be moved.

Only critical activities can shorten the project. Before recommending overtime, check the activity is on the critical path — otherwise the money buys nothing but extra float.

A latest-start schedule leaves no protection. It is worth stating as a risk: every activity becomes critical, so any delay anywhere delays the project.

Report a missed deadline as a shortfall, not as negative float. "Four days over" is a finding; "float = −4" is a symptom being reported as a number.

Name one limitation of the model — fixed durations, no resource limits, or no overlap. One sentence is enough, and it is exactly the critical evaluation the standard asks for.

Worked ExampleMeeting a deadline on the deck project

The deck project takes 26 days with critical path $A \to C \to D \to F \to G \to J$ . The client needs it finished in 22 days. The builder can shorten activities by hiring extra labour:

Activity	Duration	Can be shortened by	Cost per day saved
C (clear site)	4	2 days	$400
D (foundations)	6	3 days	$900
F (frame)	7	3 days	$700
G (decking)	5	2 days	$500

Determine the cheapest way to meet the deadline, and state what the critical path becomes.

Step 1 — Establish what must be saved

$26 - 22 = 4 \text{ days}$

Step 2 — Identify the competing paths, before shortening anything

The critical path $A\,C\,D\,F\,G\,J$ : $3 + 4 + 6 + 7 + 5 + 1 = 26$ days.

The parallel route through B and E replaces C and D with B and E:

$A\,B\,E\,F\,G\,J = 3 + 5 + 2 + 7 + 5 + 1 = 23 \text{ days}$

The route through H instead of G:

$A\,C\,D\,F\,H\,(\text{dummy})\,J = 3 + 4 + 6 + 7 + 3 + 0 + 1 = 24 \text{ days}$

$A\,B\,E\,F\,H\,J = 3 + 5 + 2 + 7 + 3 + 1 = 21 \text{ days}$

Step 3 — Rank the shortenable activities by cost

Activity	Cost per day	Max days	On which paths?
C	$400	2	the $C,D$ routes only
G	$500	2	the $G$ routes only
F	$700	3	every path
D	$900	3	the $C,D$ routes only

Step 4 — Take the cheapest saving first: C, 1 day at $400

New path lengths: $A\,C\,D\,F\,G\,J = 25$ ; $A\,C\,D\,F\,H\,J = 23$ ; $A\,B\,E\,F\,G\,J = 23$ .

Project: 25 days. Still 3 to save.

Step 5 — Take C's second day, also $400

New lengths: $A\,C\,D\,F\,G\,J = 24$ ; $A\,B\,E\,F\,G\,J = 23$ ; $A\,C\,D\,F\,H\,J = 22$ .

Project: 24 days. Running total $800. C cannot be shortened further.

Step 6 — Next cheapest is G at $500

Shorten G by 1 day. $G$ appears on both the 24-day and 23-day routes:

$A\,C\,D\,F\,G\,J = 23$
$A\,B\,E\,F\,G\,J = 22$
$A\,C\,D\,F\,H\,J = 22$ (unchanged — H route)

Project: 23 days. Running total $1,300.

Step 7 — One more day needed, and now TWO paths are at 23 and 22

Shorten G again? $A\,C\,D\,F\,G\,J$ falls to 22, but $A\,C\,D\,F\,H\,J$ is already at 22 — so the project stops at 22.

$\text{Project} = \max(22,\ 22,\ 21,\ 21) = 22 \ \checkmark$

Cost of G's second day: $500. Running total $1,800.

Step 8 — Check the alternative: use F instead of G's second day

F is on every path. Shortening F by 1 day at $700 gives:

$A\,C\,D\,F\,G\,J = 22$
$A\,C\,D\,F\,H\,J = 21$
$A\,B\,E\,F\,G\,J = 21$

Project: 22 days, at a cost of $\$800 + \$500 + \$700 = \$2{,}000$ .

Step 9 — State the cheapest solution

Step 10 — Verify by recomputing every path

Path	Original	Shortened	Length
$A\,C\,D\,F\,G\,J$	$3,4,6,7,5,1$	$3,\mathbf{2},6,7,\mathbf{3},1$	22
$A\,C\,D\,F\,H\,J$	$3,4,6,7,3,1$	$3,\mathbf{2},6,7,3,1$	22
$A\,B\,E\,F\,G\,J$	$3,5,2,7,5,1$	$3,5,2,7,\mathbf{3},1$	21
$A\,B\,E\,F\,H\,J$	$3,5,2,7,3,1$	unchanged	21

$\text{Project duration} = 22 \text{ days} \ \checkmark$

Step 11 — State the new critical path(s)

H, which previously had 2 days of float, is now critical. The handrail crew no longer has any flexibility.

Step 12 — Advise the client

Two consequences the client should know:

The project is now far more fragile. With two critical paths and only the B–E chain retaining float, a delay to almost any activity now delays the whole project. At 26 days, three activities had slack; at 22 days, only two do.
Further acceleration is expensive. To reach 21 days, both critical paths must shorten, which means shortening F (on every path) at $700, or shortening G and H together — and H has no stated shortening option at all. The $450-per-day average cost of the first four days rises to at least $700 for the fifth.