Float, critical paths and Gantt charts

Float time for each activity, critical activities and the critical path, the effect of changing a duration, drawing a Gantt chart from a network, and a spreadsheet model of a project.

The scans in 9B gave two numbers at every node. This page turns them into decisions a project manager actually makes: which jobs can be shuffled around, which ones need watching like a hawk, and what happens when the tiler calls in sick.

Float

The float (or slack) of an activity is how long it can be delayed without making the whole project finish late.

Float of an activity

$$\text{Float} = \text{LST at its end node} - \text{EST at its start node} - \text{duration}$$

Why? The activity can't start before the EST at its start node, and it must finish by the LST at its end node. That window, minus the time the activity itself needs, is the spare time. Some textbooks first find each activity's own latest start ($\text{LST at end node} - \text{duration}$) and then write float $=$ latest start $-$ earliest start. It is the same calculation.

Example 1●

Find the float of each activity in the shed network from 9B. Node ESTs: $0, 1, 4, 7, 9, 10$. Node LSTs: $0, 2, 4, 7, 9, 10$.

A: $2 - 0 - 1 = 1$ day

A runs from node 1 to node 2.

B: $4 - 0 - 4 = 0$

B runs from node 1 to node 3.

C: $4 - 1 - 2 = 1$ day

C runs from node 2 to node 3.

D: $7 - 4 - 3 = 0$, E: $9 - 7 - 2 = 0$, G: $10 - 9 - 1 = 0$

D, E and G each run between nodes whose EST and LST match, with no gap.

F: $10 - 7 - 2 = 1$ day

F runs from node 4 to node 6.

The critical path

Activities with zero float are critical activities. Joined together they form the critical path: the longest path from start to finish. Its length is the minimum completion time, because nothing can finish sooner than the longest chain of jobs that must happen one after another.

In the shed, the critical path is B–D–E–G, with length $4 + 3 + 2 + 1 = 10$ days. Check: that matches the EST at the finish node.

Changing a duration

Lengthen a critical activity by $k$ days: the whole project takes $k$ days longer.
Lengthen a non-critical activity by no more than its float: the finish date doesn't move (but the float is used up, and the activity may become critical).
Lengthen it by more than its float: the project is delayed by the extra amount, and the critical path changes.
Shorten a non-critical activity: no change at all. To finish sooner, you must shorten the critical path, and then another path may take over as critical.

Example 2●●

In the shed project, pouring the slab (C) takes $4$ days instead of $2$. What happens?

The extra $2$ days exceed the float by $1$, so the project is delayed by $1$ day.

C has a float of only 1 day, but it is 2 days longer.

Node 3: B gives $4$, C gives $1 + 4 = 5$. Then node 4: $8$, node 5: $10$, node 6: $11$.

Redo the forward scan from node 3.

Minimum completion time $11$ days. New critical path: A–C–D–E–G.

The longest path has changed.

InteractiveRenovation planner

Try this

Pick C (Rewire) and add $1$ day. The finish doesn't move. Add another day. What happened, and why after the second day but not the first? Look at the float column.

Reset, then shorten B (Order cabinets). Does the job finish any sooner? Now shorten D instead. How many days can you save before another path becomes critical? Switch to Gantt chart to see the floats as dashed boxes.

Gantt charts

A Gantt chart is a timeline. Each activity gets a row, with a bar that starts at its earliest start and is as long as its duration. The float is shown as a dashed or unshaded box after the bar: the bar can slide that far to the right without delaying the finish. Critical activities have no float box, and laid end to end they stretch across the whole chart.

For the shed, the bars would be: A days $0$ to $1$ (float $1$), B $0$ to $4$, C $1$ to $3$ (float $1$), D $4$ to $7$, E $7$ to $9$, F $7$ to $9$ (float $1$), G $9$ to $10$. A Gantt chart is how a builder actually runs the job: it shows on any given day which jobs should be underway.

A spreadsheet model

A spreadsheet can do the forward scan and update instantly when a duration changes. For the shed, put the durations of A to G in cells B2:B8, and the node ESTs in E2:E7:

Node 1: =0
Node 2: =E2+B2 (A)
Node 3: =MAX(E2+B3, E3+B4) (B or C, whichever is later)
Node 4: =E4+B5, node 5: =E5+B6
Node 6: =MAX(E5+B7, E6+B8) (F or G)

The MAX function is the "largest in" rule. Change C's duration in B4 and the completion time in E7 updates on its own, which is how Example 2 can be checked in seconds.

Your turn

In the school video network (A 1→2 $3$, B 2→3 $2$, C 2→4 $1$, D 3→4 $2$, E 4→5 $4$; ESTs $0, 3, 5, 7, 11$; LSTs $0, 3, 5, 7, 11$), find the float of C and the critical path.

Show the working

C runs from node 2 to node 4: float $= 7 - 3 - 1 = 3$ days. Every other activity has zero float, so the critical path is A–B–D–E, with length $3 + 2 + 2 + 4 = 11$ days.

Flashcards

Float of an activity

LST at end node − EST at start node − duration

Critical path

The path of zero-float activities; the longest path; its length is the minimum completion time

Delay a non-critical activity by more than its float

The project is late by (delay − float)

Practice

The float of an activity is:
A critical activity has a float of:
An activity takes $5$ days. The EST at its start node is $4$ and the LST at its end node is $12$. Find its float.
In the shed network (ESTs $0, 1, 4, 7, 9, 10$; LSTs $0, 2, 4, 7, 9, 10$), activity A ($1$ day) runs from node 1 to node 2. Find its float.
Which is the critical path of the shed network?
In the school video network, activity C ($1$ day) runs from node 2 (EST $3$) to node 4 (LST $7$). Find its float.
An activity with a float of $4$ days is delayed by $3$ days. The minimum completion time:
Exam style: In the shed project, clearing the site (A) takes $4$ days instead of $1$. Find the new minimum completion time.
Exam style: In the café network (K $2$ from node 1 to 2, L $5$ from 1 to 3, dummy 2 to 3, M $6$ from 2 to 4, N $3$ from 3 to 4), the ESTs are $0, 2, 5, 8$ and the LSTs are $0, 2, 5, 8$. Which statement is true?