Exercise set

Operations Critical Path, Schedule Risk, and Milestone Readiness Exercises

Solved operations scheduling exercises for critical path, float, PERT duration, schedule risk, buffers, rework loops and milestone release.

Branch: Industrial and Management Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises practise operations scheduling decisions: critical path, float, near-critical work, PERT duration, schedule variance, guarded milestones, buffer sizing, delay impact, rework loops, schedule compression and milestone release gates.

The goal is not to publish a date that looks precise. A schedule is only useful when dependency logic, calendars, access windows, inspections, rework paths, uncertainty and release evidence are visible enough to challenge.

Assume simplified screening data unless an exercise states otherwise. Real schedule release should also check shift calendars, outage windows, permit holds, supplier dates, inspection capacity, commissioning sequence, handover rules, abnormal findings and authority to release.

Release Evidence Notes

Schedule evidence should state the operating boundary: project, maintenance shutdown, commissioning plan, construction lookahead, production restart or recovery milestone. Calendar days, work days and staffed hours must not be mixed.

Critical path evidence should identify dependency logic, activity duration basis, calendars, near-critical paths, inspection holds and rework loops. The critical path can change when a near-critical item slips.

Risk evidence should show the margin between guarded finish and required milestone. A deterministic finish date is weak when duration uncertainty, test failures or permit delay can consume the margin.

Milestone release should be a hard-gate decision. If schedule risk, readiness evidence or hold-point closure fails, the milestone should be delayed, restricted or rebaselined.

Engineering Boundary Notes

This page covers schedule logic and milestone risk. Queue waiting time, service capacity and WIP control belong in the operations queue exercise set. Skill-hour loading, backlog recovery and shutdown resource readiness belong in the operations resource-loading exercise set.

When schedule failure is caused by a missing spare, inspection crew or craft-hour limit, record the failed resource gate separately instead of hiding it inside CPM output.

Scenario Map

Scenario	Exercises	Primary check	Engineering decision
Critical path and float	1-5	path duration, total float, free float and near-critical risk	Decide what must be protected.
Uncertainty and buffers	6-10	PERT duration, variance, z-margin and buffer size	Decide whether the milestone is credible.
Delay, rework and compression	11-15	slip impact, rework exposure, crashing value and hold-point delay	Decide whether to resequence, compress or hold.
Milestone release	16-18	readiness score, evidence closure and hard gates	Release, restrict or reject the schedule.

Exercise 1: Critical Path from a Dependency Network

A commissioning plan has activities:

Activity	Duration	Predecessors
A	4 days	none
B	6 days	A
C	5 days	A
D	3 days	B
E	4 days	B, C
F	2 days	D, E

Find project duration and critical path.

Solution

Candidate paths are:

A-B-D-F=4+6+3+2=15\ \text{days}

A-B-E-F=4+6+4+2=16\ \text{days}

A-C-E-F=4+5+4+2=15\ \text{days}

The critical path is:

A-B-E-F=16\ \text{days}

Engineering Comment

The critical path is the dependency chain that controls finish date. It is not simply the list of high-visibility tasks.

Plausibility Check

All paths start with A and end with F. The path through B and E has the largest middle duration.

Exercise 2: Total Float for a Near-Critical Activity

Activity C starts after A finishes at day $4$ . C takes:

5\ \text{days}

Activity E waits for both B and C, and B finishes at day $10$ . Find total float for C.

Solution

Earliest finish of C is:

EF_C=4+5=9\ \text{days}

C may finish as late as day $10$ without delaying E:

LF_C=10\ \text{days}

Total float:

TF_C=LF_C-EF_C=10-9=1\ \text{day}

Engineering Comment

One day of float is fragile. Failed access, inspection hold or rework can make C critical.

Plausibility Check

C finishes one day before the other predecessor, so one day of float is expected.

Exercise 3: Free Float for an Inspection Activity

An inspection activity starts at day:

ES=8

and lasts:

d=3\ \text{days}

Its successor can start at day:

ES_s=12

Find free float.

Solution

Inspection finishes at:

EF=ES+d=8+3=11

Free float is:

FF=ES_s-EF=12-11=1\ \text{day}

Engineering Comment

Free float is local. Consuming it delays the successor even if the project finish might still have some total float.

Plausibility Check

The successor starts one day after the inspection finishes, so the free float is one day.

Exercise 4: Forward-Pass Finish Date

A work package has three serial activities:

d_1=2\ \text{days},\quad d_2=7\ \text{days},\quad d_3=4\ \text{days}

The first activity starts at day $5$ . Find finish day.

Solution

Serial duration is:

D=2+7+4=13\ \text{days}

Finish day is:

F=5+13=18

Engineering Comment

The result assumes all days are valid work days. Calendars, shifts and access windows must be checked before publishing the date.

Plausibility Check

Thirteen days after day five gives day eighteen.

Exercise 5: Near-Critical Path Margin

The critical path duration is:

D_c=28.0\ \text{days}

A second path has duration:

D_n=27.2\ \text{days}

Find near-critical margin.

Solution

Margin is:

M=D_c-D_n=28.0-27.2=0.8\ \text{days}

Engineering Comment

A path within one day of critical should be actively monitored. Minor slippage can change the governing path.

Plausibility Check

The second path is slightly shorter than the critical path, so the margin is small and positive.

Exercise 6: PERT Expected Duration

An activity has optimistic, most likely and pessimistic durations:

a=4,\quad m=6,\quad b=13\ \text{days}

Estimate PERT expected duration:

t_e=\dfrac{a+4m+b}{6}

Solution

Substitute:

t_e=\dfrac{4+4(6)+13}{6}=\dfrac{41}{6}=6.83\ \text{days}

Engineering Comment

The estimate is pulled above the most likely value because the pessimistic tail is long. That tail should be tied to a real risk.

Plausibility Check

The result is above $6$ but far below $13$ , which matches a weighted PERT average.

Exercise 7: PERT Standard Deviation

For the same activity:

a=4,\quad b=13

Estimate standard deviation:

\sigma=\dfrac{b-a}{6}

Solution

Compute:

\sigma=\dfrac{13-4}{6}=1.5\ \text{days}

Engineering Comment

This simplified uncertainty should be treated as a screening value, not proof that duration follows a perfect distribution.

Plausibility Check

The full optimistic-pessimistic range is nine days; one sixth of that is one and a half days.

Exercise 8: Path Variance Combination

A path has three independent activity standard deviations:

0.6,\quad 1.0,\quad 1.5\ \text{days}

Estimate path standard deviation.

Solution

Path variance is:

\sigma_p^2=0.6^2+1.0^2+1.5^2=0.36+1.00+2.25=3.61

So:

\sigma_p=\sqrt{3.61}=1.90\ \text{days}

Engineering Comment

The result assumes independent duration uncertainty. Shared access, weather or inspection delays can make the true path risk larger.

Plausibility Check

The combined standard deviation should be greater than the largest single term but less than their direct sum.

Exercise 9: Milestone Confidence Screen

A path has expected duration:

\mu=24.0\ \text{days}

and standard deviation:

\sigma=1.90\ \text{days}

The milestone is:

T_m=27.0\ \text{days}

Compute z-margin.

Solution

Use:

z=\dfrac{T_m-\mu}{\sigma}

Substitute:

z=\dfrac{27.0-24.0}{1.90}=1.58

Engineering Comment

A z-margin of $1.58$ may be acceptable for some planning screens, but it is not a universal release rule. Critical milestones may require more margin.

Plausibility Check

The milestone is three days beyond the mean, and the standard deviation is about two days, so z near $1.5$ is expected.

Exercise 10: Buffer for a One-Sided Schedule Rule

A schedule rule uses:

z=1.28

and path standard deviation:

\sigma=1.90\ \text{days}

Find required buffer.

Solution

Required buffer is:

B=z\sigma=1.28(1.90)=2.43\ \text{days}

Engineering Comment

The buffer should be protected, not silently consumed by scope growth. Otherwise the risk calculation becomes meaningless.

Plausibility Check

The buffer is a little more than one standard deviation, so it should be a little above two days.

Exercise 11: Delay Impact on Critical Path

A critical-path activity slips by:

1.5\ \text{days}

There is no remaining float. Find the finish-date slip.

Solution

With zero float, slip transfers directly:

\Delta F=1.5\ \text{days}

Engineering Comment

Critical-path delay can be recovered only by resequencing, compression, added shifts or scope reduction. Reporting the old date without a recovery plan is weak schedule control.

Plausibility Check

No float means no buffer, so the full delay reaches project finish.

Exercise 12: Delay Impact with Float

A near-critical activity slips:

1.8\ \text{days}

Its total float is:

1.2\ \text{days}

Find project finish impact.

Solution

Float absorbs part of the slip:

\Delta F=1.8-1.2=0.6\ \text{days}

Engineering Comment

Near-critical work should be escalated before float is fully consumed. Waiting until the path becomes critical reduces recovery options.

Plausibility Check

The slip exceeds float by $0.6$ days, so only that excess affects the finish.

Exercise 13: Rework Loop Expected Duration

A test activity takes:

2.0\ \text{days}

If it fails, rework and retest add:

3.0\ \text{days}

Historical failure probability is:

p=0.25

Estimate expected duration.

Solution

Expected duration is:

E[T]=2.0+p(3.0)

So:

E[T]=2.0+0.25(3.0)=2.75\ \text{days}

Engineering Comment

Expected duration is not enough for release if the retest is a hard hold point. Tail risk and latest allowable finish should also be checked.

Plausibility Check

The rework branch happens one quarter of the time, adding $0.75$ day on average.

Exercise 14: Inspection Hold-Point Delay

A hold point has planned duration:

0.5\ \text{day}

Inspector availability adds expected wait:

1.2\ \text{days}

Find total hold-point allowance.

Solution

Total allowance is:

T=0.5+1.2=1.7\ \text{days}

Engineering Comment

Inspection wait should be modelled explicitly. Hiding it inside task duration makes accountability and recovery harder.

Plausibility Check

The wait is larger than the inspection itself, so the total is dominated by availability delay.

Exercise 15: Schedule Compression Value

A critical activity can be crashed from:

6\ \text{days}

to:

4.5\ \text{days}

The extra cost is:

9000

Find cost per day saved.

Solution

Days saved:

\Delta t=6-4.5=1.5\ \text{days}

Cost per day:

C_d=\dfrac{9000}{1.5}=6000\ \text{per day}

Engineering Comment

Crashing is valuable only if the activity is on the controlling path after compression. Otherwise the cost may not improve the milestone.

Plausibility Check

Saving one and a half days for $9000$ gives $6000$ per day.

Exercise 16: Milestone Readiness Score

A milestone package has:

N=40

required evidence items. Closed items:

N_c=37

The rule requires at least $95\%$ closure. Check readiness.

Solution

Closure is:

C=\dfrac{37}{40}=0.925

Since:

92.5\%<95\%

the readiness score fails.

Engineering Comment

Near-complete evidence can still block release if missing items are permits, inspection holds, test certificates or owner approvals.

Plausibility Check

Three missing items out of forty is $7.5\%$ missing, so closure is $92.5\%$ .

Exercise 17: Guarded Finish Margin

Expected finish is day:

22.4

The uncertainty guard is:

1.8\ \text{days}

The milestone is day:

25.0

Find guarded margin.

Solution

Guarded finish is:

T_g=22.4+1.8=24.2

Margin is:

M=25.0-24.2=0.8\ \text{days}

Engineering Comment

The schedule passes, but narrowly. The plan should protect scope and define escalation if progress consumes the margin.

Plausibility Check

The unguarded margin is $2.6$ days. The guard consumes $1.8$ days, leaving $0.8$ day.

Exercise 18: Schedule Release Gate

A milestone package has:

Gate	Requirement	Current result
critical path duration	at most $16$ days	$15.4$ days
guarded finish margin	at least $0.5$ day	$0.8$ day
evidence closure	at least $95\%$	$92.5\%$
open inspection holds	$0$	$0$

Decide whether to release.

Solution

The duration, guarded margin and hold-point gates pass:

15.4\leq16,\quad 0.8\geq0.5,\quad 0=0

Evidence closure fails:

92.5\%<95\%

The milestone is not releasable.

Engineering Comment

Schedule release should not average gates. A missing evidence gate can block release even when the dates look feasible.

Plausibility Check

Only one gate fails, but it is a hard gate. Hold or restrict release until evidence is closed.

Validation Package Checklist

A strong schedule-risk solution should check:

whether all durations use the same calendar and working-time basis;
whether dependency logic includes inspections, rework, handover and access windows;
whether near-critical paths are visible, not only the current critical path;
whether PERT or uncertainty assumptions are supported by evidence;
whether schedule buffers are protected from silent scope growth;
whether delay impact is calculated after float is consumed;
whether compression improves the controlling path;
whether readiness evidence and hold points are closed before release.

Common Release Mistakes

Common mistakes include treating the critical path as fixed, ignoring near-critical paths, mixing calendar days with staffed work days, omitting inspection wait, hiding rework loops inside optimistic durations, using expected duration as if it were a guaranteed finish, crashing noncritical work, consuming schedule buffer without approval, and releasing a milestone because the date passes while required evidence remains incomplete.

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