Exercise set

Critical Spare Parts Stockout, Repairable Pool, and Shelf-Life Exercises

Solved spare-parts exercises for reorder points, stockout risk, repairable pools, shelf life, net availability, kitting and release gates.

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

These exercises practise critical spare-parts engineering: reorder points, safety stock, stockout probability, lead-time demand, repairable pools, net available inventory, shelf-life release, supplier risk, kitting readiness, cycle-count accuracy and spare-parts release gates.

The goal is to distinguish physical inventory from usable recovery capability. A part can be counted in a store and still be unavailable because it is allocated, expired, under inspection, obsolete, incompatible, missing a kit item, waiting for repair or blocked by supplier qualification.

Assume simplified screening models unless an exercise states otherwise. Real spare-parts decisions should also check failure criticality, demand history, common usage, configuration, shelf life, repair loop quality, supplier lead time, import constraints, quarantine status, storage condition, inspection status and the downtime consequence of a stockout.

Release Evidence Notes

Critical-spare evidence should state the asset function, part number, installed configuration, demand basis, supplier lead time, repairability, shelf life and current stock status. Gross count is not enough.

Stockout evidence should use lead-time demand and tail risk, not only average demand. A low average demand can still be unacceptable when one missing spare creates high downtime, safety exposure or contractual loss.

Repairable-pool evidence should include failure demand, repair turnaround, scrap rate, inspection hold and quality release. A repairable pool does not protect the operation if returned units are unreliable or slow to clear.

Shelf-life evidence should check time to issue, installation and commissioning time, required reserve after commissioning and storage condition. A spare with short remaining life may fail release even when it is physically present.

Engineering Boundary Notes

This page covers spare-parts inventory and release capability. Reliability architecture and proof-test evidence belong in the reliability availability exercise set. Maintenance interval, P-F inspection and backlog risk belong in the maintenance interval and condition-monitoring exercise set.

If the decision is broader supplier qualification, replenishment policy or production readiness, use the supply-chain inventory and supplier release exercise set. This page is narrower: it asks whether a critical spare can protect a function when failure occurs.

Scenario Map

Scenario	Exercises	Primary check	Engineering decision
Reorder and safety stock	1-5, 12	lead-time demand, safety stock and coverage probability	Set minimum stock for critical parts.
Stockout consequence	6, 10-11, 15	downtime exposure, net available inventory and criticality priority	Decide whether counted stock is usable protection.
Repairable and shelf-life control	7-9, 13-14	repair loop size, scrap, shelf-life reserve and supplier risk	Approve pool size, last-time buy or restriction.
Release readiness	16-18	kit completeness, cycle count and hard gates	Release, restrict or escalate spare readiness.

Exercise 1: Deterministic Reorder Point

A critical spare has average demand:

d=0.7\ \text{parts/week}

Supplier lead time is:

L=8\ \text{weeks}

Safety stock is:

SS=3\ \text{parts}

Find the reorder point.

Solution

Lead-time demand is:

dL=0.7(8)=5.6\ \text{parts}

Reorder point:

ROP=dL+SS=5.6+3=8.6

Round up:

ROP=9\ \text{parts}

Engineering Comment

Critical spares should be rounded conservatively and checked against shelf life, shared demand and configuration compatibility.

Plausibility Check

Average lead-time demand is just above five parts. Adding three safety parts gives just below nine.

Exercise 2: Safety Stock from Demand Variation

Lead-time demand has standard deviation:

\sigma_L=2.4\ \text{parts}

A service rule uses:

z=1.65

Estimate safety stock.

Solution

Safety stock is:

SS=z\sigma_L

Substitute:

SS=1.65(2.4)=3.96

Round up:

SS=4\ \text{parts}

Engineering Comment

The z-based rule assumes the demand model is credible. Intermittent demand and emergency issue patterns may need direct service-level simulation.

Plausibility Check

About one and two thirds standard deviations of $2.4$ parts is about four parts.

Exercise 3: Reorder Point with Variable Demand

Average lead-time demand is:

\mu_L=11.2\ \text{parts}

Safety stock from variation is:

SS=4\ \text{parts}

Find the reorder point.

Solution

Use:

ROP=\mu_L+SS

So:

ROP=11.2+4=15.2

Round up:

ROP=16\ \text{parts}

Engineering Comment

Rounding down creates avoidable stockout risk. For critical spares, the release record should explain the rounding rule.

Plausibility Check

The reorder point must be above average lead-time demand by the safety-stock amount, so sixteen is expected.

Exercise 4: Poisson Stockout Risk

Demand during lead time follows a Poisson model with mean:

\mu=1.4

The current stock is:

s=2

Use:

P(X\leq2)=0.833

Find stockout risk.

Solution

Coverage is:

P(X\leq2)=0.833

Stockout risk is:

P(X>2)=1-0.833=0.167

So risk is:

16.7\%

Engineering Comment

Average demand is low, but tail risk may still be unacceptable when the part protects a critical function.

Plausibility Check

Two parts cover most cases with mean $1.4$ , but a stockout risk near one sixth is credible.

Exercise 5: Minimum Stock for Coverage Rule

For the same demand model:

\mu=1.4

the cumulative probabilities are:

P(X\leq2)=0.833,\qquad P(X\leq3)=0.946,\qquad P(X\leq4)=0.986

Find the minimum stock for at least $95\%$ coverage.

Solution

Two spares fail:

0.833<0.95

Three spares also fail:

0.946<0.95

Four spares pass:

0.986\geq0.95

Minimum stock is:

s=4

Engineering Comment

A small percentage gap can matter when downtime consequence is high. The decision should be based on required service level, not average usage.

Plausibility Check

Three spares almost pass, so only one more spare is needed.

Exercise 6: Expected Downtime from Stockout Risk

Stockout risk during lead time is:

p_s=0.167

If stockout occurs, expected downtime is:

D=72\ \text{h}

Estimate expected downtime exposure per lead-time cycle.

Solution

Expected downtime is:

E[D]=p_sD

Thus:

E[D]=0.167(72)=12.0\ \text{h}

Engineering Comment

Expected downtime is useful for economic screening, but high-consequence failures may require a hard maximum risk rather than average exposure.

Plausibility Check

One sixth of seventy-two hours is about twelve hours.

Exercise 7: Repairable Pool Size

A repairable module fails on average:

d=4\ \text{modules/month}

Repair turnaround is:

L=1.8\ \text{months}

Add two buffer modules. Estimate pool size.

Solution

Expected modules in repair pipeline:

N_p=dL=4(1.8)=7.2

Add buffer:

N=7.2+2=9.2

Round up:

N=10\ \text{modules}

Engineering Comment

Pool size depends on repair turnaround and release quality. A longer inspection hold or vendor delay increases the required pool.

Plausibility Check

About seven modules are in the repair loop, and the buffer pushes the need to about ten.

Exercise 8: Repair Pool with Scrap

The repair pipeline expects:

N_p=7.2

modules. Repair scrap fraction is:

f_s=0.10

Estimate effective pool requirement before adding buffer.

Solution

If $10\%$ are scrapped, usable yield is:

y=1-f_s=0.90

Required input pool is:

N=\dfrac{N_p}{y}=\dfrac{7.2}{0.90}=8.0

Engineering Comment

Scrap and no-fault-found loops must be visible. Otherwise the nominal repairable pool looks healthier than the usable pool.

Plausibility Check

A ten percent loss increases $7.2$ modules to exactly eight.

Exercise 9: Shelf-Life Release

A critical spare has remaining shelf life:

T_s=18\ \text{months}

Expected time to issue:

T_i=10\ \text{months}

Installation and commissioning require:

T_c=3\ \text{months}

The policy requires at least $4$ months reserve after commissioning. Check release.

Solution

Remaining life after commissioning:

T_r=T_s-T_i-T_c=18-10-3=5\ \text{months}

Since:

5\geq4

the spare passes the shelf-life rule.

Engineering Comment

Shelf-life release should also check storage condition, packaging, calibration and whether the part will still match the installed configuration.

Plausibility Check

Thirteen months are consumed before commissioning, leaving five months.

Exercise 10: Net Available Inventory

Stores show:

S_g=12

gross spares. Allocated to work orders:

S_a=4

Under inspection:

S_q=2

Find net available inventory.

Solution

Net available is:

S_n=S_g-S_a-S_q

So:

S_n=12-4-2=6

Engineering Comment

Gross inventory can mislead release decisions. Allocated, quarantined or inspection-hold parts should not be counted as usable protection.

Plausibility Check

Six of twelve parts are unavailable, so net usable stock is half the gross count.

Exercise 11: Shared Spare Demand

One spare type supports:

N=5

asset lines. Expected demand per line is:

d=0.3\ \text{parts/month}

Lead time is:

L=4\ \text{months}

Find average shared lead-time demand.

Solution

Total monthly demand is:

d_t=Nd=5(0.3)=1.5\ \text{parts/month}

Lead-time demand:

\mu_L=d_tL=1.5(4)=6\ \text{parts}

Engineering Comment

Shared parts require common-demand modelling. A spare that looks sufficient for one line can be weak when five lines draw from the same stock.

Plausibility Check

Five lines triple and then some the monthly demand; over four months the mean reaches six parts.

Exercise 12: Lead-Time Demand Standard Deviation

Weekly demand standard deviation is:

\sigma_w=0.9\ \text{parts/week}

Lead time is:

L=9\ \text{weeks}

Assuming independent weekly demand, estimate lead-time demand standard deviation.

Solution

For independent periods:

\sigma_L=\sigma_w\sqrt{L}

So:

\sigma_L=0.9\sqrt{9}=2.7\ \text{parts}

Engineering Comment

Independence may be weak if failures cluster during outages, campaigns or environmental events. Clustered demand requires larger protection.

Plausibility Check

Nine independent weeks triple the standard deviation because $\sqrt{9}=3$ .

Exercise 13: Supplier On-Time Delivery Effect

A supplier delivers on time:

p=0.82

of orders. A critical spare policy needs at least:

p_{min}=0.90

on-time delivery support. Find the margin.

Solution

Margin is:

M=p-p_{min}=0.82-0.90=-0.08

So the supplier is short by:

8\ \text{percentage points}

Engineering Comment

Poor delivery reliability can make a mathematically correct reorder point unsafe. Supplier performance should feed the safety-stock or alternate-source decision.

Plausibility Check

Eighty-two percent is below ninety percent by eight points.

Exercise 14: Last-Time Buy Coverage

A part is becoming obsolete. Expected annual demand is:

d=6\ \text{parts/year}

Required support horizon is:

H=5\ \text{years}

Add contingency:

C=20\%

Estimate last-time buy quantity.

Solution

Base demand:

N_b=dH=6(5)=30

With contingency:

N=N_b(1+C)=30(1.20)=36

Engineering Comment

Last-time buys should also check shelf life, storage cost, design change plan and whether repair or substitution is possible.

Plausibility Check

Five years at six per year needs thirty parts; twenty percent contingency adds six.

Exercise 15: Criticality-Weighted Stock Priority

Two spare candidates have scores:

Spare	Consequence	Stockout probability
A	$9$	$0.10$
B	$5$	$0.30$

Use priority:

P=Cp_s

Decide which has higher risk priority.

Solution

For A:

P_A=9(0.10)=0.90

For B:

P_B=5(0.30)=1.50

Spare B has higher priority by this screen.

Engineering Comment

High consequence matters, but a much higher stockout probability can dominate. Final decisions should also check safety criticality and downtime cost.

Plausibility Check

B has three times the stockout probability, which outweighs its lower consequence score.

Exercise 16: Kit Completeness

A repair kit requires:

N=14

items. Currently released and compatible:

N_c=13

The rule requires complete kits for critical repairs. Check release.

Solution

Completeness is:

C=\dfrac{13}{14}=0.929

Even though this is:

92.9\%

the kit fails because the rule requires all items.

Engineering Comment

Kit readiness is often a hard gate. A missing gasket, firmware cable, seal or certified fastener can stop a repair even when most parts are present.

Plausibility Check

One missing item out of fourteen gives high percentage completeness but not full readiness.

Exercise 17: Cycle-Count Accuracy

Inventory records show:

S_r=40

units. Physical count finds:

S_p=37

Compute record accuracy using physical count as reference.

Solution

Absolute error is:

e=|S_r-S_p|=|40-37|=3

Accuracy is:

A=1-\dfrac{e}{S_p}=1-\dfrac{3}{37}=0.919

So:

A=91.9\%

Engineering Comment

Inventory accuracy is release evidence. A critical-spares policy fails if records cannot be trusted at the point of need.

Plausibility Check

Three missing parts out of about forty creates an error near eight percent, so accuracy near ninety-two percent is expected.

Exercise 18: Critical Spare-Parts Release Gate

A spare-parts readiness package has:

Gate	Requirement	Current result
stock coverage	at least $95\%$	$98.6\%$
shelf-life reserve	at least $4$ months	$5$ months
kit completeness	$100\%$	$92.9\%$
cycle-count accuracy	at least $97\%$	$91.9\%$

Decide whether the spares package is releasable.

Solution

Coverage and shelf life pass:

98.6\%\geq95\%

5\geq4

Kit completeness fails:

92.9\%<100\%

Cycle-count accuracy also fails:

91.9\%<97\%

The package is not releasable.

Engineering Comment

A spare-parts release package needs usable, complete and trustworthy stock. Good statistical coverage does not compensate for missing kit items or weak record accuracy.

Plausibility Check

Two hard gates fail, so the correct decision is hold, correct inventory and recheck.

Validation Package Checklist

A strong critical spare-parts solution should check:

whether part number, configuration and asset function are explicit;
whether lead-time demand is based on current failure and issue data;
whether safety stock covers variability and consequence, not only averages;
whether stockout probability is tied to downtime or safety consequence;
whether gross stock is reduced for allocations, quarantine and inspection holds;
whether repairable pools include turnaround, scrap and release quality;
whether shelf life covers issue, installation, commissioning and reserve;
whether supplier delivery performance changes the replenishment policy;
whether kits are complete and inventory records are accurate before release.

Common Release Mistakes

Common mistakes include using gross inventory as usable inventory, sizing spares from average demand only, ignoring shared demand across assets, treating repairable units in transit as available, omitting scrap and inspection hold from repair loops, accepting short shelf life without commissioning reserve, trusting an obsolete supplier lead time, counting incomplete kits as ready, and releasing a spare-parts package while inventory accuracy is too weak to trust.

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