Compare expected work in pooled factory inspection

Try

A factory inspects synthetic batches for a particular defect. Direct inspection tests every item separately and costs one inspection per item. Consider another precise protocol: combine k items into a pool and test that pool once. A perfect negative result certifies all k items good. A positive result triggers individual tests of all k items, at one inspection each. There are no false results in this teaching model.

Assume each item is independently defective with probability p. Independence is an assumption about the joint distribution, not a conclusion from a shared defect rate. Pooling might save work when most pools are negative, but positive pools cost more than direct inspection. Before calculating, predict which effect dominates for small p and for p near one.

Predict

A pool has two possible costs. A negative pool costs 1; a positive pool costs 1+k. Predict the negative probability for k=2 and p=1/4. It is not 1−2p: that expression double-counts the case in which both items are defective when computing its complement.

The correct negative probability is (3/4)²=9/16. Thus positive probability is 7/16. Record the expected pool cost, and then divide by the number of items before comparing it with the direct baseline. Comparing a pool total with a per-item baseline would mix different denominators.

Experiment

The following table enumerates the four patterns for the worked pair. G means good and D means defective. The probabilities need not be equal.

Pattern	Probability	Pool result	Inspections
GG	9/16	Negative	1
GD	3/16	Positive	3
DG	3/16	Positive	3
DD	1/16	Positive	3

Expected total work is (9+9+9+3)/16=15/8 inspections per pool, or 15/16 per item. A particular positive pool still costs three inspections for two items. The expectation averages over defect configurations; it is not a guarantee of savings on every realized batch.

Use the explorer’s separate default p=0.02 and k=8. Predict whether doubling the pool size improves the cost, then compare all feasible equal-pool sizes for its fixed 64-item batch. A size that does not divide 64 still has a valid standalone-pool formula, but cannot partition that batch into equal pools without an additional remainder rule.

Compare expected inspection work

Each item independently has a defect with probability p. A perfect pool test costs one inspection; if positive, inspect every one of its k items at cost one each. This fixed protocol includes the redundant second test when k=1. These are teaching assumptions, not measured factory performance.

Defect probability p (0–1)Pool size k (whole number 1–64)Reset p=0.02, k=8

Positive-pool probability 1−(1−p)^k = 0.149237. Expected inspections per pool: 1+k·P(positive) = 2.193896. Dividing by k gives 0.274237 per item, versus direct inspection baseline 1. This is an expectation over random defects, not a guarantee for every batch.

Approximation 1/k+kp = 0.285. Its excess over the formula is 0.01076302; the mathematical bound is 0≤excess≤k(k−1)p²/2 = 0.0112. All displayed numbers are rounded floating-point evaluations. The analytic inequality, rather than this numerical comparison, supplies the guarantee; tiny roundoff in a computed excess is displayed as zero.

Pool size 8 divides 64, so it can partition a 64-item batch into equal pools.

All equal-pool sizes for 64 items at the current p; direct inspection always costs 1 per item

Pool size	Pools	Expected work per item	Compared with direct inspection
1	64	1.02	Higher
2	32	0.5396	Lower
4	16	0.3276318	Lower
8	8	0.274237	Lower
16	4	0.3387023	Lower
32	2	0.5073669	Lower
64	1	0.7411715	Lower

Smallest computed pooling cost among these divisors: 0.274237, at k=8. This is below the direct baseline. Comparisons use unrounded calculations; a displayed rounding tie does not by itself certify an exact tie.

Predict the p=0 and p=1 results before changing the probability. Then compare k=4 and 8 at p=0.02. The comparison includes only the stated two-stage protocol and equal pools; it does not optimize arbitrary inspection trees, incomplete remainder pools, noisy tests or real operating costs. No state is saved and no completion is awarded.

Try p=0 and p=1. At zero, every pool is negative and cost per item is 1/k. At one, every pool triggers individual follow-up and cost per item is 1+1/k. Also inspect k=1: the formula gives 1+p, because this stated protocol tests a positive singleton twice. Removing that redundant test would be a different protocol. Keeping its cost visible prevents an unnoticed change of rules at a boundary.

Understand

Let I be 1 if the pool is positive and 0 otherwise. Its work is 1+kI. Independence gives P(I=0)=(1−p)^k. Linearity of expectation therefore gives

Expected pool work = 1+k[1−(1−p)^k].

Dividing by k gives c(k)=1/k+1−(1−p)^k inspections per item. For a batch of N independent items split into equal pools, k must divide N. Multiplying the pool expectation by N/k gives total expected work Nc(k). Testing each item directly costs N. The comparison assumes equally costly perfect tests and ignores any extra handling costs; those assumptions define the calculation.

For small p, a useful approximation is c(k)≈1/k+kp. We can bound its error rather than merely label p “small.” Let X count defective items. For every nonnegative integer x,

0≤x−1 if x is positive, and this difference is at most x(x−1)/2; at x=0 the difference is zero.

More uniformly the middle expression is x minus the indicator of x≥1. Taking expectations gives kp−P(X≥1). The quantity X(X−1)/2 counts unordered pairs of defective items. Each pair has probability p² by independence, so its expectation is k(k−1)p²/2. Hence

0≤kp−[1−(1−p)^k]≤k(k−1)p²/2.

The approximation overestimates cost, with a stated size-dependent error. For real k>0 and p>0, the identity

1/k+kp−2√p = (1/√k−√(kp))²

shows its continuous minimum occurs at k=1/√p and has value 2√p. The factor two matters. This is not yet an exact integer decision: compare the original formula on all permitted divisors of N, and include direct inspection as a separate alternative. Approximation error can change which candidate wins.

A different defect model changes the negative probability. If a batch has exactly D defective items among N and a pool is a uniformly chosen k-subset, the negative probability is C(N−D,k)/C(N,k). Here C(n,k) counts k-subsets and is zero when k exceeds n. Randomly grouping a fixed-total population does not make its defect indicators independent. Different pools are dependent too, but symmetry and linearity still let us add their marginal expected costs. For example N=8,D=2,k=2 gives negative probability15/28 and per-item cost1/2+13/28=27/28, whereas the iid model with the same marginal p=1/4 gave15/16.

Apply

Change the inspection protocol again. Let N=2^h items occupy the leaves of a complete binary tree. Test the full root pool. Whenever a nonsingleton pool is positive, test both children. A singleton test stops, with no redundant repeat. This is a specified algorithm, not the flat protocol with its name changed.

At level ℓ≥1 there are 2^ℓ possible tests. A particular node is tested exactly when its parent is positive. A positive parent implies all its ancestors are positive, so the parent was indeed reached. Its size is N/2^(ℓ−1). By linearity,

Expected tests = 1 + sum, for ℓ=1 through h, of 2^ℓ[1−(1−p)^(N/2^(ℓ−1))].

Test this formula at N=2,p=1/2: the root costs one and both singleton children are tested with probability3/4, giving5/2 expected tests, more than direct inspection’s two. At p=0 only the root is tested; at p=1 every node is tested, totaling2N−1. For N=1 the sum is empty and the root is the sole test. Explain why using the child’s own positive probability would undercount tests: a good child is still inspected when its sibling makes their parent positive.

A changed dependence case

Suppose every item in a pool is defective together with probability p and otherwise all are good. The positive probability is p, not1−(1−p)^k, although each item still has marginal defect probability p. Flat-protocol cost per item becomes1/k+p. Shared marginals do not determine expected work without a joint model. Neither this model nor the independent model establishes measured reliability for real inspections.

Continue

In the practice, specify the test rule, defect model and feasible sizes before optimizing. Keep expected work, realized work and defect-detection reliability separate. For recursive inspection, define exactly which event triggers each node’s test, then add its indicator expectation; independence between those test events is unnecessary. The fixed numerical explorer supplies comparisons for one bounded model, while the written proofs explain which changes require a new calculation.