Under all-units pricing, crossing a break reprices every unit in the order and not only the ones above it, so the total-cost curve drops by D times the price difference at each break. It is discontinuous, drawing it as one line asserts costs no quantity has, and the minimum is found band by band: each band contributes its own economic order quantity if that lands inside it, and its floor otherwise. The candidates are then compared exactly, because some of them are irrational and two decimals that differ are not a proof.
Core ideas
Three ideas. The first is what “all-units” actually means, the second is why one candidate per band is enough, and the third is about the comparison rather than the candidates.
Buy 999 at ten each and the bill is 9990; buy 1000 at eight each and it is 8000. One more unit made the order cheaper in total, which cannot happen if price is a smooth function of quantity. So the total-cost curve has a genuine discontinuity at every break, the panel draws one segment per band, and a single polyline joining them would assert costs that no order quantity has.
Fix the band and the price is constant, so D·price is a constant and the shape is K·D/Q + h·Q/2 again: one minimum, falling to the left of it and rising to the right. If that band's own economic order quantity lies inside the band, it is the band's best. If it lies below the band, the curve is rising throughout the band and the floor is the best; if above, the ceiling is. Checking one quantity per band is therefore a complete search, and the lab still prices every whole quantity in range as a check on the claim.
A band's candidate cost is a rational purchase bill plus, where the candidate is that band's economic order quantity, an irrational √(2KDh). Deciding which of two such numbers is smaller by printing both to four places and looking is a decision that can be wrong. The lab settles same-radicand pairs by one squaring and different-radicand pairs by rational brackets refined until they separate, so the winner is proved rather than observed.
The material
The pricing rule, the shape it produces, the rule for finding the minimum, and the comparison that decides between candidates.
A price schedule is a list of breaks: a quantity at which a band starts, the unit price in that band, and the holding rate there. Under all-units pricing, an order of Q in a band is charged that band's price on every unit.
The cost per unit time becomes D·price(Q) + K·D/Q + h(Q)·Q/2. The purchase term now depends on Q — which is exactly why it could be ignored in the plain model and cannot be here.
The holding rate usually depends on the price, because holding cost is commonly charged as a fraction of the value sitting on the shelf. In the worked schedule it is 20% of price throughout: 2 at a price of ten, 9/5 at nine, 8/5 at eight. That means each band has a different economic order quantity, which is the second reason the bands must be treated separately.
Two of the three bands cannot have their own best quantity, because that quantity is smaller than the band's floor. In those bands the cost curve is rising from the floor onward, so the cheapest reachable point is the floor itself — the smallest order that still qualifies for the price. That is the usual situation, and it is why quantity discounts work as a sales device: they are offers to buy more than you want at a price that makes it worth it, or not.
The purchase bill is doing all the work. Moving from the first band to the third saves D·(10 − 8) = 2400 per unit time on purchases and costs a few hundred in a larger cycle, which is not close. A reader who compared only the cycle costs — 438.18, 546, 848 — would pick exactly the wrong plan, and would do so on figures that are all correct.
Switch the panel to the two-band schedule: the same K = 40 and D = 1200, a price of ten below 900 and 39/4 at 900 and above, with holding at 20% of price as before. The first band's candidate is 40√30 ≈ 219.09 costing 12000 + 80√30 ≈ 12438.1780; the second band's is its floor, 900, costing 75785/6 ≈ 12630.83.
The deeper price loses. Buying 900 units at a time to save a quarter each costs more in holding than it saves at the till, and the panel's exhaustive scan of 1620 whole quantities agrees, landing at 219 — below the break, where no discount applies at all.
The first band's candidate cost is 12000 + 80√30. The third band's is 10448. Those are not the same kind of object: one is a rational plus an irrational multiple of √30, the other is a whole number, and the lab compares them by bracketing each with rationals and refining the brackets until they separate. No decimal is compared with another decimal anywhere in that. The decimal 12438.1780 appears on the page beside the exact form, rounded to four places from the surd, with the sentence that names the rounding — and the winner was decided before it was printed.
This matters more than it looks. Two candidates in a real schedule can be within a fraction of a per cent of each other, and a comparison that goes through four-place decimals is a comparison that can be settled by the rounding rather than by the costs. A pair sharing a radicand is settled by one squaring — two rational totals share the radicand 1, and their comparison collapses to the sign of a difference, which is how 10448 beats 11346 — while a pair that does not share one, which here is any comparison involving the band whose candidate is its own economic order quantity, is settled by refining rational brackets. Neither route produces a decimal.
“One candidate per band” is a rule, and a rule is a claim. The panel prices every whole quantity in the drawn range under the all-units schedule — 1800 of them on the three-band example, 1620 on the two-band one — and reports whether the scan agrees with the candidate rule to the nearest unit. It agrees on both, which is the only kind of confirmation a scan of whole numbers can give about a candidate that is a surd. If it ever disagreed the panel would say so in red rather than printing the answer it preferred.
The model has been extended in one way and left alone in every other. Demand is still constant and known, the order still arrives all at once, nothing is still ever short. What has changed is that the purchase bill is now a decision variable, and the geometry of the answer changed with it: a continuous curve with one minimum became a broken curve with one candidate per piece.
Interactive
one candidate per band, and a comparison that never rounds
Method
Five steps, and the fourth is the one that separates this from the plain model.
Each band needs a floor, a price and a holding rate. If holding is charged as a fraction of value, the rate differs band by band and each band has its own economic order quantity — leaving that out is the quiet error in this model, because the answer still looks plausible.
√(2KD/h) with the band's h. Keep it as a surd. This is the quantity the band would want if the band went on forever in both directions.
If the band's EOQ is below the band, use the floor; if above, use the ceiling; if inside, use it. Within a band the curve has one minimum, so the nearest reachable point to that minimum is the band's best — this is the whole justification for one candidate per band.
D·price + K·D/Q + h·Q/2. The purchase term is the one that makes the discount worth anything and it dwarfs the other two here: 9600 against 848 in the winning band. Comparing cycle costs alone inverts the answer.
Some totals carry a √(2KDh) and some do not. Settle the comparison on the exact forms, and only then print a decimal — labelled as rounded, beside the exact form it came from.
K = 40, D = 1200, holding at 20% of price. Bands start at 0, 500 and 1000, priced 10, 9 and 8. Everything below is what the panel reports on that schedule.
The order of the three totals is the reverse of the order of the three candidate quantities, and that is the shape of the whole model: buying more is cheaper here because the saving is D times a price difference and it applies to every unit bought forever, while the extra holding is paid on half a batch. The step beats the slope, and it beats it by a wide margin.
Exactly one of the three totals is irrational, and it is the one whose candidate is a genuine economic order quantity: 12000 + 80√30, where 80√30 = √(2KDh) with this band's h = 2. The other two candidates are band floors, whole numbers, so their totals are whole numbers too. The panel prints 12438.1780 beside the exact form with the rounding named, and the comparison that chose the winner never used it.
For a rehearsal, keep K and D and change the third band's price from 8 to 19/2 with holding 19/10. The supplied first move is that the purchase saving against the first band falls from 2400 to 600 per unit time. Predict whether band 3 still wins before you type it, and say which of the two comparisons the lab would then have to settle by bracket refinement rather than by a sign.
You should be able to state what all-units pricing charges, say why that makes the cost curve discontinuous, compute each band's own economic order quantity with that band's holding rate, replace an unreachable one by the band floor or ceiling and justify the replacement, and compare candidate totals exactly — naming which of them is irrational and why.
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