Lesson · Inventory Models · The order quantity

The Economic Order Quantity

One item, a steady demand rate, and one decision. Ordering cost falls like K·D/Q and holding cost rises like h·Q/2, so the total has a minimum where neither is small, and at that minimum the two halves are exactly equal — which is the equation Q² = 2KD/h and therefore the answer. The second half of the lesson is the part textbooks assert: the cost curve really is flat around the optimum, the width of the flat part is computable, and it does not depend on K, D or h at all.

Open the interactive lesson Practice

C(Q) = K·D/Q + h·Q/2 ordering falls like 1/Q, holding rises like Q at the optimum the two halves are EQUAL: K·D/Q = h·Q/2 ⟹ Q² = 2KD/h Q* = √(2KD/h) C* = √(2KDh) both irrational unless the data are kind C(t·Q*)/C* = (t + 1/t)/2 no K, no D, no h — that ratio is the flatness within 1% of the minimum: 0.8682·Q* to 1.1518·Q* = 13.2% below, 15.2% above K = 100, D = 1200, h = 6 ⟹ Q* = 200 and C* = 1200, as 600 of each half

One cost in two halves, and the width of its flat bottom

Core ideas

What this lesson establishes

Three ideas. The first is the only modelling assumption the whole formula rests on, the second is the derivation in one line, and the third is the one that changes what you do on a Monday.

Average stock is Q/2 because the sawtooth is straight

A batch of Q arrives, demand drains it at a constant rate, it hits zero, the next batch arrives. The stock level is a straight line falling from Q to 0, over and over, so its average over a cycle is Q/2 and the holding bill is h·Q/2 per unit time. Every part of that sentence is an assumption — constant demand, instant replenishment, no shortages — and the panel draws the sawtooth so that what is being assumed is visible rather than tucked into a symbol.

The optimum is where the two halves are equal, and nothing is differentiated

K·D/Q and h·Q/2 are the two costs pulling opposite ways. Set them equal and you get Q² = 2KD/h in one line. That is not a proof on its own — the crossing has to be shown to be the minimum — and the proof is a discriminant, which is the sibling lesson “The Order Quantity as a Discriminant”. What matters here is that the answer and the reason for it are both algebra a reader can check.

Flat is a measurement, not an adjective

Ordering t times the best quantity costs (t + 1/t)/2 times the minimum. That expression contains none of K, D or h, so the penalty for being wrong by a given factor is the same in every warehouse there has ever been. Twice the best quantity and half of it both cost 5/4 of the minimum; on the worked instance that is 1500 against 1200, at Q = 400 and at Q = 100 alike.

The material

The model, the quantity, and how much it matters to miss it

What is being assumed, the one-line derivation, and then the measurement that says how much precision the answer deserves.

Definition · The economic order quantity model

One item is drawn from stock at a constant demand rate D per unit time. Replenishing costs a fixed ordering cost K per order, whatever the size, and arrives all at once the moment stock reaches zero. Carrying stock costs a holding rate h per unit per unit time.

The decision is the order quantity Q. Orders are placed every Q/D units of time, so the cost per unit time is C(Q) = K·D/Q + h·Q/2: the first term is the ordering bill, the second is the holding bill.

Read the two terms rather than the sum. K·D/Q is D/Q orders per unit time at K each, and it falls as the batch grows. h·Q/2 is the holding rate times the average stock, and it rises. Neither can be made small without making the other large, which is why there is an answer at all: a cost with only one of those terms in it is minimised by ordering everything at once, or by ordering nothing, and neither is a decision.

K = 100 per order, D = 1200 a year, h = 6 per unit per year Q ordering K·D/Q holding h·Q/2 total 100 1200 300 1500 150 800 450 1250 200 600 600 1200 equal halves 250 480 750 1230 300 400 900 1300 400 300 1200 1500 the total is least where the two columns meet, and only there

The panel samples that curve on exact rationals and draws the three lines together, because the crossing IS the optimum and a reader should watch the two halves meet rather than be told they do. Underneath it the sawtooth redraws at whatever Q you pick, so the cycle length Q/D and the average level Q/2 move while you look at them.

Where the two halves cross

Theorem · The economic order quantity

K·D/Q = h·Q/2 rearranges to Q² = 2KD/h, so the crossing is at Q* = √(2KD/h). The cost there is C* = √(2KDh), and the two halves are each √(2KDh)/2.

That the crossing is the minimum rather than merely a landmark is the content of the discriminant argument, and it is done in the next lesson rather than asserted here.

Nothing above was differentiated. This path does not have calculus and does not need it: the sibling lesson turns “is there a quantity costing at most T” into a quadratic inequality and reads the minimum off its discriminant, which is the same argument Algebra makes in “The Discriminant” and “Quadratic Inequalities”, applied to a cost.

Example · A workshop whose answer is a whole number

At K = 100, D = 1200, h = 6 the panel reports Q* = 200 and C* = 1200, split as 600 of ordering and 600 of holding. Both are rational here — 2KD/h = 40000 and 2KDh = 1440000 are perfect squares — so the panel says so and every digit it prints is exact.

This is the only kind of instance on which a reader can check the whole page by hand, which is why it is the one the lab opens with. It is also, in the arithmetic sense, a coincidence.

Example · A bench whose answer is not

Switch the worked example to K = 50, D = 600, h = 5 and 2KD/h = 12000, which is not a square. The panel now prints Q* = 20√30 and C* = 100√30, tags both as irrational, and puts 109.5445 and 547.7226 beside them with the sentence that names the rounding and the method — four places, from the exact surd, by integer square root with three guard digits.

The cheapest WHOLE quantity there is 110, and the panel finds it by pricing every whole quantity rather than by rounding: 109 really does cost more than 110, 119405/218 against 6025/11, and that is a comparison of two fractions with no decimal in it.

How flat is flat

“The cost curve is flat around the optimum” is the sentence every account of this model contains and almost none of them measures. It is measurable. Order t times the best quantity and the cost is (t + 1/t)/2 times the minimum — substitute t·Q* into C(Q) and every one of K, D and h cancels. The penalty for being out by a factor is a property of the factor alone.

t C(t·Q*)/C* on K = 100, D = 1200, h = 6: Q = t·200, cost 1/2 5/4 100 1500 3/4 25/24 150 1250 1 1 200 1200 6/5 61/60 240 1220 2 5/4 400 1500 t and 1/t give the SAME ratio, so half the batch and twice the batch cost the same 25% extra — the curve is symmetric in t, not in Q

So “within a fraction e of the minimum” is (t + 1/t)/2 ≤ 1 + e, which is t² − 2(1 + e)t + 1 ≤ 0: a second quadratic, with a second discriminant 4e(2 + e), whose roots are (1 + e) ± √(e(2 + e)). At e = 1/100 the discriminant is 201/2500 and the ends of the band are 101/100 ± √201/100. Those are irrational, and the panel prints that exact form beside the decimals rather than only the decimals.

In quantities, on the worked instance, the band runs from 173.6451 to 230.3549: you may order 13.2% too little or 15.2% too much and still be within one per cent of the best cost there is. Those two percentages are the same on every instance the panel can be given, because they come from t and t has no K, D or h in it. The two decimals are the midpoints of rational brackets under 10⁻¹³ wide, so every digit shown is a digit of the true number — a statement about the printing, not a hope about it.

Widen the tolerance and the band widens the way the ratio says it must: 5% of the minimum allows 145.9688 to 274.0312, and 25% allows exactly 100 to 400. That last one is the one case where the ends come out rational, and the panel switches its wording to say so: at e = 1/4 the roots are 1/2 and 2, which is the same 5/4 seen from the other side.

What the model left out is a list worth keeping. Demand is constant and known; the order arrives complete and instantly; nothing is ever short; the price per unit does not depend on how much is bought, so it is a constant D·price that no choice of Q can change and it is left out of C(Q) altogether. The three lessons after this one put the last three of those back, one at a time, and each one changes the answer.

Interactive

A batch too small is ordered too often; a batch too large sits in the store

the quantity where those two costs balance, and how wide the flat bottom is

Move Q off the optimum and watch how little the total cares

The two cost curves are sampled on exact rationals and drawn with their total, so the crossing is visible rather than asserted. The sawtooth beneath redraws at your Q. The shaded band is every quantity costing at most the chosen percentage more than the best one, and it comes from the discriminant of t² − 2(1 + e)t + 1 rather than from trying quantities until one is close enough. The last row prices every whole quantity in range, because a warehouse cannot order a surd.

Worked example

Ordering cost K, per order

Demand D, per unit time

Holding cost h, per unit per unit time

Order quantity Q to price200

Band: within this many per cent of the minimum1

Q* exactly—

Q* rounded—

C* exactly—

C* rounded—

Cost at your Q—

Cheapest whole Q—

The two costs cross where the total is least, and that is not a coincidence: K·D/Q = h·Q/2 rearranges to Q² = 2KD/h. Slide Q away from the optimum and watch the total barely move — the shaded band is every quantity costing at most one per cent more than the best one, computed from a discriminant rather than sampled.

Method

Working an instance from the three numbers

Four steps, and the fourth is the one that decides whether the first three were worth doing to four decimal places.

Get K, D and h onto the same clock

D is per unit time and h is per unit per unit time, and they must agree: a demand of 1200 a year with a holding rate quoted per month gives an answer wrong by a factor of √12. K has no time in it at all — it is per order — which is exactly why it is easy to misread.

Compute 2KD/h and look at it before taking the root

If it is a perfect square the answer is a whole number and everything on the page is exact. If it is not, the answer is irrational and stays a surd: 20√30, not 109.5445. The decimal is for reading, and it is only ever produced at the last moment.

Check the crossing rather than the formula

Price K·D/Q* and h·Q*/2 separately. They must be equal, and if they are not, the arithmetic is wrong — this is a one-line check that needs no second formula, and it catches a mis-typed h immediately.

Round to something you can actually order, and price that

Cases of twelve, a pallet of 200, a minimum of 500. Round the surd to the quantity you can buy, then price it: the band tells you in advance how much the rounding costs, and it is almost always less than a reader expects.

Ask which of the four assumptions your situation breaks

Constant demand, instant arrival, no shortages, a price that does not depend on the quantity. Whichever one is false is the lesson you need next, and there is one for each.

The quantity, the crossing and the flat part

Score 0 / 0

Choose the answer you can justify.

One instance priced by hand, and the band around it

K = 100 per order, D = 1200 a year, h = 6 per unit per year. Every figure below is what the panel prints on that instance, and all of them are exact.

2KD/h = 2(100)(1200)/6 = 40000 Q* = √40000 = 200 exactly 2KDh = 2(100)(1200)(6) = 1440000 C* = √1440000 = 1200 exactly check the crossing: K·D/Q* = 120000/200 = 600 h·Q*/2 = 6(200)/2 = 600 equal, as it must be the flat bottom, from the second discriminant: (t + 1/t)/2 ≤ 1 + e ⟺ t² − 2(1 + e)t + 1 ≤ 0 discriminant 4e(2 + e); at e = 1/100 that is 201/2500 ends 101/100 ± √201/100 = 0.8682… and 1.1518… in quantities: 173.6451 … 230.3549 13.2% below, 15.2% above at e = 5/100: 145.9688 … 274.0312 27.0% below, 37.0% above at e = 25/100: 100 … 400 exactly half Q* and twice Q* cheapest WHOLE quantity, by pricing every one of them: 200, at 1200

The two percentages are the whole point of the exercise. They were produced without ever mentioning K, D or h, because the ratio (t + 1/t)/2 does not contain them — so “13.2% below, 15.2% above” is the 1% band for every instance of this model, not for this one. Change the three inputs in the panel and watch those two numbers sit still while everything around them moves.

The asymmetry is worth a second look. 13.2 and 15.2 are not equal, and yet half the quantity and twice the quantity cost the same. Both facts come from the same place: the ratio is symmetric under t ⟼ 1/t, which is a symmetry of multiplication rather than of addition, and a band that is symmetric in the factor cannot be symmetric in the difference.

For a rehearsal, switch the panel to the bench example, K = 50, D = 600, h = 5. The first move is given: 2KD/h = 12000 = 400 × 30, so Q* = 20√30 and nothing about it is a whole number. Predict what happens to the two band percentages before you look, and then say why the cheapest whole quantity is 110 rather than 109.

Common mistakes

Charging holding cost on the whole batchThe batch is Q only at the instant it arrives. Averaged over the cycle the level is Q/2, and using h·Q instead of h·Q/2 reports an order quantity √2 too small and a cost √2 too large. The sawtooth in the panel is there to make the halving something you can see rather than a factor to remember.

Hearing “flat” as “anything will do”The band is finite and it is computed. Within 1% of the minimum the order quantity may move 13.2% down or 15.2% up — a real tolerance, and a good argument for rounding to a pallet — but t = 2 costs 25% more and t = 5 costs 160% more. Flat near the bottom is not flat everywhere, and the panel will draw you the difference at any tolerance you like.

Quoting the decimal as the answer109.5445 is not the economic order quantity; 20√30 is, and the decimal is that surd rounded to four places for reading. The distinction matters as soon as the number is used again: comparing two plans by their decimals can invert the comparison, which is exactly why the lab compares surd-valued costs by squaring rather than by printing.

Check your understanding

✓

Finish when you can say what it costs to be wrong, not only what the right answer is.

You should be able to write C(Q) from a described situation, derive Q* = √(2KD/h) by setting the two halves equal, say why the crossing is the minimum and where that is proved, keep an irrational answer as a surd and name what rounding does to it, and compute the band of quantities within a given percentage of the best cost — including the fact that the band does not depend on the data.

Note: The crossing is where the minimum is; it is not yet a proof that it is a minimum. The next lesson asks the question the other way round — “is there a quantity costing at most T” — and gets the whole answer, existence and optimality together, out of one discriminant.