GAM · Optimal Order Quantity — Strategy & Tricks

📐 One formula, five cousins

Order too often and you drown in paperwork; order too rarely and you drown in stock. The Economic Order Quantity is the lot size where the annual ordering bill equals the annual holding bill — the point where their sum is smallest. Everything here flows from it and a few one-liners.

Q* = √(2DS ÷ H) — optimal order size
N = D ÷ Q — orders per year
Ordering cost = (D ÷ Q) × S
Holding cost = (Q ÷ 2) × H — average stock is Q ÷ 2
At Q*: ordering cost = holding cost
TC* = √(2DSH) = Q* × H — minimum relevant cost
d = D ÷ working days (360, 250 or 365 — as stated)
ROP = d × L (+ safety stock SS)
Cycle = Q ÷ d = working days ÷ N

Not in the kit: the purchase price. With a fixed unit price the annual purchase bill is fixed too, so it cannot move Q. It enters only with a quantity discount, where you compare full* annual cost at each break.

👀 Sort it in five seconds

Every question here is one of six jobs. Name the job before you touch a number.

Optimal means the costs are equal, so do one. A Q handed to you means do both.

🧠 Label the givens, then move

Before any formula, write six labels in the margin and fill in what the passage gives. A blank is information too — it tells you which variant you're in.

1. D — demand per year: 1,500 a month → 18,000; a 900-unit lot lasting 3 months → 3,600.
2. S — cost per order, not a year's worth.
3. H — holding cost per unit per year. A percentage of price? Multiply: 10% of 40 = 4.
4. Q — is one given? Yes: cost that Q. No: compute Q* = √(2DS ÷ H).
5. L, SS, days — lead time, safety stock, working-day year (360, 250 or 365).
6. Write N = D ÷ Q and d = D ÷ days beside Q; later questions are then one line.
7. Answer exactly what was asked — Q*, one cost, N, days, ROP — then stop.

Labelling buys the whole set: Q* = √(2 × 6,000 × 30 ÷ 4) = √90,000 = 300; N = 20; d = 24; ROP = 24 × 6 + 150 = 294; cycle = 300 ÷ 24 = 12.5 days.

In a table, each row is its own EOQ: take the row the stem names and read units off the headings.

✂️ Roots without a calculator

No calculator, and the setters know it: the number under the root is nearly always a perfect square in disguise.

• Cancel before you multiply. For 2 × 18,000 × 30 ÷ 3, do 30 ÷ 3 = 10 first, then 2 × 18,000 × 10 = 360,000 — never build 1,080,000 and divide.
• Strip zeros in pairs. √360,000 = 6 × 100 = 600; √1,440,000 = 12 × 100 = 1,200. Odd zeros mean a slip — or an 'approximately' question: √2 ≈ 1.41, √5 ≈ 2.24.
• Scale, don't recompute. Q ∝ √D, √S and 1 ÷ √H: D × 4 → Q × 2; H × 4 → Q* ÷ 2; D +44% → × 1.2; S −36% → × 0.8; D and H both doubled → unchanged.
• At the optimum, one cost is enough. Holding is (Q ÷ 2) × H, ordering the same, the total double — or straight to TC = Q* × H.
• Days between orders = working days ÷ N. 360 days, 20 orders → 18 days. Q never enters.
• Penalty for a wrong lot. TC(Q) ÷ TC = 0.5 × (Q ÷ Q + Q* ÷ Q): half or double the EOQ costs exactly 25% more, 20% under costs 2.5%. That flat bottom makes 'nearest feasible lot' questions safe.

⚡ The 90-second budget

Ninety seconds is the average, not the ration: a five-question set gets about seven minutes, front-loaded.

• Read the options first. A spread like 173, 232, 600, 2,683 names the slips on offer; the clean number is usually the answer.
• Never total when a component is asked, and never do both components at Q* — they're equal.
• Round late. Carry 848.5 to the last line; options sit far enough apart that ±1 never matters.

That is the exact-versus-estimate line: a clean square is exact, anything else you bracket.

🪤 Six ways to lose a free mark

Nothing here is hard; everything is easy to get wrong at speed, and each trap has an option waiting.

1. A given Q is not optimal. 'Orders 1,000 at a time' does not equalise the costs — do both; the trap option doubles one.
2. Per unit-month vs per unit-year. 0.50 a month becomes 6 a year. '15% of unit price' means multiply, not plug in 0.15.
3. Monthly demand under the root. The answer lands √12 ≈ 3.5 times too small — always an option.
4. Units vs orders. Q is units; N = D ÷ Q is orders — the commonest mis-read here.
5. Safety stock forgotten — or added twice. ROP = d × L + SS, once; average stock with a reserve is Q ÷ 2 + SS, once.
6. The wrong year. Daily demand uses the year the passage states; switch it and every ROP option looks plausible.

🧩 Worked examples, easy to hard

Six passages built like real items — cover the steps, then compare.

🎯 Recognition drills

Recognition is the speed, not arithmetic. Say each answer out loud before the seconds run out.

🧭 Guided walk-through: one passage, four questions

1. 0:00–0:20 — Label. D = 15,000, S = 48, H = 4, days = 250, L = 5, SS = 120. Q = 1,000 appears only in Q4.
2. 0:20–0:50 — Q1, the EOQ. 2 × 48 ÷ 4 = 24; 15,000 × 24 = 360,000; √360,000 = 600.
3. 0:50–1:05 — Q2, orders a year. N = 15,000 ÷ 600 = 25; bonus, cycle = 250 ÷ 25 = 10 days.
4. 1:05–1:35 — Q3, reorder point. d = 15,000 ÷ 250 = 60; lead-time demand 60 × 5 = 300; ROP = 420. A 300 option forgets the reserve; 540 counts it twice.
5. 1:35–2:20 — Q4, cost at Q = 1,000. Not optimal, so both: ordering 15 × 48 = 720, holding 500 × 4 = 2,000, total 2,720. At Q* it would be 600 × 4 = 2,400, so a higher figure is right.
6. 2:20 — Done. Four questions in 140 seconds against a budget of 360.

📈 Progressive practice

Three untimed sets. Accuracy first — make each variant automatic before you race it.

Easy — compute Q*, N, average stock, one cost at Q* → 6 real easy questions · free, no login

Easy: D, S and H given, one line of the kit. Move on at 6 of 6.

Medium — given Q, conversions, percentage changes, two-policy comparisons → 6 real medium questions · free, no login

Medium: a hidden conversion, or a Q that isn't optimal. Move on at 5 of 6.

Hard — reverse-engineering, ROP with safety stock, cost penalties, discounts → 6 real hard questions · free, no login

Hard: two ideas chained — the hidden S, then a rescale. Move on at 4 of 6.

⏱️ Speed training

Now the clock, with targets from the budget table. Log where the seconds go: reading, labelling or the root.

Easy under 45 s each → 6 real easy questions, against the clock · free, no login

Medium under 75 s each → 6 real medium questions, against the clock · free, no login

Hard under 90 s each → 6 real hard questions, against the clock · free, no login

• Missed on easy? You're re-reading. Label once, then go back only for a number.
• Slow on medium? It's the root. Cancel first; strip zeros in pairs.
• Slow on hard? You're recomputing Q* where a ratio would do, or doing both costs at the optimum.

🏆 Exam application

This family arrives as one passage with four or five linked questions. Treat the set as a unit: budget six to seven minutes, the first on labels and Q*.

• Answer everything — nothing is deducted for a wrong one. If a root won't resolve, bracket it and take the nearest option.
• Work the set in order unless the first question is the heavy one; then bank the one-liners first.
• Match units before numbers. Units, orders, days or currency: one glance kills two options.
• Two options a factor of 2 apart? That is Q versus Q ÷ 2, or one cost versus the total.
• Skip forward if a passage adds something outside the kit — uncertain demand, backorders.

Exam simulation — 90 s per question, options only → 6 real hard questions, against the clock · free, no login

📋 Cheat sheet

Q* = √(2DS ÷ H)
N = D ÷ Q · Cycle = working days ÷ N = Q ÷ d
Ordering = (D ÷ Q) × S · Holding = (Q ÷ 2) × H
At Q*: ordering = holding · TC* = √(2DSH) = Q* × H
Average stock = Q ÷ 2 (+ SS if a reserve is held)
d = D ÷ working days (as stated) · ROP = d × L + SS
Reverse: S = HQ² ÷ 2D · H = 2DS ÷ Q² · D = HQ² ÷ 2S
Scaling: Q* ∝ √D, √S, 1 ÷ √H · D × 4 → Q* × 2
Penalty: 0.5 × (Q ÷ Q* + Q* ÷ Q) · half or double → +25%

• Labelled D, S, H, Q, L, SS before any arithmetic
• D per year, H per unit-year, S per order
• Given Q → both costs; 'optimal' → one cost
• Cancelled before multiplying; zeros stripped in pairs
• Safety stock added once; working days as stated
• Answer's units match the question's; nothing left blank