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Wilson EOQ, with the math
Step 3 of 5

Solve the first-order condition

Set dTC/dQ = 0: -D*S/Q^2 + H/2 = 0 -> Q*^2 = 2*D*S/H -> Q* = sqrt(2*D*S/H)

Rearranging: H/2 = DS/Q^2. Multiply both sides by Q^2: (H/2) Q^2 = DS. Multiply by 2/H: Q^2 = 2DS/H. Take the positive square root: Q* = sqrt(2DS/H)[1].

Square-root structure. Q* depends on the square root of (DS/H). That means doubling D inflates Q* by sqrt(2), about 41 percent. Halving H inflates Q* by sqrt(2). The non-linearity matters: Q* is forgiving of mis-estimated H but grows faster than linearly with demand.

Cycle time T*. T* = Q*/D, the time between orders. Substituting: T* = sqrt(2S/(DH)). Cycle time shrinks with both higher D and higher H, but slowly (square root) - which is why a 20 percent demand uptick rarely changes the ordering rhythm meaningfully.

The flat-curve robustness property. Total cost at Q* is TC(Q*) = sqrt(2*D*S*H). Total cost at Q = Q* * (1 + epsilon) is approximately TC(Q*) * (1 + epsilon^2/2) for small epsilon. A 20 percent error in Q produces only a 2 percent error in total cost[2]. This is why mis-estimated inputs are forgiving and why Wilson keeps ranking as a default lot-size rule even under approximate conditions.