EPQ

449. EPQ

449.1. EPQ (Economic Production Quantity)

Relax one dimension from basic EOQ: replenishment is no longer instantaneous. Instead of receiving $𝑄$ all at once, you produce at finite rate $𝑝$ units/year while demand $𝑑$ drains stock simultaneously.

449.1.1. Setup

New variables (beyond basic EOQ):

$𝑝$ = production rate (units/year), with $𝑝 > 𝑑$ (otherwise demand outpaces supply)
$𝑑$ = demand rate (units/year) — same as $𝐷$ in basic EOQ; renamed here to emphasize it’s a rate, not annual quantity

The inventory profile in each cycle:

Buildup phase (length $\frac{𝑄}{𝑝}$ ): produce at rate $𝑝$ , demand at rate $𝑑$ — net accumulation rate $𝑝 - 𝑑$ . Inventory rises to a peak $𝐼_{max}$ .
Drawdown phase (length $\frac{𝐼_{max}}{𝑑}$ ): production stops, demand continues at rate $𝑑$ — inventory drains from $𝐼_{max}$ to 0.

Peak inventory reached at end of buildup:

𝐼_{max} = (𝑝 - 𝑑) \cdot (\frac{𝑄}{𝑝}) = 𝑄 \cdot (1 - \frac{𝑑}{𝑝})

The peak is less than $𝑄$ (the case for basic EOQ where $𝑝 = \infty$ ) because some units are consumed during production.

Average on-hand inventory (triangle, height $𝐼_{max}$ ):

\bar{𝐼} = \frac{𝐼_{max}}{2} = \frac{𝑄}{2} \cdot (1 - \frac{𝑑}{𝑝})

The factor $(1 - \frac{𝑑}{𝑝}) \leq 1$ shrinks the average vs basic EOQ.

449.1.2. Cost model

Total relevant cost (drop the constant $𝑐 𝑑$ term):

TRC (𝑄) = 𝑆 (\frac{𝑑}{𝑄}) + ℎ \cdot \frac{𝑄}{2} \cdot (1 - \frac{𝑑}{𝑝})

Only the holding term changes from basic EOQ — multiplied by $(1 - \frac{𝑑}{𝑝})$ . As $𝑝 \to \infty$ (instant production), the factor $\to 1$ and we recover basic EOQ.

449.1.3. Derive $𝑄^{*}$

Same calculus as basic EOQ: differentiate, set to zero.

\frac{𝑑}{𝑑 𝑄} TRC = - \frac{𝑆 𝑑}{𝑄^{2}} + \frac{ℎ}{2} \cdot (1 - \frac{𝑑}{𝑝}) = 0

\frac{𝑆 𝑑}{𝑄^{2}} = \frac{ℎ}{2} \cdot (1 - \frac{𝑑}{𝑝})

{(𝑄^{*})}^{2} = \frac{2 𝑆 𝑑}{ℎ (1 - \frac{𝑑}{𝑝})}

Take the square root and rewrite $1 - \frac{𝑑}{𝑝} = \frac{𝑝 - 𝑑}{𝑝}$ :

𝑄^{*} = \sqrt{\frac{2 𝑆 𝑑}{ℎ}} \cdot \sqrt{\frac{𝑝}{𝑝 - 𝑑}}

The first factor is the basic EOQ; the second is a multiplier $> 1$ that grows the production batch when the production rate is close to demand (long buildup, little overlap with drawdown).

449.1.4. Final formulas

𝑄^{*} = \sqrt{\frac{2 𝑆 𝑑}{ℎ}} \cdot \sqrt{\frac{𝑝}{𝑝 - 𝑑}} 𝑇^{*} = \frac{𝑄^{*}}{𝑑} {TRC}^{*} = \sqrt{2 𝑆 𝑑 ℎ} \cdot \sqrt{1 - \frac{𝑑}{𝑝}}

Sanity check: as $𝑝 \to \infty$ (instantaneous), the multiplier $\sqrt{\frac{𝑝}{𝑝 - 𝑑}} \to 1$ and we recover basic EOQ exactly. As $𝑝 \to 𝑑^{+}$ (production barely keeps up), $𝑄^{*} \to \infty$ — you should run the line continuously.

Example

Given (shared EOQ params + a finite production rate):

Annual demand: $𝑑 = 12000$ units/year
Setup cost: $𝑆$ = $50 / setup
Holding cost: $ℎ$ = $2 / unit / year
Production rate: $𝑝 = 24000$ units/year (twice the demand rate)

Step 1 — multiplier from finite production

\sqrt{\frac{𝑝}{𝑝 - 𝑑}} = \sqrt{\frac{24000}{24000 - 12000}} = \sqrt{2} \approx 1.414

Step 2 — production batch size

𝑄^{*} = \sqrt{\frac{2 \cdot 50 \cdot 12000}{2}} \cdot \sqrt{2} = \sqrt{600000} \cdot \sqrt{2} \approx 774.6 \cdot 1.414 \approx 1095 units

Step 3 — peak inventory and cycle

𝐼_{max} = 𝑄^{*} (1 - \frac{𝑑}{𝑝}) = 1095 (1 - 0.5) \approx 548 units

𝑇^{*} = \frac{𝑄^{*}}{𝑑} = \frac{1095}{12000} \approx 0.091 years \approx 33 days

Of $𝑇^{*}$ , the buildup takes $\frac{𝑄^{*}}{𝑝} = \frac{1095}{24000} \approx 17$ days; drawdown takes the remaining $\approx 16$ days.

Total cost:

{TRC}^{*} = \sqrt{2 \cdot 50 \cdot 12000 \cdot 2} \cdot \sqrt{1 - 0.5} \approx 1549 \cdot 0.707 \approx 1095 $/year

Compare to basic EOQ on the same demand and setup costs:

Basic EOQ ( $𝑝 = \infty$ ): $𝑄^{*} = 775$ , ${TRC}^{*} \approx$ $1549
EPQ ( $𝑝 = 2 𝑑$ ): $𝑄^{*} = 1095$ (larger), ${TRC}^{*} \approx$ $1095 (29% lower)