Capacity-Constrained

441. Capacity-Constrained

Relax one dimension from basic EOQ: capacity is no longer unlimited. A shared resource — warehouse space, working-capital budget, or shelf footprint — caps the total inventory you can hold at any moment. Multiple items compete for the same budget.

441.0.1. Setup

$𝑛$ items indexed $𝑖 = 1, \dots, 𝑛$ :

$𝐷_{𝑖}$ , $𝑆_{𝑖}$ , $ℎ_{𝑖}$ as in basic EOQ (item-specific)
$𝑣_{𝑖}$ = capacity consumed per unit of item $𝑖$ (e.g., $ per unit if budget-constrained, ft³ per unit if warehouse-constrained)
$𝑉$ = total capacity available

Constraint (peak inventory after a fresh order arrives):

\sum_{𝑖 = 1}^{𝑛} 𝑣_{𝑖} 𝑄_{𝑖} \leq 𝑉

If individual EOQs already satisfy the constraint, the constraint is not binding — use them directly. The interesting case is when $\sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖}^{basic} > 𝑉$ .

441.0.2. Lagrangian formulation

Minimize total cost subject to the capacity constraint:

min_{𝑄_{1}, \dots, 𝑄_{𝑛}} \sum_{𝑖} [𝑆_{𝑖} (𝐷_{𝑖} / 𝑄_{𝑖}) + ℎ_{𝑖} (𝑄_{𝑖} / 2)] s.t. \sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖} \leq 𝑉

Introduce Lagrange multiplier $𝜆 \geq 0$ for the constraint:

𝐿 (𝑄_{1}, \dots, 𝑄_{𝑛}, 𝜆) = \sum_{𝑖} [𝑆_{𝑖} (𝐷_{𝑖} / 𝑄_{𝑖}) + ℎ_{𝑖} (𝑄_{𝑖} / 2)] + 𝜆 (\sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖} - 𝑉)

441.0.3. KKT / FOC in $𝑄_{𝑖}$

For each $𝑖$ , $𝜕 𝐿 / 𝜕 𝑄_{𝑖} = 0$ :

- 𝑆_{𝑖} \frac{𝐷_{𝑖}}{𝑄_{𝑖}^{2}} + \frac{ℎ_{𝑖}}{2} + 𝜆 𝑣_{𝑖} = 0

𝑆_{𝑖} \frac{𝐷_{𝑖}}{𝑄_{𝑖}^{2}} = \frac{ℎ_{𝑖}}{2} + 𝜆 𝑣_{𝑖} = \frac{ℎ_{𝑖} + 2 𝜆 𝑣_{𝑖}}{2}

𝑄_{𝑖}^{*} (𝜆) = \sqrt{\frac{2 𝑆_{𝑖} 𝐷_{𝑖}}{ℎ_{𝑖} + 2 𝜆 𝑣_{𝑖}}}

This is basic EOQ with an inflated holding cost $ℎ_{𝑖} + 2 𝜆 𝑣_{𝑖}$ . The Lagrange multiplier $𝜆$ raises the effective holding cost on every item in proportion to its capacity consumption $𝑣_{𝑖}$ — items that hog more capacity get penalized more, shrinking their $𝑄_{𝑖}$ .

441.0.4. Find $𝜆$ from the constraint

If the constraint binds, $\sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖}^{*} (𝜆) = 𝑉$ . This is one equation in one unknown $𝜆$ — solve numerically (or in closed form for some special cases).

Special case: budget constraint with proportional holding. If $𝑣_{𝑖} = 𝑐_{𝑖}$ (capacity is dollars), $ℎ_{𝑖} = 𝑖 𝑐_{𝑖}$ (carrying-rate model), and we substitute into $𝑄_{𝑖}^{*} (𝜆)$ :

𝑄_{𝑖}^{*} = \sqrt{\frac{2 𝑆_{𝑖} 𝐷_{𝑖}}{𝑖 𝑐_{𝑖} + 2 𝜆 𝑐_{𝑖}}} = \sqrt{\frac{2 𝑆_{𝑖} 𝐷_{𝑖}}{𝑐_{𝑖}}} \cdot \frac{1}{\sqrt{𝑖 + 2 𝜆}}

Then $𝑣_{𝑖} 𝑄_{𝑖}^{*} = 𝑐_{𝑖} 𝑄_{𝑖}^{*} = \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} 𝑐_{𝑖}} / \sqrt{𝑖 + 2 𝜆}$ . Sum:

\sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖}^{*} = (\frac{1}{\sqrt{𝑖 + 2 𝜆}}) \sum_{𝑖} \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} 𝑐_{𝑖}} = 𝑉

Solve for $𝜆$ :

\sqrt{𝑖 + 2 𝜆} = \frac{\sum_{𝑖} \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} 𝑐_{𝑖}}}{𝑉}

𝜆 = \frac{1}{2} [{(\frac{\sum_{𝑖} \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} 𝑐_{𝑖}}}{𝑉})}^{2} - 𝑖]

For other constraint forms, $𝜆$ comes out of a 1-D root-finding pass over $\sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖}^{*} (𝜆) - 𝑉 = 0$ (monotone decreasing in $𝜆$ , so bisection works).

441.0.5. Algorithm

Compute unconstrained $𝑄_{𝑖}^{basic} = \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} / ℎ_{𝑖}}$ for each $𝑖$ .
Check $\sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖}^{basic} \leq 𝑉$ . If yes, done — constraint not binding.
Else, solve for $𝜆$ (closed form above, or numerical).
Each $𝑄_{𝑖}^{*} = \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} / (ℎ_{𝑖} + 2 𝜆 𝑣_{𝑖})}$ .

441.0.6. Final formulas

𝑄_{𝑖}^{*} (𝜆) = \sqrt{\frac{2 𝑆_{𝑖} 𝐷_{𝑖}}{ℎ_{𝑖} + 2 𝜆 𝑣_{𝑖}}} where 𝜆 \geq 0 satisfies \sum_{𝑖} 𝑣_{𝑖} 𝑄_{𝑖}^{*} (𝜆) = 𝑉

Sanity check: if $𝑉 \to \infty$ (no constraint), $𝜆 \to 0$ and $𝑄_{𝑖}^{*} \to \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} / ℎ_{𝑖}}$ — basic EOQs ✓.

Example

Given (2 items sharing a budget-constrained warehouse):

Item 1: $𝐷_{1} = 12000$ , $𝑆_{1}$ = $50, $𝑐_{1}$ = $10, $𝑖 = 0.20$ → $ℎ_{1} = 2$
Item 2: $𝐷_{2} = 6000$ , $𝑆_{2}$ = $50, $𝑐_{2}$ = $20, $𝑖 = 0.20$ → $ℎ_{2} = 4$
Constraint: peak inventory value $\sum_{𝑖} 𝑐_{𝑖} 𝑄_{𝑖} \leq 𝑉$ where $𝑉$ = $10,000

Step 1 — unconstrained EOQs and constraint check

𝑄_{1}^{basic} = \sqrt{\frac{2 \cdot 50 \cdot 12000}{2}} = 775

𝑄_{2}^{basic} = \sqrt{\frac{2 \cdot 50 \cdot 6000}{4}} = 387

Peak inventory value $= 10 \cdot 775 + 20 \cdot 387 = 7750 + 7740 = 15490$ . Exceeds the $10,000 budget — constraint binds.

Step 2 — find $𝜆$ via the closed-form

\sum_{𝑖} \sqrt{2 𝑆_{𝑖} 𝐷_{𝑖} 𝑐_{𝑖}} = \sqrt{2 \cdot 50 \cdot 12000 \cdot 10} + \sqrt{2 \cdot 50 \cdot 6000 \cdot 20} = \sqrt{12000000} + \sqrt{12000000} \approx 3464 + 3464 = 6928

\sqrt{𝑖 + 2 𝜆} = \frac{6928}{10000} = 0.6928 \to 𝑖 + 2 𝜆 = 0.48

2 𝜆 = 0.48 - 0.20 = 0.28 \to 𝜆 = 0.14

Step 3 — constrained order quantities

Effective holding cost per item: $ℎ_{𝑖} + 2 𝜆 𝑣_{𝑖} = ℎ_{𝑖} + 0.28 𝑐_{𝑖}$ :

Item 1: $ℎ_{1}^{eff} = 2 + 0.28 \cdot 10 = 4.8$
Item 2: $ℎ_{2}^{eff} = 4 + 0.28 \cdot 20 = 9.6$

𝑄_{1}^{*} = \sqrt{\frac{2 \cdot 50 \cdot 12000}{4.8}} = \sqrt{250000} = 500

𝑄_{2}^{*} = \sqrt{\frac{2 \cdot 50 \cdot 6000}{9.6}} = \sqrt{62500} = 250

Step 4 — verify the constraint

$10 \cdot 500 + 20 \cdot 250 = 5000 + 5000 = 10000$ ✓ exactly hits the budget.

Step 5 — total cost and comparison to basic EOQ

Item 1 cost: $50 \cdot 12000 / 500 + 2 \cdot 500 / 2 = 1200 + 500 = 1700$
Item 2 cost: $50 \cdot 6000 / 250 + 4 \cdot 250 / 2 = 1200 + 500 = 1700$
Total constrained: $3400

Unconstrained (basic EOQ for each):

Item 1: $\sqrt{2 \cdot 50 \cdot 12000 \cdot 2} \approx 1549$
Item 2: $\sqrt{2 \cdot 50 \cdot 6000 \cdot 4} \approx 1549$
Total unconstrained: $3098

Capacity penalty: $302/year (about 10% over unconstrained). The Lagrange multiplier $𝜆 = 0.14$ acts as a shadow price — each $ of additional warehouse budget would save approximately $𝜆$ × marginal value per year.