Inventory Control

Problem Statement

Firms maintain an inventory of goods to meet future demand. Holding inventory incurs costs due to storage and tied-up capital, while replenishing inventory also incurs costs. The problem is to determine the optimal inventory management strategy to minimize total operational costs.

Economic Order Quantity (EOQ) Model

The EOQ model is a classic optimization framework used to determine the optimal order quantity that minimizes the total inventory costs.

Assumptions

• There is a steady demand of $\lambda$ units per year
• A constant cost $K$ is incurred each time an order is placed
• Acquisition cost is $c$ per unit
• Storage cost is $h$ per unit per year
• All demand must be met immediately (no back orders are allowed)
• Replenishment occurs instantaneously as soon as an order is sent
• Starting with an inventory of $Q$, it will decrease to zero in time $T=Q/\lambda$, when a replenishment of $Q$ is reordered
• This cycle repeats throughout the year

Mathematical Model

Cost per cycle: The cost per cycle consists of:

• Reorder cost: $K+cQ$
• Inventory holding cost: $\frac{hQT}{2}$ (average inventory × holding cost per unit × time)

Therefore, the total cost per cycle is: $\text{Cost per cycle}=(K+cQ)+\frac{hQT}{2}$

Annual cost: The annual cost is the cost per cycle multiplied by the number of cycles per year: $\text{Annual cost}=\text{Cost per cycle}\times \frac{1}{T}$

Since $T=Q/\lambda$, the number of cycles per year is $\lambda /Q$, giving: $\text{Annual cost}=\left(K+cQ+\frac{hQT}{2}\right)\times \frac{1}{T}=\left(K+cQ+\frac{h{Q}^2}{2\lambda }\right)\times \frac{\lambda }{Q}$

Simplifying: $\text{Annual cost}=\frac{\lambda K}{Q}+\lambda c+\frac{hQ}{2}$

Optimisation Setup

Design Variable

• $Q$ = order quantity (scalar, 1-dimensional problem)

Objective Function

We want to minimize the annual cost:

$L(Q)=\frac{\lambda K}{Q}+\lambda c+\frac{hQ}{2}$

Note that the acquisition cost term $\lambda c$ is constant with respect to $Q$ and doesn’t affect the optimization.

Solution

To find the optimal order quantity, we differentiate the objective function with respect to $Q$ and set it to zero:

$\frac{dL}{dQ}=-\frac{\lambda K}{Q^2}+\frac{h}{2}=0$

Solving for $Q$:

$\frac{\lambda K}{Q^2}=\frac{h}{2}$ $Q^2=\frac{2\lambda K}{h}$ $Q^{\ast }=\sqrt{\frac{2\lambda K}{h}}$

This result is the famous Economic Order Quantity (EOQ) formula.

Second Derivative Test

To verify this is indeed a minimum, we check the second derivative:

$\frac{d^{2}L}{d{Q}^2}=\frac{2\lambda K}{Q^3}>0$

Since the second derivative is positive, $Q^{\ast }$ is indeed a minimum.

Practical Example

Suppose:

• Annual demand ($\lambda$) = 1000 units
• Order cost ($K$) = $100 per order
• Holding cost ($h$) = $20 per unit per year

The optimal order quantity would be: $Q^{\ast }=\sqrt{\frac{2\times 1000\times 100}{20}}=\sqrt{10000}=100\text{ units}$

This means the company should order 100 units at a time, resulting in 10 orders per year.

Extensions of the Basic EOQ Model

1. Quantity Discounts: When the unit cost $c$ varies with order quantity
2. Production Quantity Model: When replenishment is not instantaneous
3. Planned Shortages: When back-ordering is allowed (at a cost)
4. Multi-Item Inventory: When multiple items share fixed ordering costs
5. Stochastic Demand: When demand is not deterministic

Key Insights

1. The EOQ model illustrates a classic trade-off in operations management between ordering costs and holding costs
2. As order quantity increases, ordering costs decrease but holding costs increase
3. The optimal policy balances these competing costs
4. This problem demonstrates how calculus techniques can be applied to find optimal business decisions
5. The model’s simplicity makes it widely applicable, despite its limiting assumptions