12.3 Deterministic Inventory Models: EOQ, EPQ, and Safety Stock

Inventory represents one of the largest capital commitments on an industrial firm's balance sheet. Holding inventory serves vital operational functions: decoupling independent stages of production, hedging against price volatility and raw material inflation, buffering against demand spikes, and capturing supplier economies of scale. However, excessive inventory ties up working capital, inflates warehousing costs, and risks spoilage and obsolescence. Industrial engineers formulate mathematical inventory models to determine precisely when to order and how much to order.

---

1. Inventory Cost Taxonomy

Total inventory cost models optimize across four primary cost categories:

1. Holding / Carrying Cost ($H$): The total annual cost incurred to maintain one unit in inventory for a full year. Expressed either as a direct dollar amount per unit-year ($H$) or as a percentage ($i$) of the unit purchase price ($C$): $$H = i \cdot C$$ Components include:
   • Opportunity Cost of Capital: The foregone interest or corporate hurdle rate (MARR) on tied-up cash (typically 10%–20%).
   • Storage & Facility Costs: Warehouse rent, utilities, automated storage/retrieval system (ASRS) operation.
   • Handling & Labor: Forklift operations, cycle counting, security.
   • Taxes & Insurance: Ad valorem property taxes on stock, fire/theft insurance.
   • Depreciation, Shrinkage & Obsolescence: Pilferage, shelf-life expiration, engineering obsolescence.
2. Ordering / Setup Cost ($S$): Fixed cost incurred every time an order is placed or a machine setup is performed, regardless of order size $Q$:
   • Procurement: Purchase requisition preparation, vendor EDI transmission, expediting, inbound delivery inspection, dock invoice processing.
   • Manufacturing: Machine tool tear-down, recalibration, line changeover, trial run quality scrap.
3. Unit Purchase / Item Cost ($C$): The purchase price paid to an external vendor or the direct material and labor cost to manufacture one unit.
4. Shortage / Stockout Cost ($C_s$): Penalties incurred when customer demand exceeds available inventory: emergency expedited freight, backorder tracking administration, contract non-performance liquidated damages, and customer loss of goodwill.

---

2. The Classic Economic Order Quantity (EOQ) Model

The Economic Order Quantity (EOQ), originally developed by Ford W. Harris, determines the order quantity that minimizes the sum of annual setup costs and annual inventory holding costs under deterministic conditions.

EOQ Sawtooth Inventory Profile
 Inventory
    ▲
  Q │    /|      /|      /|
    │   / |     / |     / |
    │  /  |    /  |    /  |      Average Inventory = Q / 2
Q/2 ┼─/───┼───/───┼───/───┼─
    │/    |  /    |  /    |  
  0 └─────┴──┴────┴──┴────┴───► Time
       ◄──T*──►

Core Assumptions of Classic EOQ

1. Annual demand $D$ is continuous, known, and occurs at a constant deterministic rate.
2. Replenishment is instantaneous (lead time is known and constant, and the entire batch $Q$ arrives all at once; replenishment rate $\to \infty$).
3. Shortages and backorders are strictly prohibited.
4. Unit purchase price $C$ is constant across all order volumes (no quantity discounts initially).
5. Order cost $S$ and unit annual holding cost $H$ are fixed and known.

Mathematical Formulation

• Number of orders placed per year: $N = \frac{D}{Q}$
• Annual Ordering Cost: $\text{AOC} = \frac{D}{Q} S$
• Average Inventory Level: $\bar{I} = \frac{Q}{2}$
• Annual Holding Cost: $\text{AHC} = \frac{Q}{2} H$
• Total Annual Inventory Cost (TC): $$\text{TC}(Q) = \frac{D}{Q} S + \frac{Q}{2} H + D C$$

Because annual purchase cost $D C$ is constant with respect to $Q$, minimizing total cost requires taking the first derivative of total variable cost with respect to $Q$ and setting it to zero:

$$\frac{d\text{TC}}{dQ} = -\frac{D S}{Q^2} + \frac{H}{2} = 0 \implies \frac{D S}{Q^2} = \frac{H}{2}$$

Fundamental EOQ Insight: At the optimal order quantity $Q^*$, Annual Ordering Cost exactly equals Annual Holding Cost: $\frac{D}{Q^*} S = \frac{Q^*}{2} H$

Solving for $Q^*$ yields the celebrated EOQ formula:

$$Q^* = \sqrt{\frac{2 D S}{H}}$$

Additional EOQ Operating Metrics

• Total Annual Variable Cost at $Q^*$ (TVC): $$\text{TVC}(Q^*) = \sqrt{2 D S H}$$
• Optimal Cycle Time (Time Between Orders, $T^*$): $$T^* = \frac{Q^*}{D} \quad (\text{years}) = \frac{Q^*}{D} \times (\text{operating days/year})$$
• Reorder Point (ROP): Under deterministic lead time $L$ and daily demand $d = D / \text{days}$: $$\text{ROP} = d \cdot L$$ (If lead time $L$ exceeds cycle time $T^*$, order cycles overlap, and $\text{ROP} = d \cdot [L \pmod{T^*}]$).

---

3. Economic Production Quantity (EPQ / POQ)

In manufacturing environments, replenishment is rarely instantaneous. A machine cell produces parts at a finite production rate $p$ while downstream assembly simultaneously consumes them at demand rate $d$ ($p > d$). This is known as the Economic Production Quantity (EPQ) or Production Order Quantity (POQ).

EPQ Inventory Build-up and Depletion Profile
 Inventory
    ▲
Imax│        /\            /\
    │       /  \          /  \
    │      /    \        /    \      Rate of build = (p - d)
    │     /      \      /      \     Rate of depletion = d
  0 └────┴────────┴────┴────────┴───► Time
         ◄─tp─►◄──td──►
         ◄─────T──────►

Dynamics of the Production Run

1. Production Phase ($t_p$): The machine operates for duration $t_p = Q / p$. Parts are manufactured at rate $p$ and simultaneously consumed at rate $d$. Net inventory accumulates at the rate $(p - d)$.
2. Maximum Inventory Level ($I_{\max}$): Occurs precisely when the production run terminates at $t_p$: $$I_{\max} = t_p (p - d) = \frac{Q}{p}(p - d) = Q \left(1 - \frac{d}{p}\right)$$
3. Depletion Phase ($t_d$): Production halts ($p = 0$). Demand consumes remaining inventory at rate $d$ over time $t_d = I_{\max} / d$ until stock reaches zero.
4. Total Cycle Length ($T$): $T = t_p + t_d = \frac{Q}{d}$.

EPQ Mathematical Formulation

• Average Inventory Level: $\bar{I} = \frac{I_{\max}}{2} = \frac{Q}{2}\left(1 - \frac{d}{p}\right)$
• Total Annual Cost: $$\text{TC}_p(Q) = \frac{D}{Q} S + \frac{Q}{2} H \left(1 - \frac{d}{p}\right) + D C$$

Setting $\frac{d\text{TC}_p}{dQ} = 0$ yields the optimal production lot size ($Q_p^*$):

$$Q_p^* = \sqrt{\frac{2 D S}{H \left(1 - \frac{d}{p}\right)}} = Q^* \sqrt{\frac{p}{p - d}}$$

Key Relationships Between EPQ and EOQ

• Because $(1 - d/p) < 1$, the denominator shrinks, which means $Q_p^* > Q^*$ (optimal production lot size is always larger than classic purchase EOQ).
• However, because inventory is consumed during production, the maximum accumulated inventory is substantially less than the batch size: $I_{\max} = Q_p^*(1 - d/p) < Q_p^*$.
• As production rate becomes extremely fast ($p \to \infty$), $d/p \to 0$, and EPQ converges mathematically to classic EOQ.

---

4. Quantity Discount Models (All-Units Discounts)

Suppliers frequently offer price discounts to incentivize large order volumes. In an all-units quantity discount, the discounted unit price $C_j$ applies to every single unit purchased in the order, provided the order quantity meets or exceeds the minimum threshold $q_j$.

Price Schedule Structure

$$\text{Unit Price } C = \begin{cases} C_1, & \text{for } 0 \le Q < q_1 \\ C_2, & \text{for } q_1 \le Q < q_2 \\ C_3, & \text{for } Q \ge q_2 \end{cases}$$

where $C_1 > C_2 > C_3$.

Quantity Discount Total Cost Curves
 Annual Cost
    ▲
    │     /           TC(C1) [No discount]
    │    /   /        TC(C2) [Discount 1]
    │   /   /   /     TC(C3) [Discount 2]
    │  /   /   / 
    │ ─── ─── ● ◄── Breakpoint q2 (Often Global Optimum!)
    │    ● ◄─────── Breakpoint q1
    │
  0 └─────────────────────────► Order Quantity (Q)

Step-by-Step Optimization Algorithm

1. Calculate $Q^*$ for the Lowest Price Tier: Start with the largest discount (lowest unit cost $C_k$, setting $H_k = i \cdot C_k$): $$Q_k^* = \sqrt{\frac{2 D S}{H_k}}$$
2. Check Feasibility:
   • If $Q_k^*$ is feasible ($Q_k^* \ge q_{k-1}$), then $Q_k^*$ is the absolute global optimum. Stop.
   • If $Q_k^*$ is infeasible ($Q_k^* < q_{k-1}$), compute total annual cost $\text{TC}$ at the minimum price break threshold $q_{k-1}$: $$\text{TC}(q_{k-1}) = \frac{D}{q_{k-1}} S + \frac{q_{k-1}}{2} H_k + D C_k$$
3. Evaluate Higher Price Tiers: Move to the next lower discount tier (unit cost $C_{k-1}$). Calculate $Q_{k-1}^*$. If feasible, compute its total cost $\text{TC}(Q_{k-1}^*)$. If infeasible, evaluate at its price break threshold.
4. Select Lowest Total Cost: Compare the total costs calculated across all candidate points (feasible EOQs and candidate price breakpoints). Select the quantity that yields the lowest Total Annual Cost (including purchase cost $D C$).

Exam Trap Alert: In quantity discount problems, you MUST include unit purchase cost $D C$ in the total cost comparison! In basic EOQ, $D C$ is constant and can be ignored; in discount problems, $D C$ changes across tiers and represents the single largest cost difference.

---

5. Stochastic Inventory Control and Safety Stock

When customer demand or supplier lead times fluctuate unpredictably, carrying only deterministic inventory leads to frequent stockouts. Industrial engineers implement Safety Stock ($SS$) as a buffer against demand surges and delivery delays during the replenishment lead time.

Lead Time Demand Distribution

Under standard stochastic assumptions, daily demand is a normally distributed random variable with mean $\bar{d}$ and standard deviation $\sigma_d$. If lead time $L$ is constant across cycles, total demand during lead time ($D_L$) is normally distributed:

$$D_L \sim \mathcal{N}\left(\bar{d} \cdot L, \; \sigma_L^2\right)$$

By the sum of independent random variables, the standard deviation of lead time demand is governed by the Square Root of Lead Time Law:

$$\sigma_L = \sqrt{\sum_{i=1}^L \sigma_d^2} = \sqrt{L \cdot \sigma_d^2} = \sigma_d \sqrt{L}$$

Safety Stock Formulation

To achieve a desired Cycle Service Level (CSL)—defined as the probability of not stocking out during a replenishment cycle ($P(D_L \le \text{ROP}) = 1 - \alpha$):

$$\text{SS} = Z \cdot \sigma_L = Z \cdot \sigma_d \sqrt{L}$$

where $Z$ is the standard normal critical value satisfying $\Phi(Z) = \text{CSL}$.

Stochastic Reorder Point (ROP)

The reorder point under uncertain demand incorporates both expected lead time demand and safety stock:

$$\text{ROP} = \bar{d} \cdot L + \text{SS} = \bar{d} \cdot L + Z \cdot \sigma_d \sqrt{L}$$

Lead Time Uncertainty Cases

• Variable Lead Time, Constant Demand ($d$): $$\sigma_L = d \cdot \sigma_{\text{LT}} \implies \text{SS} = Z \cdot d \cdot \sigma_{\text{LT}}$$
• Both Demand and Lead Time Variable (Independent): $$\sigma_{DL} = \sqrt{\bar{L} \sigma_d^2 + \bar{d}^2 \sigma_{\text{LT}}^2} \implies \text{SS} = Z \sqrt{\bar{L} \sigma_d^2 + \bar{d}^2 \sigma_{\text{LT}}^2}$$

---

6. The Single-Period Inventory Model (Newsvendor Model)

The classic EOQ model assumes inventory unsold in one period carries over seamlessly to the next. For perishable goods, seasonal merchandise, high-fashion apparel, or single-day events, unsold items cannot be held over and must be salvaged or disposed of at a steep loss. This is governed by the Newsvendor Model.

Cost Drivers

• Cost of Underage ($C_u$): The marginal profit foregone per unit of unmet demand (under-stocking penalty): $$C_u = P - C + g$$ where $P$ is retail selling price, $C$ is acquisition cost, and $g$ is customer goodwill loss.
• Cost of Overage ($C_o$): The loss incurred on each unsold unit remaining at the end of the period (over-stocking penalty): $$C_o = C - V + s$$ where $V$ is salvage/liquidation value recovered, and $s$ is disposal cost.

The Critical Fractile

To maximize expected profit, the optimal order quantity $Q^*$ occurs where the probability of satisfying all demand equals the critical fractile ($CF$):

$$P(D \le Q^*) = \frac{C_u}{C_u + C_o}$$

Critical Fractile Distribution
 Probability
    ▲
    │            Optimal Q*
    │           /│
    │       _.-' │ `-._
    │     .'     │     `.      Area to left = P(D <= Q*) = Cu / (Cu + Co)
    │   .'       │       `.
  0 └──┴─────────┴─────────┴──► Demand (D)
                 ▲
                Q*

Determining Optimal Order Quantity $Q^*$

• For Continuous Normally Distributed Demand ($D \sim \mathcal{N}(\mu, \sigma)$): Find $Z^*$ from the standard normal cumulative distribution table such that $\Phi(Z^*) = \frac{C_u}{C_u + C_o}$, then: $$Q^* = \mu + Z^* \cdot \sigma$$
• For Discrete Empirical Demand: Accumulate cumulative probabilities $P(D \le Q)$. Select the smallest order quantity $Q^*$ satisfying: $$P(D \le Q^*) \ge \frac{C_u}{C_u + C_o}$$

---

7. Step-by-Step Worked Engineering Calculations

Worked Example 12.3.1: EOQ and Quantity Discount Trade-Off

Problem: An industrial valve distributor consumes 10,000 replacement seals annually. Ordering cost is $50 per order, and the annual inventory carrying rate is $i = 20\%$. The manufacturer quotes the following all-units discount schedule:

• Tier 1 (1 to 499 units): $C_1 = \text{\textdollar}20.00$ per unit
• Tier 2 (500 to 999 units): $C_2 = \text{\textdollar}19.00$ per unit
• Tier 3 (1,000+ units): $C_3 = \text{\textdollar}18.20$ per unit

Determine the optimal order quantity $Q^*$ that minimizes total annual inventory cost.

Solution:

Step 1: Evaluate Tier 3 (Lowest Price: $C_3 = \text{\textdollar}18.20$, $q_2 = 1,000$)

• Holding cost: $H_3 = i \cdot C_3 = 0.20 \times 18.20 = \text{\textdollar}3.64\text{/unit-year}$.
• Compute candidate EOQ: $$Q_3^* = \sqrt{\frac{2 D S}{H_3}} = \sqrt{\frac{2 \times 10,000 \times 50}{3.64}} = \sqrt{\frac{1,000,000}{3.64}} = \sqrt{274,725.27} \approx 524.1\text{ units}$$
• Feasibility Check: $524.1 < 1,000$. Thus, $Q_3^*$ is infeasible in Tier 3.
• Compute Total Cost at Tier 3 price breakpoint ($Q = 1,000$ units): $$\text{TC}(1,000) = \frac{D}{Q} S + \frac{Q}{2} H_3 + D C_3$$ $$\text{TC}(1,000) = \frac{10,000}{1,000}(50) + \frac{1,000}{2}(3.64) + 10,000(18.20)$$ $$\text{TC}(1,000) = 500 + 1,820 + 182,000 = \$184,320.00$$

Step 2: Evaluate Tier 2 ($C_2 = \text{\textdollar}19.00$, $500 \le Q < 1,000$)

• Holding cost: $H_2 = i \cdot C_2 = 0.20 \times 19.00 = \text{\textdollar}3.80\text{/unit-year}$.
• Compute candidate EOQ: $$Q_2^* = \sqrt{\frac{2 \times 10,000 \times 50}{3.80}} = \sqrt{\frac{1,000,000}{3.80}} = \sqrt{263,157.89} \approx 513.0\text{ units}$$
• Feasibility Check: $500 \le 513.0 < 1,000$. Feasible!
• Compute Total Cost at $Q_2^* = 513$ units: $$\text{TC}(513) = \frac{10,000}{513}(50) + \frac{513}{2}(3.80) + 10,000(19.00)$$ $$\text{TC}(513) = 974.66 + 974.70 + 190,000 = 1,949.36 + 190,000 = \$191,949.36$$

Step 3: Compare Total Annual Costs

• $\text{TC}(1,000 \text{ units}) = \text{\textdollar}184,320.00$
• $\text{TC}(513 \text{ units}) = \text{\textdollar}191,949.36$
• Cost savings by ordering 1,000 units = $\text{\textdollar}191,949.36 - \text{\textdollar}184,320.00 = \text{\textdollar}7,629.36\text{ per year}$.
• Engineering Decision: Order $Q^* = 1,000$ units at the Tier 3 price breakpoint.

---

Worked Example 12.3.2: Economic Production Quantity (EPQ) Mechanics

Problem: A precision machine shop manufactures hydraulic couplings on a dedicated CNC lathe. Annual demand is $D = 18,000$ units. The facility operates 250 days per year, giving a daily demand of $d = 18,000 / 250 = 72$ units/day. When operating, the lathe produces couplings at a rate of $p = 300$ units/day. Setup cost per production run is $S = \text{\textdollar}300$, and unit annual holding cost is $H = \text{\textdollar}4.00$ per unit-year.

1. Calculate the optimal production lot size ($Q_p^*$).
2. Determine the maximum inventory level accumulated ($I_{\max}$).
3. Calculate the length of the production run ($t_p$) and the overall cycle length ($T$) in operating days.
4. Compute the total annual setup and holding costs.

Solution:

Step 1: Calculate $Q_p^*$

• Consumption ratio: $d/p = 72 / 300 = 0.24$.
• Production reduction factor: $1 - d/p = 1 - 0.24 = 0.76$.
• Apply the EPQ formula: $$Q_p^* = \sqrt{\frac{2 D S}{H \left(1 - \frac{d}{p}\right)}} = \sqrt{\frac{2 \times 18,000 \times 300}{4.00 \times 0.76}} = \sqrt{\frac{10,800,000}{3.04}} = \sqrt{3,552,631.58} \approx 1,884.8\text{ units}$$ Round to 1,885 units.

Step 2: Determine Maximum Inventory Level ($I_{\max}$)

$$I_{\max} = Q_p^*\left(1 - \frac{d}{p}\right) = 1,885 \times 0.76 = 1,432.6 \approx 1,433\text{ units}$$

Step 3: Determine Production Run Duration ($t_p$) and Cycle Time ($T$)

• Production run duration: $t_p = \frac{Q_p^*}{p} = \frac{1,885}{300} = 6.28\text{ operating days}$.
• Overall cycle time: $T = \frac{Q_p^*}{d} = \frac{1,885}{72} = 26.18\text{ operating days}$.
• Machine is producing for 6.28 days and idle (available for other jobs) for $26.18 - 6.28 = 19.90$ days.

Step 4: Calculate Total Annual Variable Costs

• Annual setup cost: $\frac{D}{Q_p^*} S = \frac{18,000}{1,885} \times 300 = 9.549 \times 300 = \text{\textdollar}2,864.72$
• Annual holding cost: $\frac{I_{\max}}{2} H = \frac{1,432.6}{2} \times 4.00 = 716.3 \times 4.00 = \text{\textdollar}2,865.20$
• Total Annual Variable Cost: $\text{\textdollar}2,864.72 + \text{\textdollar}2,865.20 = \text{\textdollar}5,729.92\text{/year}$ (balanced within rounding).

---

Worked Example 12.3.3: Safety Stock, Stochastic ROP, and Newsvendor Optimization

Problem (Part A - Stochastic ROP): Daily demand for a critical replacement bearing is normally distributed with mean $\bar{d} = 50$ bearings/day and standard deviation $\sigma_d = 8$ bearings/day. Supplier delivery lead time is constant at $L = 9$ days. Management mandates a 97.72% Cycle Service Level (corresponding to $Z = 2.00$). Calculate the required safety stock ($SS$) and the reorder point (ROP).

Problem (Part B - Single-Period Newsvendor): A souvenir vendor purchases commemorative jackets for a regional collegiate championship track meet. Jackets are bought for $40 each and sold to spectators for $100 each. Any unsold jackets after the meet can be liquidated to an off-price distributor for $10 each. Historical attendance models project jacket demand as a normal distribution with mean $\mu = 500$ jackets and standard deviation $\sigma = 60$ jackets. What is the critical fractile, and how many jackets should the vendor order?

Solution:

Part A: Stochastic ROP and Safety Stock

1. Calculate standard deviation of demand during lead time using the Square Root Law: $$\sigma_L = \sigma_d \sqrt{L} = 8 \sqrt{9} = 8 \times 3 = 24\text{ bearings}$$
2. Sizing Safety Stock ($SS$): $$\text{SS} = Z \cdot \sigma_L = 2.00 \times 24 = 48\text{ bearings}$$
3. Calculate Reorder Point (ROP): $$\text{ROP} = \bar{d} \cdot L + \text{SS} = (50 \times 9) + 48 = 450 + 48 = 498\text{ bearings}$$

• Operational Rule: Place an order for replenishment whenever on-hand plus on-order inventory falls to 498 bearings.

Part B: Newsvendor Critical Fractile and Stocking Quantity

1. Identify cost parameters:
   • Retail price $P = \text{\textdollar}100$, purchase cost $C = \text{\textdollar}40$, salvage value $V = \text{\textdollar}10$.
   • Cost of Underage: $C_u = P - C = 100 - 40 = \text{\textdollar}60$.
   • Cost of Overage: $C_o = C - V = 40 - 10 = \text{\textdollar}30$.
2. Calculate Critical Fractile ($CF$): $$CF = \frac{C_u}{C_u + C_o} = \frac{60}{60 + 30} = \frac{60}{90} = \frac{2}{3} \approx 0.6667$$
3. Find the standard normal $Z$-value corresponding to $\Phi(Z) = 0.6667$:
   • From the standard normal cumulative distribution table, $\Phi(0.43) \approx 0.6664$ and $\Phi(0.44) \approx 0.6700$.
   • Interpolating gives $Z^* \approx 0.431$.
4. Compute optimal stocking quantity $Q^*$: $$Q^* = \mu + Z^* \cdot \sigma = 500 + 0.431(60) = 500 + 25.86 = 525.86 \approx 526\text{ jackets}$$

• Engineering Decision: The vendor should order 526 jackets.

---

8. NCEES Reference Handbook Tips & Realistic Exam Traps

• Holding Cost Scaling Trap: When calculating $H = i \cdot C$, always verify that holding cost and ordering cost share the same time unit as demand (annual). If demand $D$ is in units/year, $H$ must be dollars/unit-year. Never mix monthly holding rates with annual demand.
• Square Root Law of Lead Time: Safety stock scales with $\sqrt{L}$, NOT $L$. If a supplier reduces lead time from 16 days to 4 days (a 75% reduction), safety stock is cut by $\sqrt{4}/\sqrt{16} = 2/4 = 50\%$ (a 50% reduction, not 75%).
• Average Inventory in EPQ: In classic EOQ, average inventory is $Q/2$. In EPQ, average inventory is $\frac{I_{\max}}{2} = \frac{Q}{2}(1 - d/p)$. Do not use $Q_p^* / 2$ when calculating EPQ holding costs!
• Quantity Discount Global Optimum: Never assume the feasible $Q^*$ is automatically the cheapest option without comparing its total annual cost to the total annual cost at every lower price breakpoint. The huge savings on unit purchase cost ($D \cdot C$) almost always outweighs the modest increase in holding cost at price break points.
• Newsvendor Critical Fractile Ratio: Do not invert the ratio! The critical fractile is $\frac{C_u}{C_u + C_o}$, NOT $\frac{C_o}{C_u + C_o}$. Underage cost (marginal profit) is in the numerator. If profit margin is high ($C_u \gg C_o$), $CF > 0.50$ and you stock more than the mean ($Q^* > \mu$).