7.3 Break-Even Analysis, Payback Period & EOQ Inventory Models

7.3 Break-Even Analysis, Payback Period & EOQ Inventory Models

Industrial mechanical engineers frequently manage factory operations, equipment selection, production scheduling, and supply chain inventory. Evaluating the operational feasibility of manufacturing processes requires quantitative techniques to establish cost structures, determine profit thresholds, evaluate capital recovery speeds, and optimize inventory holding vs. ordering expenses.

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Cost-Volume-Profit (CVP) & Break-Even Analysis

Production operations divide operating costs into two primary categories based on behavior relative to output volume $Q$:

• Fixed Costs ($FC$): Overhead expenses that remain constant regardless of production output within the relevant range (e.g., factory rent, building insurance, administrative salaries, equipment depreciation).
• Variable Costs ($VC$): Expenses that vary directly in proportion to production volume $Q$ (e.g., raw materials, direct shop labor, consumable tooling, electric power for machinery). Total variable cost is $VC(Q) = v \cdot Q$, where $v$ is the unit variable cost.
• Total Cost Function ($TC$): $TC(Q) = FC + v \cdot Q$
• Total Revenue Function ($TR$): $TR(Q) = P \cdot Q$ where $P$ is the unit selling price.
• Profit Function ($\Pi$): $\Pi(Q) = TR(Q) - TC(Q) = P \cdot Q - (FC + v \cdot Q) = (P - v) Q - FC$

1. Break-Even Quantity ($Q_{BE}$)

The break-even point occurs at the production volume $Q_{BE}$ where total revenue equals total cost, yielding zero operating profit ($\Pi = 0$):

$Q_{BE} = \frac{FC}{P - v}$

where $(P - v)$ is known as the Unit Contribution Margin ($CM$).

2. Contribution Margin Ratio ($CMR$) & Break-Even Revenue ($R_{BE}$)

$CMR = \frac{P - v}{P}$

$R_{BE} = Q_{BE} \cdot P = \frac{FC}{CMR}$

3. Margin of Safety ($MS$)

Margin of Safety measures how far expected or actual sales volume $Q_{actual}$ exceeds the break-even volume $Q_{BE}$:

$MS = \frac{Q_{actual} - Q_{BE}}{Q_{actual}} \times 100\%$

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Payback Period Analysis

Payback period measures the time required for net cash inflows from an investment project to fully recover the initial capital outlay ($C_0$).

1. Simple Payback Period ($SPP$)

Simple payback ignores the time value of money. For uniform annual net cash inflow $A$:

$SPP = \frac{C_0}{A}$

For non-uniform annual cash flows, $SPP$ is the year $n$ where cumulative net cash flow turns non-negative.

Limitations: Simple payback ignores interest rates and completely neglects all cash flows occurring after the payback horizon.

2. Discounted Payback Period ($DPP$)

Discounted payback incorporates interest rate $i$ by discounting future cash flows back to $t=0$. $DPP$ is the smallest number of periods $n$ satisfying:

$\sum_{t=1}^n \frac{CF_t}{(1+i)^t} \ge C_0$

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Inventory Control & Economic Order Quantity (EOQ)

Manufacturing plants maintain inventories of raw materials, spare parts, and components. Inventory management balances two opposing cost forces:

1. Annual Ordering Cost ($AOC$): Decreases as order quantity $Q$ increases (fewer purchase orders placed per year). $AOC(Q) = \frac{D}{Q} \cdot S$ where $D$ = annual demand (units/year) and $S$ = fixed cost per purchase order (₱/order).
2. Annual Holding / Carrying Cost ($AHC$): Increases as order quantity $Q$ increases (higher average inventory held in warehouse). $AHC(Q) = \frac{Q}{2} \cdot H$ where $H$ = annual holding cost per unit (₱/unit-year). If holding cost is given as a percentage rate $I_h$, then $H = I_h \times C_{unit}$.

Total Inventory Cost ($TIC$) Function

$TIC(Q) = \frac{D}{Q} S + \frac{Q}{2} H + D \cdot C_{unit}$

Derivation of Economic Order Quantity ($EOQ$)

Taking the first derivative of $TIC(Q)$ with respect to $Q$ and setting it to zero:

$\frac{d(TIC)}{dQ} = -\frac{D S}{Q^2} + \frac{H}{2} = 0 \implies \frac{D S}{Q^2} = \frac{H}{2} \implies Q^2 = \frac{2 D S}{H}$

$EOQ = \sqrt{\frac{2 D S}{H}}$

At the optimal $EOQ$, annual ordering cost exactly equals annual holding cost ($AOC = AHC$). The minimum total annual inventory cost is:

$TIC_{min} = \sqrt{2 D S H} + D \cdot C_{unit}$

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Reorder Point ($ROP$) & Safety Stock ($SS$)

The Reorder Point ($ROP$) defines the inventory level at which a new order of size $EOQ$ must be placed to replenish stock before depletion during lead time $L$:

$ROP = d_{daily} \times L + SS$

where:

• $d_{daily}$ = daily demand rate ($d_{daily} = D / \text{operating days per year}$)
• $L$ = replenishment lead time (days)
• $SS$ = buffer safety stock retained to guard against demand spikes or delivery delays

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Worked Industrial Engineering Problem: Valve Production & Inventory

Problem Statement

An industrial valve factory produces heavy stainless-steel control valves. The plant operations involve the following financial and manufacturing specifications:

Part A: CVP & Payback Analysis

• Fixed Costs ($FC$): $\text{₱}4,500,000$/year
• Unit Variable Cost ($v$): $\text{₱}1,200$/valve
• Unit Selling Price ($P$): $\text{₱}2,700$/valve
• Maximum Plant Capacity: $4,500 \text{ valves/year}$
• An automation machine costing $C_0 = \text{₱}6,000,000$ yields net annual cash flow $A = \text{₱}1,500,000$ at interest rate $i = 10%$.

Part B: Inventory Optimization (EOQ)

• Annual Demand ($D$): $12,000 \text{ electric motor actuators/year}$
• Order Placement Cost ($S$): $\text{₱}1,500$/order
• Holding Cost ($H$): $\text{₱}100$/unit-year
• Delivery Lead Time ($L$): $6 \text{ working days}$
• Operating Days per Year: $300 \text{ days/year}$
• Safety Stock ($SS$): $100 \text{ units}$

Perform the following calculations:

1. Compute the break-even quantity ($Q_{BE}$) and annual operating profit at full capacity (4,500 valves).
2. Compute the Margin of Safety ($MS$) at full capacity.
3. Compute the Simple Payback Period ($SPP$) and Discounted Payback Period ($DPP$) for the automation machine.
4. Compute the Economic Order Quantity ($EOQ$), number of orders per year, and Reorder Point ($ROP$) for motor actuators.

Step-by-Step Solution

Step 1: Break-Even Quantity & Operating Profit

$Q_{BE} = \frac{FC}{P - v} = \frac{4,500,000}{2,700 - 1,200} = \frac{4,500,000}{1,500} = 3,000 \text{ valves/year}$

At full capacity $Q_{actual} = 4,500$ valves:

$\Pi = (P - v) Q - FC = (1,500 \times 4,500) - 4,500,000 = 6,750,000 - 4,500,000 = \text{₱}2,250,000$

Step 2: Margin of Safety ($MS$)

$MS = \frac{4,500 - 3,000}{4,500} \times 100\% = \frac{1,500}{4,500} \times 100\% = 33.33\%$

Step 3: Simple & Discounted Payback Period For simple payback:

$SPP = \frac{C_0}{A} = \frac{6,000,000}{1,500,000} = 4.00 \text{ years}$

For discounted payback at $i = 10%$:

• Year 1 discounted cash flow: $1,500,000 / 1.10 = \text{₱}1,363,636$ (Cumulative = $\text{₱}1,363,636$)
• Year 2 discounted cash flow: $1,500,000 / (1.10)^2 = \text{₱}1,239,669$ (Cumulative = $\text{₱}2,603,305$)
• Year 3 discounted cash flow: $1,500,000 / (1.10)^3 = \text{₱}1,126,972$ (Cumulative = $\text{₱}3,730,277$)
• Year 4 discounted cash flow: $1,500,000 / (1.10)^4 = \text{₱}1,024,520$ (Cumulative = $\text{₱}4,754,797$)
• Year 5 discounted cash flow: $1,500,000 / (1.10)^5 = \text{₱}931,382$ (Cumulative = $\text{₱}5,686,179$)
• Year 6 discounted cash flow: $1,500,000 / (1.10)^6 = \text{₱}846,711$ (Cumulative = $\text{₱}6,532,890$)

At the end of Year 5, unrecovered capital is $6,000,000 - 5,686,179 = \text{₱}313,821$.

$DPP = 5 + \frac{313,821}{846,711} = 5 + 0.3706 = 5.37 \text{ years}$

Step 4: Inventory EOQ & Reorder Point

$EOQ = \sqrt{\frac{2 \times 12,000 \times 1,500}{100}} = \sqrt{\frac{36,000,000}{100}} = \sqrt{360,000} = 600 \text{ units/order}$

$\text{Orders per year} = \frac{D}{EOQ} = \frac{12,000}{600} = 20 \text{ orders/year}$

Daily demand rate $d_{daily} = \frac{12,000}{300} = 40 \text{ units/day}$.

$ROP = d_{daily} \times L + SS = (40 \times 6) + 100 = 240 + 100 = 340 \text{ units}$

Conclusion: The factory achieves break-even at 3,000 valves/year with a 33.33% margin of safety. Automation capital is recovered in 4.0 years simple (5.37 years discounted). Actuator orders should be placed in batch sizes of 600 units whenever inventory drops to 340 units.