Operations Management: Decision Making Processes and Supply Chains Note

Learning Goals

• Goal 1: Explain break-even analysis using both the graphical and algebraic approaches.

• Goal 2: Define and construct a preference matrix.

• Goal 3: Describe how to draw and analyze a decision tree.

Break-even Analysis for Evaluating Services or Products

• Primary Objective: Determining the point at which we break even. This analysis evaluates if the predicted sales volume of a service or product is sufficient to break even ($neither earning a profit nor sustaining a loss$).

• Critical Evaluation Questions:     * How low must the variable cost ($c$) per unit be to break even, based on current prices and sales forecasts?     * How low must the fixed cost ($F$) be to break even?     * How do price levels affect the break-even quantity?

Definitions and Mathematical Formulas for Break-even Analysis

• Variable Cost ($c$): The portion of the total cost that varies directly with the volume of output.

• Fixed Cost ($F$): The portion of the total cost that remains constant regardless of changes in levels of output.

• Quantity ($Q$): The number of customers served or units produced per year.

• Price ($p$): The selling price in $ per unit.

• Total Cost ($TC$): Calculated as follows:     * $TC = F + (c imes Q)$

• Total Revenue ($TR$): Calculated as follows:     * $TR = p imes Q$

• Break-even Point Condition: Occurs when Total Revenue equals Total Cost ($TR = TC$), leading to the formula:     * $F + (c imes Q) = p imes Q$     * Solving for quantity ($Q$): $Q = rac{F}{p - c}$

Example 1: Hospital Break-even Analysis

• Scenario: A hospital is considering a new procedure to be offered at $200 per patient.

• Data Provided:     * Fixed cost per year ($F$) = $100,000     * Variable costs ($c$) = $100 per patient     * Price ($p$) = $200 per patient

• Algebraic Solution:     * $Q = rac{100,000}{200 - 100} = 1,000 ext{ patients}$

• Graphic Solution Methodology:     * Two lines must be plotted: one for costs and one for revenues.     * Points for plotting ($Q=0$ and $Q=2,000$):         1. At $Q=0$: Total Annual Cost = $100,000 + (100 imes 0) = 100,000$; Total Annual Revenue = $200 imes 0 = 0$.         2. At $Q=2,000$: Total Annual Cost = $100,000 + (100 imes 2,000) = 300,000$; Total Annual Revenue = $200 imes 2,000 = 400,000$.     * Cost Line: Draw through points $(0, 100,000)$ and $(2,000, 300,000)$.     * Revenue Line: Draw through points $(0, 0)$ and $(2,000, 400,000)$.     * Intersection: The two lines intersect at $1,000$ patients, which is the break-even quantity (BEQ).

• Contribution Margin Analysis:     * Contribution Margin definition: Total Revenue minus Total Cost ($TR - TC$).     * Scenario Analysis: If the most pessimistic sales forecast for the proposed service was $1,500$ patients, the total contribution per year is calculated as:         * $p imes Q - (F + c imes Q)$         * $200 imes (1,500) - [100,000 + 100 imes (1,500)] = 300,000 - 250,000 = 50,000$         * The total contribution to profit and overhead is $50,000.

Evaluating Processes (Make-or-Buy Decisions)

• Process Selection: Managers must choose between two internal processes or between maintaining an internal process and buying services/materials from an outside source.

• Underlying Assumption: The decision choice does not affect revenues; only costs are compared.

• Analytical Goal: Find the quantity for which the total costs of the two alternatives are equal.

• Formula Components:     * $F_b$: The fixed cost (per year) of the buy option.     * $F_m$: The fixed cost of the make option.     * $c_b$: The variable cost (per unit) of the buy option.     * $c_m$: The variable cost of the make option.

• Total Cost Functions:     * Total Cost to Buy = $(F_b + c_b imes Q)$     * Total Cost to Make = $(F_m + c_m imes Q)$

• Break-even Quantity Equation: Set the two functions equal and solve for $Q$:     * $F_b + c_b imes Q = F_m + c_m imes Q$     * $Q = rac{F_m - F_b}{c_b - c_m}$

Example 2: Fast-Food Salad Make-or-Buy Analysis

• Scenario: A fast-food hamburger restaurant is adding salads to the menu. They compare making them in-house versus buying preassembled salads.

• Data Provided:     * Make Option: Fixed costs ($F_m$) = $12,000; Variable costs ($c_m$) = $1.50 per salad.     * Buy Option: Fixed costs ($F_b$) = $2,400 (installation and refrigeration); Variable costs ($c_b$) = $2.00 per salad.     * Expected Demand: 25,000 salads per year.

• Break-even Quantity Calculation:     * $Q = rac{12,000 - 2,400}{2.00 - 1.50}$     * $Q = rac{9,600}{0.50} = 19,200 ext{ salads}$

• Process Comparison (Costs at Different Quantities):     * At $Q=0$: Make Cost = $12,000; Buy Cost = $2,400.     * At $Q=38,400$: Make Cost = $12,000 + (1.5 imes 38,400) = 69,600$; Buy Cost = $2,400 + (2.0 imes 38,400) = 79,200$.

Decision Making under Risk

• Context: The manager can list possible events and estimate their probabilities. This environment provides less information than decision making under certainty, but significantly more information than decision making under uncertainty.

• Expected Value Rule: This is the most widely used rule for decision making under risk.

Example 3: Payoff Matrix and Expected Value

• Scenario: Choose the best alternative between a small facility, a large facility, or doing nothing, given two possible demand conditions.

• Probabilities:     * Probability of Low Demand = $0.4$     * Probability of High Demand = $0.6$

• Payoff Matrix (Values in $):     * Small Facility: Low Demand = $200; High Demand = $270.     * Large Facility: Low Demand = $160; High Demand = $800.     * Do Nothing: Low Demand = $0; High Demand = $0.

• Expected Value (EV) Calculations:     * Small Facility: $EV = 0.4 imes (200) + 0.6 imes (270) = 80 + 162 = 242$     * Large Facility: $EV = 0.4 imes (160) + 0.6 imes (800) = 64 + 480 = 544$

• Conclusion: The Large Facility is the best alternative based on the expected value rule.

Decision Trees

• Definition: Schematic models of available alternatives and possible consequences.

• Applicability: Useful for probabilistic events and sequential decisions.

• Symbolic Notation:     * Square Nodes: Represent decision points.     * Circular Nodes: Represent chance/event points.

• Characteristics:     * Events leaving a circular chance node must be collectively exhaustive.     * Conditional payoffs for every alternative-event combination are shown at the end of each branch.

• General Rules for Operation:     * Drawing: Proceed from left to right.     * Solving: Calculate expected payoffs and move from right to left (backward induction).

Example 4: Retailer Multi-Stage Decision Tree

• Scenario: A retailer must choose between building a small or large facility.

• Initial Estimates:     * Probability of Small Demand = $0.4$     * Probability of High Demand = $0.6$

• Decision 1: Small Facility Logic:     * Low Demand scenario Payoff = $200,000.     * High Demand scenario presents a second decision: Do not expand (Payoff = $223,000) or Expand (Payoff = $270,000).     * Choice at high demand node: Expand ($270,000 is higher than $223,000).     * Small Facility Expected Value: $0.4 imes (200) + 0.6 imes (270) = 80 + 162 = 242$.

• Decision 2: Large Facility Logic:     * Low Demand scenario presents a second decision: Do nothing (Payoff = $40,000) or Advertise.     * Advertising Events: Modest response (Prob = $0.3$, Payoff = $20,000) or Sizable response (Prob = $0.7$, Payoff = $220,000).     * Expected Value of Advertising: $0.3 imes (20) + 0.7 imes (220) = 6 + 154 = 160$.     * Choice at low demand node: Advertise ($160 is higher than $40).     * High Demand scenario Payoff = $800,000.     * Large Facility Expected Value: $0.4 imes (160) + 0.6 imes (800) = 64 + 480 = 544$.

• Final Decision: The expected value for the Large Facility ($544,000) is greater than the Small Facility ($242,000), making the Large Facility the optimal choice.

Exercise: White Valley Ski Resort Lift Operation

• Scenario: Planning whether to install one or two ski lifts. Each lift accommodates 250 people per day.

• Operational Parameters:     * Season Length = 14 weeks (7 days/week) = 98 days.     * Lift Ticket Price = $20 per customer.     * Total Revenue at 100% capacity for one lift: $250 imes 98 imes 20 = 490,000$.

• Alternatives and Conditions:     * One Lift Option:         * Bad Times (0.3 prob): Util = 0.9; Installation = $50,000; Operation = $200,000.         * Normal Times (0.5 prob): Util = 1.0; Installation = $50,000; Operation = $200,000.         * Good Times (0.2 prob): Util = 1.0; Installation = $50,000; Operation = $200,000.     * Two Lifts Option:         * Bad Times (0.3 prob): Util = 0.9; Installation = $90,000; Operation = $200,000.         * Normal Times (0.5 prob): Util = 1.5; Installation = $90,000; Operation = $400,000.         * Good Times (0.2 prob): Util = 1.9; Installation = $90,000; Operation = $400,000.

• Payoff Calculations (Revenue - Cost):     * One Lift:         1. Bad: $0.9 imes (490) - (50 + 200) = 191$         2. Normal: $1.0 imes (490) - (50 + 200) = 240$         3. Good: $1.0 imes (490) - (50 + 200) = 240$         * One Lift EV: $0.3 imes (191) + 0.5 imes (240) + 0.2 imes (240) = 225.3$     * Two Lifts:         1. Bad: $0.9 imes (490) - (90 + 400) = 151$         2. Normal: $1.5 imes (490) - (90 + 400) = 245$         3. Good: $1.9 imes (490) - (90 + 400) = 441$         * Two Lifts EV: $0.3 imes (151) + 0.5 imes (245) + 0.2 imes (441) = 256.0$

• Conclusion: The resort should purchase two lifts as the expected payoff of $256,000 exceeds the $225,300 payoff of one lift.