Decision Theory and the Normal Distribution Module 3 Decision Theory and the Normal Distribution To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-1

Learning Objectives Students will be able to Understand how the normal curve can be used in performing break-even analysis. Compute the expected value of perfect information (EVPI) using the normal curve. Perform marginal analysis where products have a constant marginal profit and loss. To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-2

Module Outline M3.1 Introduction M3.2 Break-Even Analysis and the Normal Distribution M3.3 EVPI and the Normal Distribution To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-3

Normal Distribution for Barclay’s Demand Break-even point (Units) Fixed Cost Price/Unit - Variable Cost/Unit = 15 Percent Chance Demand is Less Than 5,000 Games Demand Exceeds 11,000 Games Mean of the Distribution, µ 5,000 11,000 µ=8,000 Demand (Games) X Z = Demand - µ  To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-4

Barclay’s Opportunity Loss Function In general, the opportunity loss function can be computed by: Opportunity loss K (Break-even point - X) for X < Break-even $0 for X > Break-even = where K = the loss per unit when sales are below the break-even point X = sales in units. To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-5

Barclay’s Opportunity Loss Function 6,000 µ Demand (Games) X Normal Distribution Slope = 6 Loss ($) Opportunity loss $6 (6,000 - X) for X < 6,000 games $0 for X > 6,000 games = µ = 8,000  = 2,885 Break-even point (XB) Profit Loss To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-6

Expected Value of Perfect Information EVPI = EOL = K N(D) µ = mean sales N(D) = the value for the unit normal loss integral given in Appendix B, for a given value of D. Where EOL = expected opportunity loss, K = loss per unit when sales are below the break-even point  = standard deviation of the distribution To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-7

Expected Value of Perfect Information – cont. K = $6  = 2,885 Therefore EOL = K N(.69) = ($6)(2885)(.1453) = $2,515.14 EVPI = $2515.14 N(.69) = .1453 To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna M3-8