决策理论与正态分布分析

Module 3Decision Theory and the Normal Distribution

Learning ObjectivesStudents will be able to:

1. Understand how the normal curve can be used in performing break-even analysis.

2. Compute the expected value of perfect information (EVPI)

using the normal curve.

3. Perform marginal analysis where products have a constant marginal profit and loss.

Module Outline

M3.1Introduction

M3.2Break-Even Analysis and the

Normal Distribution

M3.3 EVPI and the Normal

Distribution

Introduction

The normal distribution can be used when there are a large

number of states of nature and/or alternatives.

Note : it would be impossible to develop a decision table or tree if there were 50, 100, or even more states and/or alternatives!

Break -Even Analysis and the Normal Distribution Break-even analysis, also called cost-volume analysis, answers managerial questions relating the effect of a decision to overall revenues or costs.Break-even point (units) =

Fixed Costs

Price/unit –Variable Costs/unit

Barclay Brothers New Product DecisionThe Barclay Brothers Company is a large manufacturer of adult parlor games. They are considering introducing Strategy, a new game. Their fixed cost for Strategy = $36,000 and their variable costs are $4 per game produced. They intend to sell the game for $10. What is the break-even point?

$10 -$4= 6,000 games

Barclay Brothers New Product DecisionThe Barclay Brothers expect demand to be 8,000. They believe there is a 15% chance it will be less than 5,000 and a 15% chance it will be greater than 11,000 ~ what is the probability they will lose money by not selling 6,000 games?Hint : the normal distribution can be used if the Barclay Brothers believe demand is normally distributed.

Normal Distribution for Barclay’s Demand5,00011,000Demand (Games)µ=8,000

Demand -µZ =σÎUse the normal table to find Z

Normal Distribution for Barclay’s DemandLook up 0.85 in the body of the normal table and find the associated Z:

Z = 1.04; so

1.04 = 11,000 -8,000σ= 2,885 unitsWhat is the probability of selling less than 6,000 –8,0006,000 units????2,885

l e b a t e h t F r o m = -.69

Z = .7549 so, 1 -.7549 = .2451Or a 24.51% chance of losing $$$ !

Barclay ’s Break-Even Analysis

P(loss) = P(demand

= 24.51%

P(profit) = P(demand > breakeven) = 0.7549

= 75.49%

Should Barclay bring its Strategy game to the market?

Expected Value of Perfect Information and the Normal DistributionThe opportunity loss functiondescribes the loss that would be suffered by making the wrong decision. The opportunity loss function can be computed by:Opportunity K (Break-even point -X) =loss for X

$0for X > Breakeven

where

K = the loss per unit when sales are below thebreak-even point

X = sales in units.

Barclay ’s Opportunity Loss Function

Expected Opportunity Loss

The expected opportunity lossis the sum of the opportunity losses multiplied by the probability of that demand occurring.

In the Barclay Brothers example, there are a large number of possible sales volumes (0,1,2,…5,999), so the unit normal loss integral is used.

N(D) is the table value for the normal loss integral for a given value of D.

Expected Value of Perfect InformationThe expected value of perfect information is equivalent to the minimum EOL.

EVPI = EOL = K σN(D)

Where

EOL = expected opportunity lossK = loss per unit when sales are belowthe break-even point

σ= standard deviation of the distribution

μ−breakeven D =σ

µ=mean sales

N(D) = the table value for the unit normal loss integral

Expected Value of Perfect Information (continued)

K = $6 σ= 2,8858, 000−6, 000D ==0. 69=0. 60+0. 092, 885

N (.69) = .1453

Therefore

EOL = KσN(.69)

= ($6)(2885)(.1453) = $2,515.14EVPI = $2515.14