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  • 00:05

    DAVID SCHWEIDEL: So what are the constraintsthat we're going to face?Well, for now, we're going to assumethat the failure rate of a productis 0.5 incidents per year.And we're going to assume that it comes from a Poissondistribution.We'll talk about that in a moment.And we're going to assume that the retailer marks up--

  • 00:28

    DAVID SCHWEIDEL [continued]: or rather that there is a markup on repair services.So let's assume that the retailer can get repairs donefor 50% of what a third-party company isgoing to charge consumers.But that's something that we're goingto want to manipulate, because what if I can find even a lowerprice?What if I can get repairs done for 25% of the cost?

  • 00:50

    DAVID SCHWEIDEL [continued]: If I can find someone that does repairs for a lower pricepoint, if I can manage to do that,I can keep some of those savings and pass some of those savingson to the consumer.All right.So what do we need to determine?What's the price for one year of warranty coverage?What's the additional cost for a second year of coverage?What's the deductible going to be?

  • 01:10

    DAVID SCHWEIDEL [continued]: And that might be zero or some number greater than zero.And what's the quota going to be for the contract?Now, if we take the most general case--warranty at a fixed price, optional second yearof coverage, a deductible that might be zero,

  • 01:32

    DAVID SCHWEIDEL [continued]: and a quota that might be infinite--that's the most general scenario.Our special cases where there is no deductibleand there is no quota, or where thereis a deductible but no quota, are all nested within that.But we're going to work on building this up.So we're going to start by saying,let's build out the case where I pay

  • 01:54

    DAVID SCHWEIDEL [continued]: for the warranty as a consumer and there's no deductible,no quota.Then we'll say, let's bring in that deductible.Then, on top of that, we're goingto say, well, OK, let's bring in this quota.So we want to build as general a decision-support toolas we can.Now, the piece that we don't know with certainty

  • 02:14

    DAVID SCHWEIDEL [continued]: is, how many incidents are we going to haveto plan for for each customer?So depending on what I'm selling--let's say I'm selling flat screen TVs--how many incidents can I expect?Is it going to be zero incidents?One incident during the life of the contract?Two incidents over the life of the contract?

  • 02:35

    DAVID SCHWEIDEL [continued]: How many are we talking about?That's the unknown for us.So that's where we're going to use the Poissondistribution to characterize the number of failuresduring the course of the contract.Very similar to how we use the binomial distributionto characterize the number of no-shows.All right.So just to give you a little bit of background

  • 02:56

    DAVID SCHWEIDEL [continued]: on the Poisson distribution.And I've got a small Excel file that you can download,where some of this is already set up for you.The Poisson distribution, very common for what'sreferred to as count data within marketing.So count data-- so if we think about how many units somebodybuys during a fixed period of time,

  • 03:17

    DAVID SCHWEIDEL [continued]: that's the count that we're referring to.You go to the grocery store--how many bottles of soda did you buy?How many bags of bagels did you buy?How many oranges did you buy on this trip?Well, that's going to be--that's a count that we're trying to model.That's where the Poisson distribution gets used.A feature of the Poisson distribution to bear in mind

  • 03:39

    DAVID SCHWEIDEL [continued]: is the average and the variance of the Poisson distribution,characterized by the same parameter.So when the mean and the varianceare equal to each other, the Poisson distributiontends to fit the data very well.If this assumption doesn't hold, well, thisisn't going to be the right distribution for you.For this particular exercise, let'sassume that we know the Poisson distribution is appropriate

  • 04:00

    DAVID SCHWEIDEL [continued]: and we know that the failure rate is going to be 0.5--0.5 incidents per year.One characteristic of the Poissonthat's going to serve us well in this particular case--if I know that I'm dealing with a failure rate of 0.5 incidentsper year-- that's my expected number of failures in a given

  • 04:21

    DAVID SCHWEIDEL [continued]: year--well, what's my expected number of failures over two years?Well, let's say my X is the number of failuresin the first year.My Y is going to be the number of failures in the second year.X plus Y is going to be my combined number of failuresduring that two-year period.

  • 04:42

    DAVID SCHWEIDEL [continued]: Well, I can just add up the two parameters, lambda 1and lambda 2.And so in this case, if I have a 0.5 expectedfailures in each year--lambda 1 is 0.5, lambda 2 is 0.5--I add those up, the number of product failuresduring a two-year period is going to--

  • 05:03

    DAVID SCHWEIDEL [continued]: or expected failures, is going to be 1.All right.So let's take a look at this illustrative worksheet.What I've done here is calculated the probabilityof a particular event occurring, given the value of lambdaunder the Poisson distribution.So for a lambda of 0.5-- again, this is our one-year--

  • 05:23

    DAVID SCHWEIDEL [continued]: what can we expect as far as the number of failuresthat someone is going to experience?According to the Poisson distribution,there is a 60% chance of zero failures occurring,30% chance of one failure occurring,7.5% chance of two failures, and so forth.And we see, the number of failures,

  • 05:43

    DAVID SCHWEIDEL [continued]: as illustrated in this plot, looks nothinglike a normal distribution.So again, this is where it's important for usto match the probability distribution to the contextthat we're working with.What about if we deal with over a two-year period?We said that lambda for each year is 0.5.

  • 06:04

    DAVID SCHWEIDEL [continued]: So over a two-year period, we're expecting lambda of 1.Notice that our probability of not having any failuresdrops considerably.So over a two-year period, it goes from--we were at around 60% when lambda was 0.5.It goes from just over 60% to now

  • 06:25

    DAVID SCHWEIDEL [continued]: around 37% for zero failures.Same percentage chance for only one failure.So we're expecting, actually, more failuresover a two-year period than over a one-year period.So you can play around using that tool,just to get a sense for how the failure

  • 06:45

    DAVID SCHWEIDEL [continued]: rate or the rate at which service incidents occursis going to affect our decision-making.

Video Info

Series Name: Managing Uncertainty in Marketing Analytics

Episode: 11

Publisher: Emory University

Publication Year: 2017

Video Type:Tutorial

Methods: Probability distribution, Marketing research, Poisson regression

Keywords: decision making; deductible; marketing research; poisson distribution; prediction (methodology); probability; quantitative data analysis; quotas; Statistical distributions; uncertainty; warranties ... Show More

Segment Info

Segment Num.: 1

Persons Discussed:

Events Discussed:

Keywords:

Abstract

David Schweidel, PhD, Professor of Marketing at Emory University, discusses company constraints and decisions plans when designing extended service plans and introduces the Poisson distribution in part 11 of the Managing Uncertainty in Marketing Analytics course.

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Application: Designing Extended Service Warranty Plans: Company Constraints & Decisions

David Schweidel, PhD, Professor of Marketing at Emory University, discusses company constraints and decisions plans when designing extended service plans and introduces the Poisson distribution in part 11 of the Managing Uncertainty in Marketing Analytics course.

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