L NM O QP

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L NM O QP

901 Notes Maloney⎯Fall ‘98

Assume:

P= price of the washing machine;

w= price of a repair

n= the number of repairs per period

u= the life of the machine in periods

Cleaning Up Dirty Laundry:

The Washing Machine Problem

Let the repair rate be given by n = u β /a, where β>1 and a>0. Let the interest rate be zero. The

cost of repairs in any period is wn.

The most important question in this problem is: What is the consumer’s objective function? In

the absence of an explicit statement, we might hypothesize that consumers attempt to minimize

the average cost of the washing machine including the purchase price and maintenance by the

choice of the length of time that they keep the machine before buying a new one. This is wrong,

but to demonstrate such, let’s first develop this AC minimization model.

AC Minimization

The FOC:

min { }

u AC P

= +

u

wu

a

β−1 β−2

∂AC

P wu

= − + ( β − 1) = 0

2

∂u

u a

β

( β − 1) wu = Pa

u~

=

L

NM

Pa

w

1

( β −1)

The Comparative Statics of interest are the effect of machine durability on the optimal life:

O

QP

1

β

∂ ln ∂

∂ β β ∂

~

u 1 1 ln P

= +

ln a ln a

We can think of this problem as a short-run⎯long-run phenomenon. In the short run, P does not

change as a changes. For instance, when a manufacturer brings out a new model that is more

durable, at first it cannot raise the price because consumers will not trust that the machine is truly

more durable. In that case the response of consumers is simply 1/β. This says that consumers do


901 Notes Maloney⎯Fall ‘98

run the machine longer; the response is inelastic. However, after the machine is proven, the

manufacturer will increase its price. The question is: How much?

In the AC minimization model it seems reasonable to assume that a given manufacturer will

increase its price (relative to its competition) by an amount that keeps AC constant. Hence the

optimal price from the manufacturer’s perspective is given by taking the derivative of AC w.r.t.

a.

β−1

∂AC

∂P

1 1 wu

= − = 0

∂a

∂a

u a a

Rewriting and collecting terms gives:

∂P

a w β

∂a

P Pa u =

Substituting for u from the consumer’s behavior equation:

L

MF

N

HG

I

KJ

O

P

Q

∂P

a w Pa β

=

∂a P Pa M w β − β M

P = 1 1

1 −1

Therefore, the full effect of a change in a on the life of the washing machine is:

∂ ln

∂ β β β β

~

u

= +

ln a − = 1 1 1 1

1 −1

As we would expect, the consumers response in the long run (that is, after the manufacturer

responds with a higher price) is more elastic

Value Maximization

As noted in the introduction, AC minimization while simple is not the right way to think about

this problem. Consumers do not minimize average cost, they maximize value. The problem here

is that value is not explicitly defined. However, that can be rectified.

Assume that the implicit shadow price of washing services is given by R. That is, the washing

machine yields some gross value of services each period. Maybe this is the equivalent of what

would be spent at the laundromat or how much you have to spend on flowers to get your mother

to do it. Regardless, there is some value; call it R. The net value of washing services is R minus

the repair cost.

The consumer objective is to maximize the value of washing services obtainable from the

machine. Hence, we can write:

1

β

2


901 Notes Maloney⎯Fall ‘98

L

NM

u*

max { u}

PV R wu z

β

= −

0

The derivative of an integral w.r.t. the upper bound is simply the integrand evaluated at the upper

bound. Hence, the FOC is

R wu

β

=

a

which can be solved for the optimal life of the machine. This gives:

u

Ra

w

* = F HG I

KJ 1

β

However, this behavior function is not revealing because it expresses machine life as a function

of R, which is unknown to us. This can be resolved. Let’s assume, as we did in the AC

minimization problem, that there is a competitive equilibrium operating. The washing machine

manufacturers P as high as possible subject to competitive pressures. In this case, the

competitive boundary is given by R. Let’s assume that P is set equal to the maximized present

value of the optimal use of the machine; i.e., P = PV(u*). Hence, we can write:

z

L

NM

O

L

NM

a

O

QP

du

u*

β β β

P R wu

du Ru

a

wu

u*

u

wu*

u*

= − = −

= Ru*


0 QP a β + 1 a β + 1

0

Solving for the R implied by P gives:

Substituting into the FOC

which reduces to:

u*

=

P 1

R = +

u*

β + 1

FL

G

GNM

HG

O

QP

wu*

a

P 1 wu*

+

u*

β + 1 a

w

u*

=

F

HG

Pa

w

β + 1

β

I

β

O

QP

KJ +

1

β 1

a

I

J

KJ

1

β β

3


901 Notes Maloney⎯Fall ‘98

Notice that the behavior function for the consumer is different under value maximization than

average cost minimization. So, too, are the comparative statics.

F

HG

∂ln

u*

1 ∂ln

P

= +

∂ln a β + 1 ∂ln

a

1

As in the AC minimizing problem, we can consider a short run and long-run response to a

change in the washing machine durability factor, a. In the short run there is no price response by

the manufacturer. Hence, that term is zero. The result says that consumers keep their washing

machines longer. The elasticity is one over one plus beta, which is more inelastic than the

response in the AC minimization model.

To determine the effect of durability on the price of the machine, we return to the price equation,

P=PV(u*). Taking the derivative w.r.t. a gives:

β

∂P

wu * u*

=

∂a a a(

β + 1)

Rewriting to put the expression in elasticity terms and substituting from the FOC, we have:

∂P

a u

R

= R

∂a P P β + β P u

= * 1 1

1 + 1 / *

However from the price equilibrium condition and the FOC we can substitute R back out:

So,

P Ru wu

β

* u*

= −

a β + 1

P

u

= R − R

R

P / u*

β

1

+ 1

+

= β 1

β

∂ ln P 1

=

∂ln a β

which is the same for optimal life in the long run:

∂ln

u*

1

=

∂ln a β

I K J

4


901 Notes Maloney⎯Fall ‘98

Comparison of the Models

Note that in both the short and long run, the value maximizing model predicts more inelastic

consumer responses to changes in durability. This is symptomatic of the error in the AC

minimization model. To fully appreciate the inaccuracy of that model, let’s consider some actual

values for the parameters. Let:

P = $1200

R = $400

w = $100

β = 1.5

These seem about right. Beta is the elastic of repair trips with respect to the age of the machine.

The problem assumes that it is greater than 1, but it shouldn’t be too large. The price of a new

machine is high, but I am assuming that this is a washer/dryer combo (we are not in France). The

cost of a repair trip is a little low, but the way the problem is set up, there are more trips per year

as the machine ages. Hence, we can alternatively consider the number of trips to be constant, but

the cost of the trips to increase.

At all events, given R and w, from the value maximizing model we know the optimal n, which is

4. From the other equations describing value maximization, we can derive u* = 5 and a = 2.8.

Now, given these parameters, what does AC minimization predict about consumer behavior. It

says that optimal life is 16.5 years and that the repair cost in the last year are $2400, or twice the

cost of a new machine. This is obviously absurd. The problem with the average cost

minimization idea is that it is based on an objective that is only spuriously associated with utility

maximization. However, following this behavior pattern causes the consumers to be crucified on

a cross of repair bills.

One final point to recognize, the value maximization model is built on the idea that the decision

to buy the machine is a discrete one. That is, if the maximized present value of the machine over

its optimal life is greater than the cost of the machine, then you buy the machine. If not, you

don’t. A more complete model would be one like the replanting problem in which the decision to

buy the machine is also to buy and replace it as it wears out forever into the future. I have

ignored that problem here by assuming that the machine is priced so that the consumer is

indifferent between replacing the machine or adopting the alternative (that is, whatever

determines R).

5

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