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# How much do cars cost a person over his lifetime?

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Q. How much do cars cost a person over his lifetime?  Suppose a
person buys his first car at the age of 20.  He keeps it eight years
and then sells it and buys another.  He keeps the second car eight
years and then sells it and buys another.  And he does the same thing
with the third car and fourth car, etc..  He keeps each car eight
years then sells it and buys another.  He does this all his life
until he is 80 years old.  How much have his cars cost him over his

A. We could answer this question by simply adding up our estimates of
all the money the person would have spent on the cars over his
lifetime.  That is not what we are going to compute, however.  We are
going to try to also include the interest he has lost out on from all
the money he has paid out on cars.  A person can take either of two
routes: 1. He can own a car  2. He can do without a car.  It is his
choice.  Doing without a car has its drawbacks.  And owning a car has
its costs.  A person could consciously say to himself when he is
young, "I am not going to own cars.  I am going to forgo the
convenience of cars in my life and put all the money I would have
spent on cars into an account and let it sit there and accumulate
interest."  Suppose a person did this.  The money that accumulates in
this account is the figure we are going to compute.  It represents
the loss the person has incurred by taking the route of owning cars.

So we obtain the answer to our question by doing the following:  we
make the supposition that each time the person spends a certain amount
of money on a car he instead puts that money into an account where it
accumulates at some fixed interest rate (i.e. we take all his car
related expenses as they occur over the years and put them into this
account).  We then ask how much money would be in the account after
60 years (when he was 80 years old).  The problem thus reduces to
that of computing the amount of an annuity stretched over 60 years.

For the sake of simplicity let us assume that the person buys each
car new and that each, when new, costs him \$18700 (i.e. we assume no
increase in car prices over the years, no inflation --- an admittedly
unrealistic assumption).  Below is a table titled "Estimated Cost
of Operating a 1992 Chevrolet Caprice Station Wagon".  We assume each
car the person bought was a Chevrolet Caprice station wagon and use
the cost estimates given in this table as the costs for each of the
cars owned (with no adjustments for inflation).  We thus generate a
sequence of payments:

10280, 5534, 5346, 5297, 5532, 2015, 1992, 1927
8055, 5534, 5346,  ...  .   .     ....  , 1927
8055, 5534, ....
8055,  ...
......
..........

We feed these into a computer program that computes the amount of an
annuity, supply it with an interest rate at which the money is
assumed to accumulate, and it computes the answer.  Let us assume the
money accumulates at an annual interest rate of 6%.  The answer for
the value of the account at the end of 60 years?  \$2,590,583!  Now
most Americans have two cars, not one car.  The husband and wife each
have a car.  How much are the cars in these two-car families costing
them?  If they always buy cars comparable in price to a Chevrolet
station wagon one can see how much it really costs them.

Suppose the person in the study we just did kept each car only six
\$2,833,845.

I am driving a 1979 Chevrolet station wagon that is 13 years old
with a wholesale value of no more than \$200 - \$300.  The total
operating cost for it is around \$1500 /year.  Suppose a person only
drove such cars all his life, incurring a driving cost of \$1500
/year.  Suppose he did this from age 20 to age 80.  How much would it
cost him assuming the money accumulated in the account at an interest
rate of 6% ?  The answer is \$799,692.

***********************************************************************

ESTIMATED COST OF OPERATING A 1992 CHEVROLET CAPRICE STATION
WAGON

Purchase price:  \$18700.

Year  Year-end   Depreciation  Tax  Loan      Repair  Cost of
wholesale                     payment   costs   Operation
value
_________________________________________________________________

1    \$15,025       \$3675     \$686   \$3569      \$0    \$10280
2     11,825        3200      540    3569     100      5534
3       7700        4125      352    3569     100      5346
4       6625        1075      303    3569     100      5297
5       5200        1425      238    3569     400      5532
6       4150        1050      190             500      2015
7       3650         500      167             500      1992
8       2225        1425      102             500      1927
_________

Total    37923

Notes and assumptions.
Insurance: \$500 per year
Gasoline: \$750 per year
Auto registration, county sticker, safety inspection, etc: \$75 /yr
Personal property tax rate: 4.57%
Loan: \$14,000 at 10% for 5 years
Down payment: \$4700
Year-end wholesale values: from Oct 92 bluebook
Repair costs: based on the yearly repair costs for my own 1979
Chevrolet station wagon

Cost of operation = Loan payment (principal + interest) + Tax + Repair
costs + Insurance + Gasoline + Auto registration, county
sticker, safety inspection, etc.

Note.  In this method of computing the cost of operation Depreciation
Expense is not included in our formula.  What we do is include all
money we put into the car in a particular year.  The car is sold
at the end of the last year and the proceeds from it's sale (i.e.
year-end wholesale value) are deducted from the down payment of
the next car.  This method does not give us an accurate figure for
the cost of operation in a particular year but for the computation
we are doing (summing costs over many years) it gives us a more
accurate figure.

Oct 1992