Website owner: James Miller
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 lifetime? 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 years instead of eight. What would the cost be then? Answer: $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 More from SolitaryRoad.com:
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