Question

Suppose you want to buy a new car and trying to choose between two models:

Model A: costs $17,000 and its gas mileage is 20 miles per gallon and its insurance is $200 per year.

Model B: costs $25,000 and its gas mileage is 35 miles per gallon and its insurance is $400 per year.

If you drive approximately 40,000 miles per year and the gas costs $3 per gallon:

Find a formula for the total cost of owning Model A where the number of years is the independent variable.

Find a formula for the total cost of owning Model B where the number of years is the independent variable.

Find the total cost for each model for the first five years.

If you plan to keep the car for four years, which model is more economical? How about if you plan to keep it for six years?

Find the number of years in which the total cost to keep the two cars will be the same.

Identify the number of months where neither car holds a cost of ownership advantage.

What effect would the cost of gas doubling have on cost of ownership? Graph or show hand calculations.

If you can sell neither car for 40% of its value at any time, how does the analysis change? Graph or show hand calculations.

Answer 1

Let x be the number of years and y the total cost of owning Model A or, B.

i. The insurance cost for model A is $ 200x and the cost of gas is $ 3x*40000/20 = $ 6000x. Hence the total cost of owning Model A for x years is $ 17000+$ 200x+$ 6000x = $ ( 6200x+17000).

ii. The insurance cost for model A is $400x and the cost of gas is $ 3x*40000/35=$ 3428.57x (on rounding off to 2 decimal places). Hence the total cost of owning Model B for x years is $ 25000 + $ 400x+$ 3428.57x = $( 25000+ 3828.57x).

iii. The total cost of owning Model A for 5 years is $ ( 6200*5+17000) = $ 48000.

iv. The total cost of owning Model B for 5 years is $( 25000+ 5* 3828.57) = $ 44142.86 (on rounding off t0 the nearest cent).

v. The total cost of owning Model A for 4 years is $ ( 6200*4+17000) = $ 41800 and the total cost of owning Model B for 4 years is $( 25000+ 4* 3828.57) = $ 40314.29(on rounding off t0 the nearest cent). Hence if we plan to keep the car for four years, then model B is cheaper.

vi. If the total cost to keep the two cars is the same, then 6200x+17000 = 25000+ 3828.57x or, (6200- 3828.57)x = 8000 or, 2371.43x = 8000 so that x = 8000/2371.43 = 3.37(on rounding off to 2 decimal places). Hence, the total cost to keep the two cars will be the same if we keep the car for 3.37 years.

vii. The number of months where neither car holds a cost of ownership advantage is 3.37*12 = 40.44.

viii. If the cost of gas doubles to $ 6 per gallon , then the total cost of owning Model A for x years is $ 17000+$ 200x+ 2*$ 6000x = $ (12200x+17000) and the total cost of owning Model B for x years is $ 25000 + $ 400x+ 2 *$ 3428.57x = $ (7657.14x+25000). Thus, the total cost to keep the two cars will be the same if 12200x+17000 = 7657.14x+25000 or, 4542.86 x = 8000 or, x = 8000/4542.86 = 1.76 (on rounding off to 2 decimal places). Thus, if we keep the cars for 1.76 years, then the total cost of both the cars will be same and if we keep the car for more than 1.76 years, then model B will be cheaper, otherwise model A will be cheaper.

ix. If we sell neither car for 40% of its value after x years, then the net cost of owning Model A for x years is $ 0.60 *17000+$ 200x+$ 6000x = $ ( 6200x+10200) and the net cost of owning Model B for x years is $( 3828.57x + 0.6*25000) = $ (3828.57x + 15000).

The net cost of keeping model, A or, model B will be same if 6200x+10200= 3828.57x + 15000 or, 2371.43x = 4800 or, x = 4800/2371.43 = 2.02 (on rounding off to 2 decimal places).

Hence if we sell the car after 2.02 years, then the net cost of keeping model A or, model B will be same. If we sell the car after more than 2.02 years, then model B will be cheaper, otherwise model A will be cheaper.