Question

A truck is purchased for $40,000. The interest rate is 14%. Consider 2 alternatives.

A1: The truck is driven for 10 years, and then sold for $6000. Routine maintenance is $600 per year. A $3000 repair occurs in year 8.

A2: The truck is driven for 5 years and then sold for $18000. Routine maintenance is $600 per year. Repairs covered by warranty.

a. Show the money flow for each alternative (use diagram similar to figure 8.5 or table of disbursements /receipts) .

      b. Compute the annual equivalent cost of the $3000 repair that occurs in year 8 of Alternative A1.

c. Calculate the annual equivalent cost of A1 (keeping the truck for 10 years)

d. Calculate the annual equivalent cost of A2 (keeping the truck for 5 years)

Answer 1

Answer a

For Alternative A1:

Cashflow at year 0 will be -$40000

Cashflow for year 1 to 7 will be -$600 for Routine Maintenance

Cashflow for year 8 will be -3000-600=-$3600

Cashflow for year 9 will be -$600 for Routine Maintenance

Cashflow of year 10 will be -600+6000=$5400

Hence Cashflow table will be as follows:

Year	0	1	2	3	4	5	6	7	8	9	10
Cashflows	-40000	-600	-600	-600	-600	-600	-600	-600	-3600	-600	5400

For Alternative A2:

Cashflow at year 0 will be -$40000

Cashflow for year 1 to 4 will be -$600 for Routine Maintenance

Cashflow of year 5 will be -600+18000=$17400

Hence Cashflow table can be drawn as follows:

Year	0	1	2	3	4	5
Cashflows	-40000	-600	-600	-600	-600	17400

Answer b

PV at the end of year 0 of $3000= 3000/(1+14%)^8 = 3000/1.14^8=3000/2.8526=$1051.68

Annuity of $1051.68 for 10 years will be given as: PV=A*(1-(1+r)^-n)/r

or, 1051.68=A*(1-(1+14%)^-10)/14%

or, 1051.68=A*(1-(1.14)^-10)/0.14

or, 1051.68=A*(1-0.2697)/0.14

or, 1051.68=A*0.7303/0.14

or, A = 1051.68*0.14/0.7303

or, A = 201.62

Hence Equivalent Annual cost of $3000 is $201.62

Answer c

Equivalent Annual Cost of Initial cost will be given by: PV=A*(1-(1+r)^-n)/r

or, 40000=A*(1-(1+14%)^-10)/14%

or, 40000=A*(1-(1.14)^-10)/0.14

or, 40000=A*(1-0.2697)/0.14

or, 40000=A*0.7303/0.14

or, A = 40000*0.14/0.7303

or, A = $7668.54

Equivalent Annual Cost of Initial Cost = $7668.54

Equivalent Annual Cost of Salvage Value will be given by: FV=A*((1+r)^n-1)/r

or, 6000=A*((1+14%)^10-1)/14%

or, 6000=A*((1.14)^10-1)/0.14

or, 6000=A*(3.7072-1)/0.14

or, 6000=A*2.7072/0.14

or, A = 6000*0.14/2.7072

or, A=310.28

Equivalent Annual Cost of Salvage Value = $310.28

Hence Net Equivalent Annual Cost (EAC)= EAC of Initial Cost+ EAC of Routine Maintenance+ EAC of Repair cost- EAC of Salvage Value = 7668.54+600+201.62-310.28=8159.88

Hence EAC of A1=$8159.88

Answer d

Equivalent Annual Cost of Initial cost will be given by: PV=A*(1-(1+r)^-n)/r

or, 40000=A*(1-(1+14%)^-5)/14%

or, 40000=A*(1-(1.14)^-5)/0.14

or, 40000=A*(1-0.5194)/0.14

or, 40000=A*0.4806/0.14

or, A = 40000*0.14/0.4806

or, A = $11651.34

Equivalent Annual Cost of Initial Cost = $11651.34

Equivalent Annual Cost of Salvage Value will be given by: FV=A*((1+r)^n-1)/r

or, 18000=A*((1+14%)^5-1)/14%

or, 18000=A*((1.14)^5-1)/0.14

or, 18000=A*(1.9254-1)/0.14

or, 18000=A*0.9254/0.14

or, A = 18000*0.14/0.9254

or, A=2723.10

Equivalent Annual Cost of Salvage Value = $2723.10

Hence Net Equivalent Annual Cost (EAC)= EAC of Initial Cost+ EAC of Routine Maintenance- EAC of Salvage Value = 11651.34+600-2723.10=9528.24

Hence EAC of A2=$9528.24