Question

You can choose between Machine A or B. Your annual interest rate is 7%. You need a machine for 6 years (required service period). 1. Machine A costs $54,000 and lasts for 3 years. It has no salvage value and costs an additional $18,000 each year to operate. 2. Machine B costs $92,000 and lasts for 6 years. It has a salvage value of $18,000 and costs $13,000/year to operate. Assume both machines can be purchased again for the same costs. What is the Annual Equivalent Cost of the machine that you should purchase?

Answer 1

Answer:

M1 EAC =  38,576.91

M2 EAC = 29,784.89

Explanation:

The equivalent annual cost is the PMT of the present worh of the machine/investment.

Machine A

54,000 at year 0 then 54,000 at beginning of year 4th

and 18,000 per year

We need to bering into present the 54,000 of the fourth year

the 18,000 are already split into each year.

$\frac{Maturity}{(1 + rate)^{time} } = PV$  

Maturity  54,000

time   3 (it is done at the beginning of the 4th year not at the end of it)

rate  0.07

$\frac{54000}{(1 + 0.07)^{3} } = PV$  

PV   44,080.09

54,000 + 44,080.09 = 98,080.09

Then we calculate the PMT

$PV \div \frac{1-(1+r)^{-time} }{rate} = C$

PV  $98,080.09

time 6 years

rate         0.07

$98080.09 \div \frac{1-(1+0.07)^{-6} }{0.07} = C$

C  $ 20,576.791

Now we add the annual cost of 18,000

getting 38,576.79 as annual equivalent cost ofr machine 1

For machine B

anual cost of 13,000

purchase of 92,000

and 18,000 salvage value at end of year 6:

$\frac{Maturity}{(1 + rate)^{time} } = PV$  

Maturity  18,000.00

time   6.00

rate  0.07

$\frac{18000}{(1 + 0.07)^{6} } = PV$  

PV   11,994.16

This is positive as is a cash inflow.

net worth: 92,000 - 11,994.16

net worth: 80.005,84‬

Now, we solve for PMT:

$80005.84 \div \frac{1-(1+0.07)^{-6} }{0.07} = C$

C  $ 16,784.889

add the yearly maintenance cost of  13,000

Equivalent Annual Cost: 29,784.89