Question

The following costs and useful life data are associated with two new machines being considered at Arun Tech Inc.

Data A B
Useful Life (Years) 11 19
First Cost $250,000 $460,000
Salvage Value $13,000 $26,000
Annual Profit $92,000 $103,000

The company interest rate (MARR) is 12%. Which machine should be purchased?

Answer 1

While Machine B has an slightly higher net present value his annual worth is much lower than machine A therefore, the company should purchase machine A which yield better annual return

Machine A

Net Present Value: 300,005.00

Anual worth $ 50,525.463

Machine B

Net Present Value: 301.693,8‬

Anual worth $ 40,958.857

Step-by-step explanation:

We calculate the present value of each machine

and also, the annual worth of each one to get a fair comparison considering their useful life differ

Machine A

`C * (1-(1+r)^(-time) )/(rate) = PV\\`

C 92,000.00

time 11

rate 0.12

`92000 * (1-(1+0.12)^(-11) )/(0.12) = PV\\`

PV $546,268.3202

`(Maturity)/((1 + rate)^(time) ) = PV`

Maturity $13,000.00

time 11.00

rate 0.12000

`(13000)/((1 + 0.12)^(11) ) = PV`

PV 3,737.1894

Net Present Value:

$546,268.32 + $3,737.19 - $250,000 = 300.005,51‬

Annual worth:

`PV / (1-(1+r)^(-time) )/(rate) = C\\`

PV 300,005.00

time 11

rate 0.12

`300005 / (1-(1+0.12)^(-11) )/(0.12) = C\\`

C $ 50,525.463

Machine B

`C * (1-(1+r)^(-time) )/(rate) = PV\\`

C 103,000.00

time 19

rate 0.12

`103000 * (1-(1+0.12)^(-19) )/(0.12) = PV\\`

PV $758,675.0165

`(Maturity)/((1 + rate)^(time) ) = PV`

Maturity $26,000.00

time 19.00

rate 0.12000

`(26000)/((1 + 0.12)^(19) ) = PV`

PV 3,018.7762

Net present value

$758,675.02 + $3,018.78 - 460,000 = 301.693,8‬

Annual worth

`PV / (1-(1+r)^(-time) )/(rate) = C\\`

PV 301,693.80

time 19

rate 0.12

`301693.8 / (1-(1+0.12)^(-19) )/(0.12) = C\\`

C $ 40,958.857

Answer 2

Machine B has a higher NPV therefore should be produced

Step-by-step explanation:

The machine with the higher Net Present Value (NPV) should be produced .

NPV of Machine A

PV of cash flow

PV of annual profit = A × (1- (1+r)^*(-n)/r

A- 92,000, n- 11, r- 12%

PV = 92,000 × (1- (1.12^(-11)/0.12 = 546268.32

PV of salvage value = 13,000× 1.12^(-11)= 3737.189

NPV = 546268.320 + 3737.189 -250,000 = $300,005.50

NPV of Machine B

A- 103,00, n- 19, r- 12%

PV = 103,000 × (1- (1.12^(-19)/0.12= 758675.0165

Pv of salvage value = 26000× 1.12^(-19)= 3018.776199

NPV =758675.0165 + 3018.77 -460,000 = $301,693.79

Machine B has a higher NPV , therefore should be produced.