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

Answer:

Machine B has a higher NPV therefore should be produced

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.

Answer 2

Answer:

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

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

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

C 92,000.00

time 11

rate 0.12

$92000 \times \frac{1-(1+0.12)^{-11} }{0.12} = PV$

PV $546,268.3202

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

Maturity  $13,000.00

time  11.00

rate  0.12000

$\frac{13000}{(1 + 0.12)^{11} } = PV$  

PV   3,737.1894

Net Present Value:

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

Annual worth:

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

PV 300,005.00

time 11

rate 0.12

$300005 \div \frac{1-(1+0.12)^{-11} }{0.12} = C$

C  $ 50,525.463

Machine B

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

C 103,000.00

time 19

rate 0.12

$103000 \times \frac{1-(1+0.12)^{-19} }{0.12} = PV$

PV $758,675.0165

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

Maturity  $26,000.00

time  19.00

rate  0.12000

$\frac{26000}{(1 + 0.12)^{19} } = PV$  

PV   3,018.7762

Net present value

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

Annual worth

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

PV 301,693.80

time 19

rate 0.12

$301693.8 \div \frac{1-(1+0.12)^{-19} }{0.12} = C$

C  $ 40,958.857