Engineering Economic Analysis 12th Edition 980 A machine cos

Engineering Economic Analysis 12th Edition: 9-80: A machine costs $5240 and produces benefits of $1000 at the end of each year for 8 years. Assume an annual interest rate of 10%.

a) What is the payback period (in years), answer is 6%, how do I arrive at this?

b) What is the break even point (in years), answer is 7.8%, how do I achieve this?

Thanks!

Solution

Pay Back Period is Number of years taken to recover by benefits initial cost

We have initial cost in terms of Machine cost =$5240 and have benefits of $1000 for each year

then it will take 6 years to have sum of benefits equal to $6000 which exceeds initial cost of $5240 in 6th year therefore payback period is 6 years

Ans b)

Breakeven Point is number of years at which discounted cash inflows(benefits) equalize to cost

1000/(1.1)+1000/(1.1)^2+1000/(1.1)^3+1000/(1.1)^4+1000/(1.1)^5+1000/(1.1)^6+1000/(1.1)^7=$4868.4

We need to find at what point of time this difference will be earned during 8th year 5240-4868.4=371.6

Assume y is the breakeven point

$5334.92 is Present Worth of Project which is achieved in 8 years & $4868 in present worth of proejct achieved in 7 years and we are finding point in number of years at which Present worth becomes $5240

then we can set an equation such as (5334.92-5240)/(5240-4868)=(8-y)/(y-7)

94.92/372=(8-y)/(y-7)

94.92(y-7)=(8-y)*372

94.92y-664.44=2976-372y

466.92y=3640

y=7.79 years approx. 7.8 years

y=7.79 years

Break even point is 7.8 years