Question

This question about " Engineering Economics ".

Question:

A growing machine shop wishes to set aside money now to invest over the coming years in automating its customer service department. The company can earn 10% on a lump sum amount deposited now, and it wishes to do the following:

* Year 1 - no expenses

* Year 2 - withdraw $25,000 purchase computers and database software

* Year 3 - withdraw $3,000 to purchase additional software

* Year 4 - deposit $1,000 from the income

* Year 5 - withdraw $5,000 to purchase software updates

How much money must be deposited now in order to cover the expected payments over the next five years? Assume that all the transactions are made at the beainning of each year.

O -$27,870.36

O -$29,372.98

O $27,870.36

O $29,372.98

Answer 1

All amounts are in $

Let the amount to be invested today be Y

Deposit today Y

At end of 1st year = Y x 1.1 = 1.1Y

We will withdraw 25,000 at start of year 2

So Balance at end of 2nd year

= (1.1Y - 25,000) x 1.1

= 1.21Y - 27,500

At beginning of year 3, we withdraw 3,000

So balance at end of year 3

= (1.21Y - 27,500 - 3,000) x 1.1

= 1.331Y - 33,550

At start of year 4, 1000 is deposited

Balance at end of year 4 will be

= (1.331Y - 33,550 + 1,000) x 1.1

= 1.4641Y - 35,805

At start of year 5 the balance must be equal to 5,000 (payment to be made)

1.4641Y - 35,850 = 5,000

1.4641Y = 5,000 + 35,805

Y = (40,805/1.4641)

Initial deposit = $27,870.36 (positive)

answered by: anonymous