NE 364 Engineering Economy - cloudecampus.org · Uniform Series Necessary Conditions: • P occurs...

NE 364

Engineering EconomyLecture 4

Money-Time Relationships and Equivalence

(Part 2: Uniform Series)

1NE 364 Engineering Economy

A

Revision


Uniform Series

Necessary Conditions:

• P occurs one Interest Period before the first A (uniform amount)

• F occurs at the same time as the last A, and N periods after P,

• A occurs at the end of periods 1 through N, inclusive


Uniform Series Function and Proof

4

Proof:F=

A(F/P,i%,N−1)+A(F/P,i%,N−2)+…+A(F/P,i%,1)+A(F/P,i%,0)

= A((1+i)N-1

+ (1+i)N-2

+(1+i)N-3

+ ….+ (1+i)1

+ (1+i)0)

This is a geometric series of a form

Where b=(1+i)-1

, a1= (1+i)N-1

, and aN=(1+i)0

NE 364 Engineering Economy

Example 1

5

How much will you have in 40 years if you

save $3,000 each year and your account

earns 8% interest each year?


Interest Tables


Example 2A recent government study reported that a college degree

is worth an extra $23,000 per year in income (A)

compared to what a high-school graduate makes. If the

interest rate (i) is 6% per year and you work for 40 years

(N), what is the future compound amount (F) of this extra

income?


Finding A when given F


Example 3

10

How much would you need to set aside each

year for 25 years, at 10% interest, to have

accumulated $1,000,000 at the end of the 25

years?


Finding P when given A

11

and

It results:

Dividing both sides by (1+i)N

, hence:

From :


Example 4

12

How much is needed today to provide an

annual amount of $50,000 each year for 20

years, at 9% interest each year?


Example 5If a certain machine undergoes maintenance now, its

output can be increased by 20% - which translates into

additional cash flow of $20,000 at the end of each year for

five years. If i=15% per year, how much can we afford to

invest to maintain this machine?


Finding A when given P


Example 6

16

If you had $500,000 today in an account

earning 10% each year, how much could you

withdraw each year for 25 years?


Example 7You borrow $15,000 from your credit union to purchase a

used car. The interest rate on your loan is 0.25% per

month and you will make a total of 36 monthly payments.

What is your monthly payment?


What if i is unknown?


Example 5 (Finding i)After years of being poor, debt-encumbered college student, you decide that you want to pay for your dream car in cash. Not having enough money now, you decide to specifically put money away each year in a "dream car" fund.

The car you want to buy will cost $60,000 in eight years. You are going to put aside $6,000 each year (for eight years) to save for this.

At what interest rate must you invest your money to achieve your goal of having enough to purchase the car after eight years?


Solution The car you want to buy will cost $60,000 in eight years

means F8=$60,000.

You are going to put aside $6,000 each year (for eight

years) means A=$6,000 for 8 years.

So,

F8= A * (F/A, i%, 8)

$60,000=$6,000 * (F/A, i%, 8)

10= (F/A, i%, 8)


Solution cont. So we are looking for a factor (F/A, i%, 8) which is

equal to 10.

We know that we have to look in the F/A column and in

the 8th row, but we don’t know which interest rate

(which interest table).

We have searched all the tables that we have at the

F/A column and 8th row and found two close values in

the 6% and 7% tables.


Solution cont. The interest rate we are searching for is between 6%

and 7%.

The equation for F/A factor is a non-linear but we can

approximate it to a linear equation


Solution cont. (interpolation)The dashed curve is what we are linearly

approximating. The answer, i', can be

determined by using the similar triangles

dashed in the figure.


Solution cont.

So if you can find an investment account that will earn

at least 6.28% interest per year, you'll have the $60,000

you need to buy your dream car in eight years.


What if N is unknown?


Example 7Joe borrowed $100,000 from a local bank, which charges

him an interest rate of 7% per year. If Joe pays the bank

$8,000 per year, how many years will it take to pay off the

loan?


Solution Joe borrowed $100,000 from a local bank means

P0=$100,000

If Joe pays the bank $8,000 per year means

A=$8,000 for N years but N is unknown.

So,

P0 = A * (P/A, 7%, N)

$100,000=$8,000 * (P/A, 7%, N)

12.5 = (P/A, 7%, N)


Solution cont. The factor at 30 years = 12.4090 not 12.5

The factor at 35 years = 12. 9477 not 12.5

This means that if Joe paid for 30 years $8,000, this will not cover his $100,000 loan. And if he paid for 35 years $8,000 it will be more than his loan.

So, let’s assume he will pay for 31 years and calculate the amount of the loan he will cover.

The row 31 is not calculated in the 7% table and therefore we will use the equation of the P/A factor.


Solution cont. P0new= $8,000 * (P/A, 7%, 31)

= $8,000 * ((1.07)31 – 1 ) / (0.07 * (1.07)31)

= $100,254.51

This is more than he owes the bank. This means that his last payment on the 31st year should not be $8,000 but less. How much less?

We should calculate how much the extra $254.51 are worth in the 31st year and subtract it from the $8,000 payment


Solution cont. F31 = $254.51 * (F/P, 7%, 31)

= $254.51 * (1.07)31

= $2,073

So, the last payment on the 31st year should not be

$8,000 but should be

$8,000 – $2,073 = $5,927 only


There are specific spreadsheet

functions to find N and i.

The Excel function used to solve for i is

RATE(nper, pmt, pv, fv), which returns a fixed interest rate for

an annuity of pmt that lasts for nper periods to either its present

value (pv) or future value (fv).

The Excel function used to solve for N is

NPER(rate, pmt, pv), which will compute the number of

payments of magnitude pmt required to pay off a present

amount (pv) at a fixed interest rate (rate).


EXCEL

Solving for N Solving for i


Thank You

NE 364 Engineering Economy 36

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