1. NE 364 Engineering Economy
Lecture 4
Money-Time Relationships
and Equivalence
(Part 2: Uniform Series) A
NE 364 Engineering Economy

2. Revision
NE 364 Engineering Economy 2

3. 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
NE 364 Engineering Economy

4. Uniform Series Function and Proof
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

5. Example 1
How much will you have in 40 years if you save $3,000 each year and your account earns 8% interest each year?
NE 364 Engineering Economy

6. Interest Tables
NE 364 Engineering Economy

7. Example 2
A 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?
NE 364 Engineering Economy

8. NE 364 Engineering Economy

9. Finding A when given F
NE 364 Engineering Economy

10. Example 3
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?
NE 364 Engineering Economy

11. Finding P when given A From : and It results:
Dividing both sides by (1+i)N , hence:
NE 364 Engineering Economy

12. Example 4
How much is needed today to provide an annual amount of $50,000 each year for 20 years, at 9% interest each year?
NE 364 Engineering Economy

13. Example 5
If 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?
NE 364 Engineering Economy

14. NE 364 Engineering Economy

15. Finding A when given P
NE 364 Engineering Economy

16. Example 6
If you had $500,000 today in an account earning 10% each year, how much could you withdraw each year for 25 years?
NE 364 Engineering Economy

17. Example 7
You 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?
NE 364 Engineering Economy

18. NE 364 Engineering Economy

19. NE 364 Engineering Economy19

20. What if i is unknown?
NE 364 Engineering Economy

21. 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?
NE 364 Engineering Economy

22. 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)
NE 364 Engineering Economy

23. 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.
NE 364 Engineering Economy

24. 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
NE 364 Engineering Economy

25. 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.
NE 364 Engineering Economy

26. 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.
NE 364 Engineering Economy

27. What if N is unknown?
NE 364 Engineering Economy

28. Example 7
Joe 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?
NE 364 Engineering Economy

29. 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)
NE 364 Engineering Economy

30. NE 364 Engineering Economy

31. Solution cont. The factor at 30 years = 12.4090 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.
NE 364 Engineering Economy

32. 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 $ are worth in the 31st year and subtract it from the $8,000 payment
NE 364 Engineering Economy

33. 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
NE 364 Engineering Economy

34. 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).
NE 364 Engineering Economy

35. EXCEL
Solving for N
Solving for i
NE 364 Engineering Economy

36. Exam Dates Reminder 7th Week Assessment 12th Week Assessment
Exam (20pt.) + Quizzes (10pt)
Exam on Monday at 4 PM and the location will be announced at exam time.
Quizzes in the tutorial.
12th Week Assessment
Exam (15 pt.) + Quizzes (5 pt.)
Continuous Assessment (10 pt.)
Participation in Lecture + Attendance
NE 364 Engineering Economy