1. Chapter 4 More Interest Formulas
Faculty of Applied Engineering and Urban Planning
Civil Engineering Department
Engineering Economy
Chapter 4
More Interest Formulas
Lecture 1
Week 1
2nd Semester 20015/2016

2. Chapter 4: More Interest Formulas
Uniform Series
Arithmetic Gradient
Geometric Gradient
Nominal and Effective Interest
Continuous Compounding

3. Uniform Series

4. Uniform Series In Chapter 3 we dealt with single payments:
F = P (1+i)n F = P (F/P,i%,n)
P = F (1+i)-n P = F (P/F,i%,n)

5. Uniform Series
Quite often we have to deal with uniform (equidistant and equal-valued) cash flows during a period of time:

6. Deriving Uniform Series Formula
We want an expression for the present worth P of a stream of equal, end-of-period cash flows.
Write a present worth value for each period individually, and add them.

7. Deriving Uniform Series Formula

8. Deriving Uniform Series Formula

9. Deriving Uniform Series Formula

10. Uniform Series
Example 4-1: You deposit $500 in a bank at the end of each year for five years. The bank pays 5% interest, compounded annually. At the end of five years, immediately following your fifth deposit, how much will you have in this account?

11. Uniform Series
Example 4-2:How much money do you put in bank every month to have $1,000 at the end of the year. Assume you will put the same amount in the bank each month and the bank pays ½ % interest monthly?
Solution:
A = 1000 (A/F, ½%,12) = 1000 (0.0811) = $81.10/month1

12. Uniform Series
Example 4-3: Suppose on January 1 you deposit $5,000 in a bank paying 8% interest, compounded annually. You want to withdraw all the money in five equal end-of year sums, beginning December 31stof the first year.
Solution:
Given: P = $5000 n = 5 i= 8% A = unknown
A = P (A/P,8%,5) = P{[i(1 + i)n]/[(1+i)n–1]}
= 5000 ( ) = $1,252.28
The withdrawal amount is about $1,252
Note: This is the source of the $1,252 in Plan C from Chapter 31

13. Uniform Series Examples 4-4 & 4-5:
An investor holds a time payment purchase contract on some machine tools. The contract calls for the payment of $140 at the end of each month for a five-year period. The first payment is due in one month. He offers to sell you the contract for $6,800 cash today. If you otherwise can make 1% per month on your money, would you accept or reject the investor’s offer?
Solution:
Given: A =$140, n=60, i=1%
P = A (P/A,i,n) = 140 (P/A,1%,60) = 140 (44.955) = $6,293.70
This is less than the required $6,800 => Reject offer

14. Uniform Series Example 4-6:
Compute the value of the following cash flows at at the end of year 5 given i= 15%.
The Sinking Fund Factor diagram is based on the assumption the withdrawal coincides with the last deposit. This does not happen in this example. he end of year 5 given i= 15%.

15. Uniform Series

16. Uniform Series
Second Approach: Compute the future values of each deposit then add them.
F = F1+ F2+ F3
= 100(F/P,15%,4) + 100(F/P,15%,3) + 100(F/P,15%,2)
= 100 (1.749) (1.521) (1.322) = $459.20

17. Uniform Series
Example 4-7: For diagram below and given i= 15%, find value of P?
This diagram is not in a standard form
Approach 1:Compute the present value of each flow then add them:
P = P1+ P2+ P3
= 20 (P/F,15%,2) + 30 (P/F,15%,3) + 20 (P/F,15%,4)
= 20 (0.7561) + 30 (0.6575) + 20 (0.5718) = $46.28

18. Uniform Series
Approach 2: Compute the future value, F, of the flows at the end of year 4. Then compute the present value of the future value, F.
Approach 3:Compute the present values of the flows at the end of year 1, P1. Then compute P, the present value of P1.
Approach 4: How about this one:
P = 20 (P/A,15%,3) (P/F,15%,1) + 10 (P/F,15%,3)
= 20 (2.283) (0.8696)+ 10 (0.6575) = $46.28
Other approaches will also work.

19. Arithmetic Gradient Series

20. Arithmetic Gradient Series
Sometimes cash flows increase/decrease by a uniform fixed amount G every subsequent period.
Write a future worth value for each period individually, and add them
F = G(1+i)n-2 + 2G(1+i)n-3+ … + (n-2)G(1+i)1+ (n- 1)G(1+i)0
= G [(1+i)n-2 + 2(1+i)n-3+ … + (n-2)(1+i)1+ n-1]

21. Arithmetic Gradient Series

22. Arithmetic Gradient Series
Arithmetic Gradient Present Worth -(P/G, i%, n):
Arithmetic Gradient Future Worth -(F/G, i%, n):
Arithmetic Gradient Uniform Series -(A/G, i%, n):

23. Arithmetic Gradient Series
Example 4-8: Suppose you buy a car. You wish to set up enough money in a bank account to pay for standard maintenance on the car for the first five years. You estimate the maintenance cost increases by G = $30 each year. The maintenance cost for year 1 is estimated as $120. i= 5%.
Thus, estimated costs by year are $120, $150, $180, $210, $240.

24. Arithmetic Gradient Series

25. Arithmetic Gradient Series
Example 4-9: Maintenance costs of a machine start at $100 and go up by $100 each year for 4 years. What is the equivalent uniform annual maintenance cost for the machinery if i= 6%.
Solution:
This is not in the standard form:
Because year-one cash flow is not zero
We need to reformulate the problem

26. Arithmetic Gradient Series

27. Arithmetic Gradient Series
Example 4-10: With i= 10%, n = 4, find an equivalent uniform payment A for the following CFD
This is a problem with decreasing costs instead of increasing costs.
Solution:
The cash flow can be rewritten as the DIFFERENCE of the following two diagrams: (1) the standard form we need for arithmetic gradient, and (2) a series of uniform payments.

28. Arithmetic Gradient Series

29. Arithmetic Gradient Series
Example 4-11: Find P for the following CFD with i= 10%.
Solution:
 This is not in the standard form for the arithmetic gradient.
 If we insert a “present value” J at the end of year 2, the diagram from that point on is in standard form.
J = 50 (P/G,10%,4) = 50 (4.378) =
P = J (P/F,10%,2) = (0.8264) = $180.9

30. Geometric Gradient Series
Instead of constant amount of increase, sometimes cash flows increase by a uniform rate of increase g (constant percentage amount) every subsequent period.
At the end of year i, i= 1, ..., n;
we incur a cost Ai= A(1+g)i-1
Write a future worth value for each period individually, and add them up
P= A(1+i)-1+A(1+g)1(1+i)-2+A(1+g)2(1+i)-3+…
+A(1+g)n-2(1+i)-n+1+A(1+g)n-1(1+i)-n

31. Geometric Gradient Series

32. Geometric Gradient Series
Example: Suppose you have a vehicle. The first year maintenance cost is estimated to be $100. The rate of increase in each subsequent year is 10%. You want to know the present worth of the cost of the first five years of maintenance, given i= 8%.
Solution: Repeated Present-Worth (Step-by-Step) Approach:

33. Geometric Gradient Series

34. Nominal and Effective Interest Rates

35. Multiple Compounding
The time standard for interest computations is One Year.
Many banks compound interest multiple times during the year.
E.g.: 12% per year, compounded monthly (1% interest is paid monthly)
E.g.: 8% per year, compounded semi-annually (4% interest is paid twice a year or once every 6 months)
Compounding is not less important than interest. You have to know all the info to make a good decision.
You need to pay attention to the following terms:
Time Period –The period over which the interest is expressed (always stated).
E.g.: “6% per year”
Compounding Period (sub-period)–The shortest time unit over which interest is charged or earned.
E.g.: If interest is “6% per year, compounded monthly”, compounding period is one month
Compounding Frequency –The number of times (m) that compounding occurs within time period.
Compounding semi-annually: m = 2; Compounding quarterly: m = 4
Compounding monthly: m = 12; Compounding weekly: m = 52; daily = 365

36. Nominal and Effective Interest Rate
Two types of interest are typically quoted:
Nominal interest rate, r, is an annual interest rate without considering the effect of (sub-period) compounding.
Effective interest rate, i, is the actual rate that applies for a stated period of time which takes into account the effect of (sub-period) compounding.
The effective rate is commonly expressed on an annual basis denoted as “ia”.
Sometimes one interest rate is quoted, sometimes another is quoted. If you confuse the two you can make a bad decision.
Effective interest is the “real” interest rate over a period of time; nominal rate is just given for simplicity (per year)
All interest formulas use the effective interest rate to properly account for the time value of money.

37. Nominal and Effective Interest Rate
To get effective rate per compounding period, i:
To get effective rate per year, ia:
Example: Given an interest rate of 12% per year, compounded quarterly:
Nominal rate = 12%
Effective (Actual) rate = 12%/4 = 3% per quarter
Effective rate per year = [1+(0.12/4)]4-1= =12.55%
Investing $1 at 3% per quarter is equivalent to investing $1 at 12.55% annually

38. Nominal and Effective Interest Rate
Example: A bank pays 1.5% interest every three months. What are the nominal and effective interest rates per year?
Solution:
Nominal interest rate per year is
r = 4 *1.5% = 6% a year
Effective interest rate per year:
ia= (1 + r/m)m–1 = (1.015)4–1 =
6.14% a year

39. Nominal and Effective Interest Rate
Example: $10K is borrowed for 2 years at an interest rate of 24% per year compounded quarterly. If the same sum of money could be borrowed for the same period at the same interest rate of 24% per year compounded annually, how much could be saved in interest charges?
Interest charges for quarterly compounding:
$10,000(1+24%/4)2x4-$10,000 = $
Interest charges for annually compounding:
$10,000(1+24%)2-$10,000 = $
Savings:
$ $5376 = $562.48

40. Nominal and Effective Interest Rate
Example 4-15: A loan shark lends money on the following terms. “If I give you $50 on Monday, then you give back $60 the following Monday.”
1.What is the nominal rate, r ?
The loan shark charges i= 20% a week:
60 = 50 (1+i) » i= 0.2 [Note we have solved 60 = 50(F/P,i,1) for i]
We know m = 52, so r = 52 *i= 10.4, or 1,040% a year
2.What is the effective rate, ia?
ia= (1 + r/m)m–1 = (1+10.4/52)52–1 ≈13,104
This means about 1,310,400 % a year (can you believe this!!)

41. Nominal and Effective Interest Rate
Words of Warning: When all various compounding periods in a problem match, it makes calculations much simpler. When they do not match, life is more complicated.
Example 4-16: You deposit $5,000 in a bank paying 8% nominal interest, compounded quarterly. You want to withdraw the money in five equal yearly sums, beginning Dec. 31 of the first year. How much should you withdraw each year? Note effective interest is i= 2% = r/4 = 8%/4 quarterly, and there are 20 periods.

42. Nominal and Effective Interest Rate
Note: The withdrawal periods and the compounding periods are not the same. If we want to use the formula
A = P (A/P,i,n)
then we must find a way to put the problem into an equivalent form where all the periods are the same.
Solution 1: Suppose we withdraw an amount A quarterly. (We don’t, but suppose we do.). We compute
A = P (A/P,i,n) = 5000 (A/P,2%,20) = 5000 (0.0612) = $306

43. Nominal and Effective Interest Rate
Now consider the following:
Consider a one-year period:
This is now in a standard form that repeats every year.
W = A(F/A,i,n) = 306 (F/A,2%,4) = 306 (4.122) = $1,260
You should withdraw $1,260 at the end of each year.

44. Nominal and Effective Interest Rate
Solution 2: (Probably the easiest way)
ia= (1 + r/m)m–1 = (1 + i)m–1 = (1.02)4–1 =
8.24%
Now use:
W = P(A/P,8.24%,5) = P {[i(1+i)n]/[(1+i)n –1]}
= 5000(0.252) = $1,260 per year

45. Continuous Compounding

46. Continuous Compounding

47. Continuous Compounding
Continuous compounding can sometimes be used to simplify computations, and for theoretical purposes.
The previous equation illustrates that er-1 is a good approximation of (1 + r/m)m for large m (i.e., ∞).
This means there are continuous compounding versions of the formulas we have seen earlier.
Summary:
F = P ernis analogous to F
= P (F/P,r,n): (F/P,r,n)∞= ern
P = F e-rnis analogous to P
= F (P/F,r,n): (P/F,r,n)∞= e-rn

48. Summary: Notation
i: effective interest rate per interest period (stated as a decimal) n: number of interest periods P: present sum of money F: future sum of money: an amount, n interest periods from the present, that is equivalent to P with interest rate i A: end-of-period cash receipt or disbursement amount in a uniform series, continuing for n periods, the entire series equivalent to P or F at interest rate i. G: arithmetic gradient: uniform period-by-period increase or decrease in cash receipts or disbursements g: geometric gradient: uniform rate of cash flow increase or decrease from period to period r: nominal interest rate per interest period (usually one year) ia: effective interest rate per year (annum) m: number of compounding sub-periods per period

49. Summary: Formulas
Single Payment formulas: Compound amount: F = P (1+i)n= P (F/P,i,n) Present worth: P = F (1+i)-n = F (P/F,i,n) Uniform Series Formulas: Compound Amount: F = A{[(1+i)n–1]/i} = A (F/A,i,n) Sinking Fund: A = F {i/[(1+i)n–1]} = F (A/F,i,n) Capital Recovery: A = P {[i(1+i)n]/[(1+i)n–1] = P (A/P,i,n) Present Worth: P= A{[(1+i)n–1]/[i(1+i)n]} = A (P/A,i,n) Arithmetic Gradient Formulas: Present Worth: P= G {[(1+i)n–in –1]/[i2(1+i)n]} = G(P/G,i,n) Uniform Series: A= G {[(1+i)n–in –1]/[i(1+i)n–i]} = G (A/G,i,n) Geometric Gradient Formulas: If i≠g, P = A {[1 –(1+g)n(1+i)-n]/(i-g)}= A (P/A,g,i,n) If i= g,P= A [n (1+i)-1] = A (P/A,g,i,n)

50. Summary: Formulas
Nominal interest rate per year, r : the annual interest rate without considering the effect of any compounding
Effective interest rate per year, ia:
ia= (1 + r/m)m–1 = (1+i)m–1 with i=r/m
Continuous compounding, ∞:
r: one-period interest rate, n: number of periods
(P/F,r,n)∞ = e-r n
(F/P,r,n)∞ = ern

51. Next Class
Read Chapter 4 of textbook Try to do some of the problems at end of Chapter 4 Be ready for Midterm Exam