1 EGGC4214 Systems Engineering & Economy Lecture 4 Interest Formulas

2 INTEREST FORMULAS Example Jane deposits $500 in a credit union at the end of each year for five years. The CU pays 5% interest, compounded annually. At the end of five years, immediately following her fifth deposit, how much will Jane have in her account? A AAAA F AFAAAA Jane’s point of view: CU’s point of view:

3 UNIFORM SERIES FORMULA DERIVATION n = 5 periods F = A (1+i) 4 + A (1+i) 3 + A (1+i) 2 + A (1+i) + A F = A (1+i) 4 + A (1+i) 3 + A (1+i) 2 + A (1+i) + A F = A [(1+i) 4 + (1+i) 3 + (1+i) 2 + (1+i) + 1 ] F = A [(1+i) 4 + (1+i) 3 + (1+i) 2 + (1+i) + 1 ] (1+i) F = A [(1+i) 5 + (1+i) 4 + (1+i) 3 + (1+i) 2 + (1+i)] (1+i) F = A [(1+i) 5 + (1+i) 4 + (1+i) 3 + (1+i) 2 + (1+i)] (1+i) F – F = A [(1+i) 5 – 1] (1+i) F – F = A [(1+i) 5 – 1] F = A [(1+i) 5 – 1]/i F = A [(1+i) 5 – 1]/i A = F i/[(1+i) 5 – 1] F = A [(1+i) 5 – 1]/i A = F i/[(1+i) 5 – 1] F = A [(1+i) 5 – 1]/i Uniform Series Factors: (A/F,i,n)=i/[(1+i) n – 1] (F/A,i,n)=[(1+i) n – 1]/i AFAAA A

4 UNIFORM SERIES Uniform series compound amount factor : (F/A,i,n) = [(1+i) n – 1]/i, i > 0 (F/A,i,n) = [(1+i) n – 1]/i, i > 0 Uniform series fund (A/F,i,n) = i/[(1+i) n – 1] (A/F,i,n) = i/[(1+i) n – 1] Example 4-1. Jane deposits $500 in a credit union at the end of each year for five years. The CU pays 5% interest, compounded annually. At the end of five years, immediately following her fifth deposit, how much will Jane have in her account? F = A (F/A,i,n) = A [(1+i) n – 1]/i = $500[(1.05) 5 – 1]/(0.05) = $500 (5.5256) = $2,  $2,763. = $500 (5.5256) = $2,  $2,763.

5 UNIFORM SERIES Example A lot of land is available for $1000. Jane will put the same amount in the bank each month to have $1,000 at the end of the year. The bank pays 6% annual interest, compounded monthly. How much does Jim put in the bank each month? Solution A = 1000 (A/F, ½ %,12) = 1000 (0.0811) = $81.10/month A = 1000 (A/F, ½ %,12) = 1000 (0.0811) = $81.10/month Example Suppose on January 1 Jane deposits $5000 in a credit union paying 8% interest, compounded annually. She wants to withdraw all the money in five equal end-of year sums, beginning December 31 st of the first year AAAAA P

6 UNIFORM SERIES We know A = F i/[(1+i) n – 1] F = P (1 + i) n. F = P (1 + i) n. Hence A = P{[i (1 + i) n ]/[(1+i) n – 1]} =P(A/P,i,n) This is a formula to determine the value of a series of end-of-period payments, A, when the present sum P is known. (A/P,i,n) is called the uniform series capital recovery factor.

7 UNIFORM SERIES Example Suppose on January 1 Jim deposits $5000 in a credit union paying 8% interest, compounded annually. He wants to withdraw all the money in five equal end-of year sums, beginning December 31 st of the first year. Example Suppose on January 1 Jim deposits $5000 in a credit union paying 8% interest, compounded annually. He wants to withdraw all the money in five equal end-of year sums, beginning December 31 st of the first year. P = $5000 n = 5 i = 8% A = unknown A = P (A/P,8%,5) = P{[i (1 + i) n ]/[(1+i) n – 1]} A = P (A/P,8%,5) = P{[i (1 + i) n ]/[(1+i) n – 1]} = 5000 ( ) = $ = 5000 ( ) = $ The withdrawal amount is about $ AAAAA P

8 UNIFORM SERIES Example 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 $6800 cash today. If you otherwise can make 1% per month on your money, would you accept or reject the investor ’ s offer? A =140, n=60, i=1% …………. A = P (A/P,i,n) P = A / (A/P,i,n) = = A (P/A,i,n) = 140 (P/A,1%,60) = 140 (44.955) = $ = A (P/A,i,n) = 140 (P/A,1%,60) = 140 (44.955) = $ Imagine 60 +ve arrows

9 UNIFORM SERIES The Reverse Point of View. We can solve A = P{[i (1 + i) n ]/[(1+i) n – 1]} for P to obtain: P = A {[(1+i) n – 1]/[i (1 + i) n ]} = A (P/A,i%,n) In this case (P/A,i%,n) = {[(1+i) n – 1]/ [i (1 + i) n ]} is called the uniform series present worth factor.

10 UNIFORM SERIES Uniform series compound amount factor : (F/A,i,n) = [(1+i) n – 1]/i, i > 0 Uniform series sinking fund : (A/F,i,n) = i/[(1+i) n – 1] Uniform series capital recovery : (A/P,i,n) = [i (1 + i) n ]/[(1+i) n – 1] (A/P,i,n) = [i (1 + i) n ]/[(1+i) n – 1] Uniform series present worth : (P/A,i,n) = [(1+i) n – 1]/[i (1 + i) n ] (P/A,i,n) = [(1+i) n – 1]/[i (1 + i) n ]

11 UNIFORM SERIES Example Compute the value of the following cash flows at the first of year 5. Compute the value of the following cash flows at the first of year 5. Year Cash flow F 5 - F The Sinking Fund Factor diagram is based on the assumption that the withdrawal coincides with the last deposit. This does not happen in this example. Note the zero

12 UNIFORM SERIES The standard approach is: A AA -F AAA What we have is: -F First Approach: AAA F1 F Use the “standard” approach to compute F 1. Then compute the future value of F 1 to get F. F 1 = 100 (F/A,15%,3) = 100 (3.472) = $ F = F 1 (F/P,15%,2) = (1.322) = $459.00

13 UNIFORM SERIES Second Approach: Compute the future values of each deposit and add them. F= F 1 + F 2 + F 3 = 100(F/P,15%,4) + 100(F/P,15%,3) + 100(F/P,15%,2) = 100 (1.749) (1.521) (1.322) = $ = A AA F A F1F1 A F2F2 A F 3 ++

14 UNIFORM SERIES Example i = 15% Example i = 15% This diagram is not in a standard form. Approach 1. Compute the present value of each flow and add them. P = P 1 + P 2 + P 3 = 20 (P/F,15%,2) + 30 (P/F,15%,3) + 20 (P/F,15%,4) = 20 (0.7561) + 30 (0.6575) + 20 (0.5718) = $ P

15 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, P 1. Then compute P, the present value of P 1. Other approaches will also work.

16 Relationships Between Compound Interest Factors F = P(1 + i) n = P(F/P,i,n) P = F/(1 + i) n = F (P/F,i,n) P = A[(1+i) n – 1]/[i(1 + i) n ] =A(P/A,i,n) A = P[i(1 + i) n ]/[(1 + i) n – 1] = P(A/P,i,n) = P(A/P,i,n) F = A{[(1 + i) n – 1]/i} = A(F/A,i,n) A = F{i/[(1 + i) n – 1]} = F(A/F,i,n) Present Worth factor Compound Amount factor (F/P,i,n) = 1/ (P/F,i,n) (A/P,i,n) = 1/(P,A,i,n) Uniform Series Capital Recovery Factor (A/F,i,n) = 1/ (F/A,i,n) Uniform Series Compound Amount Factor Uniform Series Sinking Fund Factor Uniform Series Present Worth Factor

17 P = A(P/A,i,n) = A[ (1+i) -1 + (1+i) -2 + … + (1+i) -n ] = A[(P/F,i,1)+(P/F,i,2)+...+(P/F,i,n)]  F = A (F/A,i,n) = A[ 1 + (1+i) + (1+i) (1+i) n-1 ] = A[ 1 + (F/P,i,1) + (F/P,i,2) (F/P,i,n-1) ] (P/A,i,n) = [(P/F,i,1) + (P/F,i,2) (P/F,i,n)] n-1 (F/A,i,n) = 1 + (F/P,i,1) + (F/P,i,2) (F/P,i,n-1) 0123n A F AAA … n AAAA P Relationships Between Compound Interest Factors

18 (A/P,i,n) = [i(1 + i) n ]/[(1+i) n – 1] (A/F,i,n) = {i/[(1+i) n – 1]} We start with an identity: i (1+i) n = i + i (1+i) n – i = i + i [(1+i) n – 1] Now divide by {(1+i) n – 1} to get: [i (1+i) n ]/[(1+i) n – 1] = i/[ (1+i) n – 1] + i This gives: (A/P,i,n) = (A/F,i,n) + i Relationships Between Compound Interest Factors

19 Arithmetic Gradient 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. Thus, estimated costs by year are $120, $150, $180, $210, $ $120 $150 $180 $210 $240

20 We break up the cash flows into two components: and G = A = 120 P1P1 P2P2 P 1 = A (P/A,5%,5) = 120 (P/A,5%,5) = 120 (4.329) = $ P 2 = G (P/G,5%,5) = 30 (P/G,5%,5) = 30 (8.237) = $ P = P 1 + P 2 = $ Note: 5 and not 4 Standard form diagram for an Arithmetic Gradient: (n) periods and (n-1) non-zero flows in an increasing order Arithmetic Gradient 12345

21 F = G(1+i) n-2 + 2G(1+i) n-3 + … + (n-2)G(1+i) 1 + (n-1)G(1+i) 0 F = G [(1+i) n-2 + 2(1+i) n-3 + … + (n-2)(1+i) 1 + n-1] (1+i) F = G [(1+i) n-1 + 2(1+i) n-2 + 3(1+i) n-3 + … + (n-1)(1+i) 1 ] iF= G [(1+i) n-1 + (1+i) n-2 + (1+i) n-3 + … + (1+i) 1 – n + 1] iF= G [(1+i) n-1 + (1+i) n-2 + (1+i) n-3 + … + (1+i) 1 – n + 1] = G [(1+i) n-1 + (1+i) n-2 + (1+i) n-3 + … + (1+i) 1 + 1] – nG = G (F/A, i, n) - nG = G [(1+i) n -1]/i – nG F= G (F/G, i, n) = G [(1+i) n -in-1]/i 2 F= G (F/G, i, n) = G [(1+i) n -in-1]/i 2 P= G (P/G, i, n) = G [(1+i) n -in-1]/[i 2 (1+i) n ] P= G (P/G, i, n) = G [(1+i) n -in-1]/[i 2 (1+i) n ] A= F (A/F, i, n) A= F (A/F, i, n) = G [(1+i) n -in-1]/i 2 × i/[(1+i) n -1] A= G (A/G, i, n) A= G (A/G, i, n) = G [(1+i) n -in-1]/[i(1+i) n -i] 0123….n G 2G …… (n-1)G 0 F Arithmetic Gradient

22 Arithmetic Gradient Uniform Series Arithmetic Gradient Present Worth (P/G,5%,5) = {[(1+i) n – i n – 1]/[i 2 (1+i) n ]} = {[(1.05) 5 – 0.25 – 1]/[ (1.05) 5 ]} = {[(1.05) 5 – 0.25 – 1]/[ (1.05) 5 ]} = / = = / = (P/G,i,n) = { [(1+i) n – i n – 1] / [i 2 (1+i) n ] } (A/G,i,n) = { (1/i ) – n/ [(1+i) n –1] } (F/G,i,n) = G [(1+i) n – in – 1]/i 2 =1/(G/P,i,n) =1/(G/A,i,n) =1/(G/F,i,n) Arithmetic Gradient

23 Example 4-6. 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% AAAA This is not in the standard form for using the gradient equation, because the first year cash flow is not zero. We reformulate the problem as follows: Arithmetic Gradient

24 = + A 1 =100 G = The second diagram is in the form of a $100 uniform series. The last diagram is now in the standard form for the gradient equation with n = 4, G = 100. A = A 1 + G (A/G,6%,4) = (1.427) = $ Arithmetic Gradient

25 Example Example With i = 10%, n = 4, find an equivalent uniform payment for: This is a problem with decreasing costs instead of increasing costs. The cash flow can be rewritten as the DIFFERENCE of the following two diagrams, the second of which is in the standard form we need, the first of which is a series of uniform payments Arithmetic Gradient

26 = _ A 1 = A 1 A 1 3G 2G G A = A 1 – G(A/G,10%,4) = – 6000 (A/G,10%,4) = – 6000(1.381) = $ G = 6000 Arithmetic Gradient

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

28 Geometric Gradient 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%. Repeated Present-Worth (Step-by-Step) Approach:

29 Geometric Gradient Geometric Gradient. At the end of year j, j = 1,..., n, we incur a cost A j = A(1+g) j-1. P = A(1+i) -1 +A(1+g) 1 (1+i) -2 +A(1+g) 2 (1+i) -3 + … 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 … +A(1+g) n-2 (1+i) -n+1 +A(1+g) n-1 (1+i) -n P(1+g) 1 (1+i) -1 =A(1+g) 1 (1+i) -2 +A(1+g) 2 (1+i) -3 + … P(1+g) 1 (1+i) -1 =A(1+g) 1 (1+i) -2 +A(1+g) 2 (1+i) -3 + … … +A(1+g) n-1 (1+i) -n +A(1+g) n (1+i) -n-1 P - P(1+g) 1 (1+i) -1 = A(1+i) -1 - A(1+g) n (1+i) -n-1 P (1+i-1-g) = A (1 - (1+g) n (1+i) -n ) i ≠ g:P = A (1 - (1+g) n (1+i) -n )/(i-g) i = g:P = A n(1+i) ….n A A(1+g) 1 A(1+g) 2 A(1+g) n-1 (P/A,g,i,n)

30 Geometric Gradient Example n = 5, A 1 = 100, g = 10%, i = 8%. (P/A,g,i,n) = (P/A,10%,8%,5) = P = A(P/A,g,i,n) = 100 (4.8042) = $480.42

31 Ten thousand dollars is borrowed for two years at an interest rate of 24% per year compounded quarterly. If this 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) 2*4 -$10,000 = $ interest charges for annually compounding: $10,000(1+24%) 2 - $10,000 = $ Savings: $ $5376 = $ Compounding is not less important than interest. You have to know all the information to make a good decision Nominal and Effective Interest Rate

32 Sometimes one interest rate is quoted, sometimes another is quoted. If you confuse the two you can make a bad decision. A bank pays 5% compounded semi-annually. If you deposit $1000, how much will it grow to by the end of the year? The bank pays 2.5% each six months. You get 2.5% interest per period for two periods  1000(1.025) = 1,025  1025(1.025) = $1, With i = 0.05/2, r = 0.05, P  (1 + i) P  (1+r/2) 2 P = ( /2) 2 P = ( ) P Nominal and Effective Interest Rate

33 Terms the example illustrates: r = 5% is called the nominal interest rate per one year i = 2.5 % is called the effective interest rate per interest period i a = % is called the effective interest rate per one year In the example:m = 2 is the number of compounding sub-periods per time period. i a = ( ) – 1 = (1.025) 2 – 1 = (1+0.05/2) 2 – 1 r = nominal interest rate per year m = number of compounding sub-periods per year i = r/m = effective interest rate per compounding sub-period. i a = (1 + r/m) m – 1 = (1 + i) m – 1 The term i we have used up to now is more precisely defined as the effective interest rate per interest period. If the interest period is one year (m = 1) then i = r. Nominal and Effective Interest Rate

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

35 Nominal and Effective Interest Rate

36 Example Joe lends money on the following terms: “ If I give you 50 dollars on Monday, then you owe me 60 dollars the following Monday. ” 1.What is the nominal rate, r? We first note Joe charges i = 20% a week, since 60 = (1+i)50  i = 0.2. Note we have solved F = 50(F/P,i,1) for i. since 60 = (1+i)50  i = 0.2. Note we have solved F = 50(F/P,i,1) for i. We know m = 52, so r = 52  i = 10.4, or 1040% a year. We know m = 52, so r = 52  i = 10.4, or 1040% a year. 2. What is the effective rate, i a ? From i a = (1 + r/m) m – 1 we have i a = (1+10.4/52) 52 – 1  This means exactly % per year. Nominal and Effective Interest Rate

37 Suppose Joe can keep the $50, as well as all the money he receives in payments, out in loans at all times? How much would Joe have at the end of the year? We use F = P(1+i) n to get F = 50(1.2) 52  $655,232 Words of Warning. When the various compounding periods in a problem all match, it makes calculations much simpler. When they do not match, life is more complicated. Recall Example. We put $5000 in an account paying 8% interest, compounded annually. We want to find the five equal EOY withdrawals. We used A = P(A/P,8%,5) = 5000  (0.2505)  $1252 Suppose the various periods are not the same in this problem. Nominal and Effective Interest Rate

38 Example: Sally deposits $5000 in a CU paying 8% nominal interest, compounded quarterly. She wants to withdraw the money in five equal yearly sums, beginning Dec. 31 of the first year. How much should she withdraw each year? Note: effective interest is i = 2% = r/4 = 8%/4 quarterly, and there are 20 periods. Solution: $5000 i = 2%, n = 20 W WWWW Nominal and Effective Interest Rate

39 Nominal and Effective Interest Rate 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 compute: A = P (A/P,i,n) = 5000 (A/P,2%,20) = 5000 (0.0612) = $ These withdrawals are equivalent to P = $5000 $5000 i = 2%, n = 20

40 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) = $ Sally should withdraw about $1261 at the end of each year. A W W WW W i = 2%, n = 4 Nominal and Effective Interest Rate

41 Nominal and Effective Interest Rate Solution 2 i a = (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( ) = $ per year i a = 8.24%, n = 5 W $5000

42 Continuous Compounding

43 r= nominal interest rate per year r= nominal interest rate per year m= number of compounding sub-periods per year i= r/m = effective interest rate per compounding sub-period. Continuous compounding can sometimes be used to simplify computations, and for theoretical purposes. The table illustrates that e r - 1 is a good approximation of (1 + r/m) m for large values of m. This means there are continuous compounding versions of the formulas we have seen earlier. For example, F = P e rn is analogous to F = P (F/P,r,n): (F/P,r,n) inf = e rn P = F e -rn is analogous to P = F (P/F,r,n): (P/F,r,n) inf = e -rn Continuous Compounding i a = (1 + i) m – 1 = (1 + r/m) m – 1

44 i: effective interest rate per interest period (stated as a decimal) i: effective interest rate per interest period (stated as a decimal) n: number of interest periods n: number of interest periods P: present sum of money 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 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. Summary Notations

45 G: arithmetic gradient: uniform period-by-period increase or decrease in cash receipts or disbursements 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) i a : effective interest rate per year (annum) i a : effective interest rate per year (annum) m: number of compounding sub-periods per period Summary Notations

46 Single Payment 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 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) Summary Formulas

47 Arithmetic Gradient Formulas Arithmetic Gradient Formulas Present Worth,P = G {[(1+i) n – i n – 1]/[i 2 (1+i) n ]} = G (P/G,i,n) Uniform Series,A = G {[(1+i) n – i n – 1]/[i (1+i) n – i]} = G (A/G,i,n) Geometric Gradient Formulas 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) Summary Formulas

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