Matakuliah: D0762 – Ekonomi Teknik Tahun: 2009 Factors - Extra Problems Course Outline 3

Outline Problems - Single Payment (P and F) Problems - Uniform Series Problems – Gradient Problems – Nominal and Effective Interest References : -Engineering Economy – Leland T. Blank, Anthoy J. Tarquin p Engineering Economic Analysis, Donald G. Newman, p Engineering Economy, William G. Sulivan, p , p next

Mathematical Factor Single Payment Compound Formula If you put P in the bank now at an interest rate of i% for n years, the future amount you will have after n years is given by F = P (1+i) n The term (1+i) n is called the single payment compound factor. The factor is used to compute F, given P, and given i and n. Handy Notation. (F/P,i,n) = (1+i) n F = P (1+i) n = P (F/P,i,n). Single Payment Present Worth Formula P = F/(1+i) n = F(1+i) -n 3

Problems – Single Payment Define the symbol for the problems : 1.What amount would have to be invested for five years, earning an annual interest rate of 10 %, to have $805.26? In other words, what is the present value of discounted back five years at an annual rate of 10 %. 2.If you put the money $45,000 at the Bank right now, earning 10% of annual interest, what would you have in 10 years? 4 answer no. 1 -F = $805.26, n = 5, I 10%, P? ; P = F/(1+i) n = F(1+i) -n Answer no.2 P = $45,000, n =10years, i = 10%, F? F = P (1+i) n

Problems – Single Payment 3. Suppose at year 0 (now) you offered a piece of paper that guaranteed you would be paid $400 at the end of three years and $600 at the end of five years. How much would you be willing to pay for this piece of paper if you wanted your money to produce a 12% interest? 5 Answer F n=3 = $400; F n=5 =$600, i =12%, P? P = (1+i) -3 + (1+i) -5

Problem-Single Payment If you were to deposit $2000 in a bank that pays 5% nominal interest, compounded continuously, how much would be in account a the end of two years? t6 F = Pe rn R = nominal interest rate = 0,05 N = number of years = 2 F = 2.000e (0,05x2 ) = 2000(1,1052)

Problem-Single Payment A Bank offers to sell saving certificates that will pay the purchaser $5000 at the end of ten years but will pay nothing to the purchaser in the meantime. If interest is computed at 6%, compounded continuously, at what price is the bank selling the certificate? 7 P = Fe -rn P = 5000e -(0,06x10) =5000(0,5488) =$2744

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] Uniform series present worth: (P/A,i,n) = [(1+i) n – 1]/[i (1 + i) n ] 8 Given F, Find A Given A, Find F Given A, Find P Given P, Find A

Uniform Series Problem If you will receive $20 in the end year two, $30 in the end of year three, and $20 in the end of year 4, compute the cash flow value in year 0 (annual interest = 15%) 9 Answer Alternative 1 P = F n=2 (P/F,i,2) + F n=3 (P/F,Ii,3) + F n=4 (P/F,i,4) Alternative 2 First - Find total F for the cash flow and then Find P P = Total F (P/F,i,4) = [P n=2 (F/P,I,2) + P n=3 (F/P,I,3), P n=4 ](P/F,i,4) Alternative 3 Find A, and then Find P P= [A n=2,3,4 (P/A,i,3) + F n=2 (P/F,i,2)](P/F,i,1) P = [20(P/A,15%,3) + 10(P/F,15%,2)](P/F,15%,1)

Uniform Series Problem If company agrees to pay a machine for $12,000 in five equal annual payment, what would be the payment each year? (i=4%) If immediately after the second payment, the terms of the agreement are changed to allow the balance due to be paid in a single payment the next year, what is the final single payment? 10 Answer A = P(A/P,5,4%) Answer Final payment = A + A(P/A,4%,2)

Problem – Uniform Series A man deposited $500 per year into a credit union that paid 5% interest, compounded annually. At the end of five years, he had $2763 in the credit union. How much would he have if they paid 5% interest compounded continuously? 11 F=A(F/A,r,n)

Problem - Gradient Find the present worth value of the cash flow on the tables below : Table 1 Table 2 Table 3 12 year Value year Value year Value

Problem – Nominal & Effective Interest A department store charges 1 ¾ % interest per month, compounded continuously, on its customer’s charge accounts. What the nominal interest rate? What is the effective interest rate? 13 Answer : Nominal = 1 ¾ % x 12 month = 21% Effective = (1+i) m – 1 = ( ) = = 23,14%,

If someone borrow $2000 and repay $51 for the next fifty months, beginning thirty days after receiving the money- compute the nominal annual interest for this loan. What is the effective interest rate? 14 Problem – Nominal & Effective Interest Annual nominal = (51:2000) x 12 month = 0,306 ≈ 30,6% Effective : [1+(51:2000)] = 0,356 ≈ 35,6%

Problem – Nominal & Effective Interest If bank A offers 5% interest, compounded annually for deposit, and bank B pays 5% interest compounded quarterly, and you have $3000 and want to put the in a savings account and leave the money for two year, compare how much additional interest would you obtain from bank A and bank B? 15 Bank A : F = P (1+i) n = 3000(1+5%) 2 Bank B : F = P (1+i) n = 3000(1+(5%/3)) 6

What are the nominal and the effective interest rate for 6% compounded continuously? 16 Nominal : 6% Effective : e r -1 = e = 6.18%

Continuous Compounding 17 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 above illustrates that e r - 1 is a good approximation of (1 + r/m) m for large 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 i a = (1 + i) m – 1 = = (1 + r/m) m – 1 We will pay little attention to continuous compounding in this course. You are supposed to read the material on continuous compounding in the book, but it will not be included in the homework or tests.

18 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) i a : effective interest rate per year (annum) m: number of compounding sub-periods per period

19 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 WorthP = G {[(1+i) n – i n – 1]/[i 2 (1+i) n ]} = G (P/G,i,n) Uniform SeriesA = G {[(1+i) n – i n –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)

20 Summary: Formulas Nominal interest rate per year, r: the annual interest rate without considering the effect of any compounding Effective interest rate per year, i a : i a = (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) inf = e -rn (F/P,r,n) inf = e rn