Matakuliah : D0762 – Ekonomi Teknik Tahun : 2009

1. Matakuliah : D0762 – Ekonomi Teknik Tahun : 2009 Factors - Extra ProblemsCourse Outline 3

2. Outline next next next next • 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.44-99 • Engineering Economic Analysis, Donald G. Newman, p. 41-86 • Engineering Economy, William G. Sulivan, p.137-194, p. 135-140 2

3. 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

4. Problems – Single Payment 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 Define the symbol for the problems : • 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 805.26 discounted back five years at an annual rate of 10 %. • 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

5. Problems – Single Payment Answer Fn=3 = $400; Fn=5=$600, i =12%, P? P = (1+i)-3 + (1+i)-5 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

6. Problem-Single Payment F = Pern R = nominal interest rate = 0,05 N = number of years = 2 F = 2.000e(0,05x2) = 2000(1,1052) 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

7. Problem-Single Payment P = Fe-rn P = 5000e-(0,06x10)=5000(0,5488) =$2744 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

8. Uniform Series Given F, Find A Given A, Find F Given A, Find P Given P, Find A 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

9. Uniform Series Problem Answer Alternative 1 P = Fn=2 (P/F,i,2) + Fn=3(P/F,Ii,3) + Fn=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) = [Pn=2 (F/P,I,2) + Pn=3(F/P,I,3), Pn=4](P/F,i,4) Alternative 3 Find A, and then Find P P= [An=2,3,4(P/A,i,3) + Fn=2(P/F,i,2)](P/F,i,1) P = [20(P/A,15%,3) + 10(P/F,15%,2)](P/F,15%,1) • 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

10. Uniform Series Problem Answer A = P(A/P,5,4%) Answer Final payment = A + A(P/A,4%,2) 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

11. Problem – Uniform Series F=A(F/A,r,n) 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

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

13. Problem – Nominal & Effective Interest Answer : Nominal = 1 ¾ % x 12 month = 21% Effective = (1+i)m – 1 = (1+0.0175)12-1 = 0.2314 = 23,14%, 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

14. Problem – Nominal & Effective Interest Annual nominal = (51:2000) x 12 month = 0,306 ≈ 30,6% Effective : [1+(51:2000)]12-1 = 0,356 ≈ 35,6% • 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

15. Problem – Nominal & Effective Interest Bank A : F = P (1+i)n = 3000(1+5%)2 Bank B : F = P (1+i)n = 3000(1+(5%/3))6 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

16. Nominal : 6% Effective : er-1 = e0.06-1 = 6.18% • What are the nominal and the effective interest rate for 6% compounded continuously? 16

17. Continuous Compounding 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. 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 er - 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 ern is analogous to F = P (F/P,r,n): (F/P,r,n)inf= ern P = F e-rn is analogous to P = F (P/F,r,n): (P/F,r,n)inf= e-rn ia = (1 + i)m – 1 = = (1 + r/m)m – 1 17

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

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 Worth P = G {[(1+i)n – i n – 1]/[i2 (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 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) 19

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, 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)inf= e-rn (F/P,r,n)inf= ern 20