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Engineering Economics

Engineering Economics. Introduction. Why Engineering Economics?. Accreditation requirement in Ontario Engineers have to understand financial implications Communicate with the bean counters. Method of Instruction. 1 Lecture per week

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Engineering Economics

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  1. Engineering Economics Introduction

  2. Why Engineering Economics? • Accreditation requirement in Ontario • Engineers have to understand financial implications • Communicate with the bean counters

  3. Method of Instruction • 1 Lecture per week • Lecture PLUS (Participation, Learning, Understanding, Success) • Participation - in class • Learning - together • Understanding - individual • Success - individual A B C

  4. Demonstration - Lecture PLUSQuick Checks You are buying a new car and have three choices: Choice A B C Auto 57 Chevy 87 Honda 82 Mercedes Price ($) 12,000 7,000 20,000 Operation($) 200/mo. 50/mo. 150/mo. Resale($) 13,000 6,000 20,000 Which would you choose?

  5. Module 4 Outline • Intro to Engineering Economic Decisions • Read Chapter 1 of the text • Study the Chapter Summary (pp. 36) • The time value of money • Equivalence and Interest Formulas ( Chapter 2) • Real world examples (Chapter 3) • Analysis of independent Investments (Chapter 4)

  6. Why Pay Interest? • The borrower pays the lender for: • administrative costs associated with the loan • compensation for the risk of default • opportunity cost associated with not investing the money elsewhere

  7. Key Notation Interest ……………………………… I nominal interest rate……………… r effective interest rate (per period)… i Present Value (or Worth) ………… P Total Number of interest Periods…… N Future Value………….…………. F Annuity amount …………………….. A A discrete payment or receipt at the end of an interest period …………An

  8. Interest Rate Example(Quick Check) If you borrow $1000 from me and agree to pay the $1000 plus $125 extra at the end of one year, what is the interest rate? A. 8 % B. 12.5 % C. .125 %

  9. Simple Interest: Interest is paid only on the original principal amount, not on any accumulated interest In = Pin Fn = P + In = P(1+in) Rarely used

  10. Simple Interest:Example (QC) How much interest is due for a loan of $10,000 at 10% simple interest: 1) after 1 yr? I = Pin = 10000(.10)(1) = $1000 2) after 2 yrs? A) $1,000 B) $1,100 C) $2,000

  11. Compound Interest The interest accumulates interest F1= P(1+i) where F1 = future value an the end of period 1 F2= F1 (1+i) = P(1+i) (1+i) = P(1+i)2 F3= P(1+i)3 : Fn= P(1+i)n where, (1+i)n = (F/P,i,n),is called thecompound amount factor

  12. Compound Interest (QC) How much borrowed today will have to be repaid by $10,000 in three periods(i =10%) ? A) $9 000 B) $9 091 C) $7 513

  13. Compound Interest Factors • There are three categories of factor: • those that convert a single payment to a present or future amount • those that convert a series of payments to a present or future amount • those that convert a series of payments that increase or decrease by some constant amount G (for Gradient series) every period to an equivalent uniform series

  14. Factor Formula Name Description Converts a single Future amount to a Present amount Present worth Factor (P/F,i,n) Converts a single present amount to a single future amount Compound amount factor (F/P,i,n) Single Payment Factors

  15. Single Payment F i Yrs 1 2 3 4 5 P Two amounts P, F

  16. Example (QC) F=? i = 6% yrs 1 2 3 4 5 P=200 A. $267.65 B. $267.64 C. $268.00

  17. Uniform Series Payment Factors Factor Formula Name Description Converts a series of uniform payments to a single future amount Uniform series compound amount factor (F/A,i,n) Converts a single future amount into a series of equal payments that would be necessary to accumulate F in n periods Sinking fund factor (A/F,i,n)

  18. More Uniform Series Factors Factor Formula Name Description Converts a series of equal payments to a present amount Series present worth factor (P/A,i,n) Finds the series of payments necessary to recover (repay) a present amount in a fixed number of periods. Capital recovery factor (A/P,i,n)

  19. Uniform Series (Annuity) A A A A A yrs 5 0 1 2 3 4 i F Example, RRSP

  20. Uniform Series A A A A A yrs 5 1 2 3 4 i P Example - bank loan/mortgage

  21. Example - Loan repayment You take out a mortgage (150 000$) for a term of 20 yrs, with i=5.25% per yr. What are your annual payments? A A A A A 1 2 3 19 20 i = 5.25 P = 150 000

  22. Solution by formula A A A A A 1 2 3 19 20 i = 5.25 A = P (A/P,i,n) P = 150 000 A = 12 292,83$

  23. Solution by interpolation A = P (A/P,i,n) A = 150000(A/P,5.25,20) (A/P, 5, 20) = 0.0802 (A/P, 6, 20) = 0.0872 0.0872 ? (A/P, 5.25%,20) 0.0802 5.25% 5% 6% (A/P, 5.25%,20)= =0.08195 A = 150000(0.08195) = 12 292,50$

  24. Gradient present worth factor Factor Formula Description Converts an increasing or decreasing series of payments by a constant amount G to a single present value. (P/G,i,n) Arithmetic gradient to annuity conversion factor (A/G,i,n)

  25. Arithmetic Gradient 4G 3G 2G G Yrs 5 1 2 3 4 i P

  26. Example (QC) Find the Present value P=? i = 10% 7 500 150 500/yr 2000 A. $2 018 B. $3 804 C. $3 208

  27. Formula Summary • See pp. 94 • Complete Assignment 1(from Chapter 2)

  28. Chapter 2 Review Problems • The following questions from Chapter 2 are recommended: • Level 1: 2.1 to 2.8 (answers on pp. 905), 2.20 to 2.22, 2.33 to 2.35 • Level 2: 2.40 to 2.42, 2.50 to 2.51 • Level 3: 2.65, 2.67 a only

  29. Nominal and Effective Interest • Generally interest rates are quoted on an annual basis (annual percentage rate - APR); but the contract may specify that compounding occurs more frequently: • monthly • quarterly (every 3 months) • daily • etc

  30. Nominal Vs. Effective • What is the effect of compounding more frequently than once per yr? • Given: • r = nominal interest per yr (APR - annual percentage rate) • m= number of times per year (sub-periods) that interest is compounded • is = r/m = effective interest rate per period

  31. Example $10 000 is deposited in the bank at a nominal rate of 10% per year, compounding is quarterly.. Given: r = 10% m = 4 periods per yr. Therefore: is = r/m = 10/4 = 2.5% per quarter

  32. Quick Check Do you think the corresponding effective annual interest rate will be: A. Less B. Equal C. More than the nominal rate?

  33. Example Cont’d We have P=10,000, is=2.5% Therefore: F3mo=P(1+ is) = 10 000(1.025)= 10 250 F6mo = F3mo(1+ is)= 10250(1.025)= 10 506 F9mo = F6mo(1+ is)= 10 506(1.025) = 10 769 F12mo= F9mo(1+ is) = 11 038 Therefore the accumulated interest at the end of the yr = $1,038 which is more than the $1,000 That we would receive with simple interest The Effective annual rate : ie =(F-P)/P= (11038-10000)/10000 = 10.38%

  34. Formula for calculating ia for 1 yr ia = (1 + r/m)m -1 where ia = effective annual rate r = nominal interest rate per year m = number of compounding periods per yr Note: 1. This formula applies when compounding is more frequent than once per yr. 2. If m=1, ia = r 3. The factor (1+r/m)m increases with m,therefore the more frequent the compounding, the more interest will accumulate.

  35. Quick Check Choose between: A) borrowing at 12% compounded monthly B) borrowing at 13% compounded semi- annually C) borrowing at 11.5% compounded daily.

  36. Example Would you prefer to receive: $200 after 1 yr (case 1) $150 in 1 yr and another $100$ after 2 yrs (case 2)? The interest rate at the bank is 15% compounded monthly.

  37. Solution 200$ Case 1 i = 15/12 = 1.25% 12 0 P = 200 (P/F,1.25,12) = 172,30$ 150$ 100$ i = 15/12 = 1.25% Case 2 12 24 0 P = 150 (P/F,1.25,12) + 100(P/F,1.25,24) = $203.45 By calculating a single present amount that is equivalent to each cash flow we can compare the 2 cases directly.

  38. Quick Check A student borrowed $10,000 from the bank to buy a new car. The bank charges 10% compounded annually but give the student three repayment options. Which would you choose? End Yr Plan A Plan B Plan C 1 $2638 - $1800 2 2638 - 1800 3 2638 - 1800 4 2638 - 1800 5 2638 16105 6916 Total $13190 $16105 $14116

  39. Effective Interest rate – any payment period Effective interest are usually based on the payment period, but can Be calculated for any desired period. For example, if cash flow payments Occur quarterly, but interest is compounded monthly we may wish To calculate an effective quarterly interest rate. i = (1 + r/m)C -1 = (1 + r/CK)C -1 Where M=The number of compounding periods per year C= The number of compounding periods per payment period K=The number of payment periods per year Note that M=CK

  40. Example – effective rate per period Suppose you make quarterly deposits in a savings account which earns 9% interest compounded monthly. Calculate the effective interest rate per quarter. r=9%, C = 3 compounding periods per quarter K=4 quarterly payments per year, and M=12 compounding periods per year. Or, M=CK = 3x4=12 i = (1 + r/m)C -1 i = (1 +0.09/12)3 –1 = 2.27%

  41. Fourth Quarter First Quarter Second Quarter Third Quarter 0.75% 0.75% 0.75% 2.27% 0.75% 0.75% 0.75% 2.27% 0.75% 0.75% 0.75% 0.75% 0.75% 0.75% 2.27% 2.27% Example explained r = 9% compounded monthly, i=9%/12 = 0.75% i = (1 +0.09/12)3 –1 = 2.27% Effective interest rate per quarter ia = (1 + r/m)m –1 = (1+.0075)12-1 = (1+.0227)4-1 = 9.38%

  42. Equivalence and demand loans A demand loan is a loan that either the lender or borrower can decide to have terminated by immediate payment of the outstanding balance. In practice, borrowers often choose to pay off a loan in order to save on interest charges. Using the principal of equivalence, the remaining balance of a loan can be determined in two ways. Example: Julie borrowed $1 000 from the bank at 9% compounded monthly. She agreed to repay the loan in 6 monthly payments but since this is a demand loan, she can clear the balance at any time. Immediately following her second payment, Julie wins big in the lottery and she decides to clear off her debt. How much must she pay to clear the loan?

  43. Solution - Method 1 Payments: A = 1000(A/P,9/12,6) = 1000(.1711) = $171.10/mo We find an equivalent value of all preceding transactions just after the second payment: B2 = 1000(F/P,.75,2) - 171.1(F/A,.75,2) = 1000(1.015)-1.71(2.008) = $671.76 is the remaining balance after the second payment

  44. Solution - Method 2 We simply discount the remaining payments to the time at which the loan is to be paid off. I.e., we calculate the PW of the remaining payments. B2 =171.1(P/A,.75,4) =171.1(3.9261) = $671.76

  45. Loan Tables capital recovered 163.60 164.83 166.06 167.30 168.56 169.80 A balance 1 000.00 836.40 671.57 505.51 338.20 169.63 0.00 B payments 171.10 171.10 171.10 171.10 171.10 171.10 A(.0075) interest 7.50 6.27 5.04 3.79 2.54 1.27 End Yr 0 1 2 3 4 5 6

  46. Calculations involving compound interest • The frequency of payments is not always the same as the compounding period • Usually we have 3 of {P,A,F,i,n} and we have to find the forth • We generally assume that P occurs at the beginning of the first period (or at the end of period 0) and that F and A occur at the end of the period (end of period convention)

  47. F=1600 i=? n=6 P=1000 Example - Single Transaction You have $1000 available to invest. You also know that 6 years from now you have a requirement for $1600. What rate of return i is necessary achieve $1600?

  48. Solution - single transaction F=1600 i=? n=5 P=1000 F = P(F/P,i,n) therefore, (F/P,i,6) = F/P = 1.6 = (1+i)6  (1+i) = 1.6(1/6) = 1.08148 therefore, i = 0.08148 = 8.15%/yr Therefore, we require an internal rate of return of 8.15%/ yr. To meet the $1600 requirement.

  49. Multiple TransactionQuick Check In general, most projects have a combination of individual transactions and annuities. Example: An investment pays $10,000 immediately and $1,000 at the end of each yr for a period of 5 yrs and also has an individual payment of $2,000 at the end of the 5th year. If i=7%, what is the present worth of the transactions?

  50. Special Cases • There are 3 special cases that prevent the direct application of the compound interest factors: • Compounding is more frequent than payments • The interest rate is not constant for the whole period • Annuities occur at the start of the period

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