(3-4) Basic Concepts & Annual Compounding.pptx

Economic

1. Basic concepts Prepared by: engr. Owen francis a. maongat es 123 instructor

2. Objectives: To know the definition, the key concepts in engineering economics, principles of engineering economy, role of engineers in economic decision, and design process in an Engineering Economy

3. Basic concepts Interest Interest rate Simple interest Compound interest Time value of money Inflation Taxes cashflows

4. Basic concepts Interest It is a fee that is charged for the use of someone else's money. The size of the fee will depend upon the total amount of money borrowed and the length of time over which it is borrowed.

5. Basic concepts interest: Example

6. Basic concepts Interest rate it is the given amount of money is borrowed for a specified period of time (typically, one year), a certain percentage of the money is charged as interest. This percentage is called the interest rate.

7. Basic concepts interest rate: Example

8. Basic concepts 3. Simple interest It is defined as a fixed percentage of the principal (the amount of money borrowed), multiplied by the life of the loan. Thus, I = nip

9. WHERE: I = total amount of simple interest n = life of the loan I = interest rate (expressed as a decimal) P = principal It is understood that n and i refer to the same unit of time (e.g., the year). Normally, when a simple interest loan is made, nothing is repaid until the end of the loan period; then, both the principal and the accumulated interest are repaid. The total amount due can be expressed as F = P + I = P ( l + n i )

10. Basic concepts simple interest: Example F = P(1+ni) F = $3,000[1+2(0.055)] F = $3,330

11. Basic concepts 4. compound interest When interest is compounded, the total time period is subdivided into several interest periods (e.g., one year, three months, one month). Interest is credited at the end of each interest period, and is allowed to accumulate from one interest period to the next. Notice that F, the total amount of money accumulated, increases exponentially with n, the time measured in interest periods.  

12. Basic concepts compound interest: Example F = $2,012.20     Compound Interest Simple Interest   F = P(1+ni) F = $1000[1+12(0.06)] F = $1,720.00 Thus, the student's original investment will have more than doubled over the 12-year period

13. Basic concepts 5. Time value of money Since money has the ability to earn interest, its value increases with time. For instance, $100 today is equivalent to five years from now if the interest rate is 7% per year, compounded annually. We say that the future worth of $100 is $140.26 if i = 7% (per year) and n = 5 (years). Since money increases in value as we move from the present to the future, it must decrease in value as we move from the future to the present. Thus, the present worth of $140.26 is $100 if i = 7% (per year) and n = 5 (years)  

14. Basic concepts time value of money: Example P = $4,258.07     Compound Interest   The present worth of $5,000 is $4,258.07 if I = 5.5%, compounded annually, and n=3

15. Basic concepts 6. Inflation National economies frequently experience inflation, in which the cost of goods and services increases from one year to the next. Normally, inflationary increases are expressed in terms of percentages which are compounded annually. Thus, if the present cost of a commodity is PC, its future cost, FC, will be    

16. Basic concepts inflation: Example FC = $133.82    

17. Basic concepts inflation: Example   Where F is the future worth, measured in today’s dollars, of a present amount P OR   In a inflationary economy, the value (buying power) of money decreases as costs increase. Thus,

18. Basic concepts inflation: Example F = $74.76     Thus, $100 in five years will be worth only $74.73 in terms of today's dollars. Stated differently, in five years $100 will be required to purchase the same commodity that can now be purchased for $74.73.

19. If interest is being compounded at the same time that inflation is occurring, then the future worth can be determined by combining (1.3) and (1.5):     (1.3) (1.5)   Or, defining the composite interest rate,

20.   We have    

21. Basic concepts inflation: Example F = $74.76 A composite interest rate can be determined from this equation;     Substituting this value, we obtain  

22. Basic concepts 7. TAXES In most situations, the interest that is received from an investment will be subject to taxation. Suppose that the interest is taxed at a rate t, and that the period of taxation is the same as the interest period (e.g., one year). Then the tax for each period will be , so that the net return to the investor (after taxes) will be I’ = I – T = (1 – t) iP T = tiP

23. If the effects of taxation and inflation are both included in a compound interest calculation, may still be used to relate present and future values, provided the composite interest rate is redefined as  

24. Basic concepts taxes: Example F = $9,236.61 Let us assume that the engineer is able to invest the entire $10,000 in a savings certificate and that the tax bracket includes all federal, state, and local taxes. By; Substituting this value, we obtain      

25. Because of the combined effects of inflation and taxation, I3 is negative, and the engineer ends up with less real purchasing power after 15 years than he has today. (To make matters worse, the engineer will most likely have to pay taxes on the original $10 000, substantially reducing the amount of money available for investment.)

26. Basic concepts 7. Cash flows A cash flow is the difference between total cash receipts (inflows) and total cash disbursements (outflows) for a given period of time (typically, one year). Cash flows are very important in engineering economics because they form the basis for evaluating projects, equipment, and investment alternatives. The easiest way to visualize a cash flow is through a cash flow diagram, in which the individual cash flows are represented as vertical arrows along a horizontal time scale. Positive cash flows (net inflows) are represented by upward-pointing arrows, and negative cash flows (net outflows) by downward-pointing arrows; the length of an arrow is proportional to the magnitude of the corresponding cash flow. Each cash flow is assumed to occur at the end of the respective time period.

27. Basic concepts cash flows: Example

28. In a lender-borrower situation, an inflow for the one is an outflow for the other. Hence, the cash flow diagram for the lender will be the mirror image in the time line of the cash flow diagram for the borrower.

29. Prepared by: engr. Owen francis a. maongat es 123 instructor Annual compounding

30. 1. Single-payment, compound-amount factor     IT IS THE RATIO OF F/P

31. Annual compounding single-payment, compound-amount factor: Example F = $2,012.20 We wish to solve for F, given P , i , and n . Thus,      

32. 2. Single-payment, present-worth factor   IT is the reciprocal of the single-payment, compound amount factor:

33. Annual compounding single-payment, present-worth factor: Example P = $2,792.00 We wish to solve for P, given F , i, and n . Thus,      

34. 3. UNIFORM-SERIES, COMPOUND-AMOUNT factor Let equal amounts of money, A, be deposited in a savings account (or placed in some other interest-bearing investment) at the end of each year, as indicated in Fig. 2-1. If the money earns interest at a rate i , compounded annually, how much money will have accumulated after n years?

35. 3. UNIFORM-SERIES, COMPOUND-AMOUNT factor

36. 3. UNIFORM-SERIES, COMPOUND-AMOUNT factor  

37. Annual compounding UNIFORM-SERIES, COMPOUND-AMOUNT factor: Example F = $7,908.48 We wish to solve for F, given A , i, and n . Thus,      

38. 4. UNIFORM-SERIES, SINKING-FUND factor   IT is the reciprocal of the uniform-series, compound-amount factor:

39. Annual compounding UNIFORM-SERIES, SINKING-FUND factor: Example A = $7,908.48 We wish to solve for A, given F , i, and n . Thus,      

40. 5. UNIFORM-SERIES, CAPITAL-RECOVERY factor Let us now consider a somewhat different situation involving uniform annual payments. Suppose that a given sum of money, P, is deposited in a savings account where it earns interest at a rate i per year, compounded annually. At the end of each year a fixed amount, A, is withdrawn (Fig. 2-2). How large should A be so that the bank account will just be depleted at the end of n years?

41. 5. UNIFORM-SERIES, CAPITAL-RECOVERY factor

42. 5. UNIFORM-SERIES, CAPITAL-RECOVERY factor     IT IS THE PRODUCT OF UNIFORM-SERIES, SINKING-FUND factor and Single-payment, compound-amount factor

43. Annual compounding UNIFORM-SERIES, CAPITAL-RECOVERY factor: Example A = $6,795 We wish to solve for A, given P , i, and n . Thus,      

44. 6. UNIFORM-SERIES, PRESENT-WORTH factor     It is the reciprocal of the uniform-series, capital-recovery factor:

45. Annual compounding UNIFORM-SERIES, PRESENT-WORTH factor: Example P = $83,839 We wish to solve for P, given A , i, and n . Thus,      

46. 7. GRADIENT SERIES FACTOR IT is a series of annual payments in which each payment is greater than the previous one by a constant amount, G.

47. 7. GRADIENT SERIES FACTOR

48. 7. GRADIENT SERIES FACTOR

49. 7. GRADIENT SERIES FACTOR   IT is a series of annual payments in which each payment is greater than the previous one by a constant amount, G.  

50. Annual compounding gradient-SERIES factor We wish to solve for P, given Ao , G, i, and n . We first obtain a series withdrawals A’ equivalent to the series of gradients       A = $5,926

51. Annual compounding gradient-SERIES factor We can now calculate P as follows: A’ = Ao + A A’ = $5,000 + $5,926 A’ =$ 10,926     P = $106,116.59

52. Annual compounding gradient-SERIES factor A more concise, and the preferable way, to solve this problem is to write P = $106,116.59     P = Ao x (P/A, i %, n) + G x (A/G, i %, n) x (P/A, i %, n)