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CHAPTER-3

CHAPTER-3. Interest & Equivalence. Computing Cash Flows. Example -3.1 The manager has decided to purchase a new $30,000 mixing machine. The machine may be paid for in one of two ways: 1. Pay the full price now minus a 3% discount.

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CHAPTER-3

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  1. CHAPTER-3 Interest & Equivalence

  2. Computing Cash Flows Example -3.1 The manager has decided to purchase a new $30,000 mixing machine. The machine may be paid for in one of two ways: 1. Pay the full price now minus a 3% discount. 2. Pay $5000 now; at the end of one year, pay $8000; at the end of each of the next four years, pay $6000. List the alternatives in the form of a table of cash flows.

  3. Computing Cash Flows SOLUTION: The first plan represents a lump sum of $29,100 now, the second one calls for payments continuing until the end of the fifth year. Those payments total $37,00, but the $37,000 value has no clear meaning, since it has added together cash flows occurring at different times. The second alternative can only be described as a set of cash flows. Both alternatives are shown in the cash flow table, with disbursements given negative signs.

  4. Computing Cash Flows This can be presented as a cash flow diagram: 0 1 2 3 4 5 0 1 2 3 4 5 $5,000 $6,000 $6,000 $6,000 $6,000 $29,000 $8,000

  5. Computing Cash Flows Example – 3.2 A man borrowed $1000 from a bank at 8% interest. He agreed to repay the loan in two end-of-year payments. At the end of the first year, he will repay half of the $1000 principal amount plus the interest that is due. Compute the borrower's cash flow.

  6. Computing Cash Flows SOLUTION: In engineering economic analysis, we normally refer to the beginning of the first year as "time 0.“ At this point the man receives $1000 from the bank. (A positive sign represents a receipt of money and a negative sign, disbursement.) Thus, at time 0, the cash flow is +$1000. At the end of the first year, the man pays 8% interest for the use of $1000 for one year. The interest is 0.08 x $1000 = $80. In addition, he repays half the $1000 loan, or $500. Therefore, the end-of-year-I cash flow is -$580. At the end of the second year, the payment is 8% for the use of the balance of the principal ($500) for the one-year period, or 0.08 x 500 = $40. The $500principal is also repaid for a total end-of year-2 cash flow at -$540.

  7. (continued) Example – 3.2 The cash flow is: $1,000 0 1 2 $540 $580

  8. Example – 3.3 You have agreed to loan a friend $5000 for 5 years at a simple interest rate of 8% per year. How much interest will you receive from the loan'? How much will your friend pay you at the end of 5 years? SOLUTION: Total interest earned= P i n = ($5000) (0.08) (5 yr) = $2000 Amount due at end of loan = P + P i n = 5000 + 2000 = $7000

  9. Example – 3.4 To highlight the difference between simple and compound interest, rework Example 3.3 using an interest rate of 8% per year compound interest. How will this change affect the amount that your friend pays you at the end of 5 years? SOLUTION: Original loan amount (original principal) = $5000 Loan term = 5 years Interest rate charged =8% per year compound interest

  10. (Continued) Example- 3.4 In the following table we calculate on a year-to-year basis the total dollar amount due at the end of each year. Notice that this amount becomes the principal upon which interest is calculated in the next year (this is the Compounding effect). Total Principal(p) Interest owed at end of year Total amount Due at end of year year on which interest n from year n’s n ,new Total principal for year is calculated in year n total principal n+1 1 $5000 $5000 x 0.08 = 400 $5000+ 400 =5400 2 5400 5400 x 0.08 = 432 5400 + 432 = 5832 3 5832 5832 x 0.08 = 467 5832+ 467 = 6299 4 6299 6299 x 0.08 = 504 6299 + 504 = 6803 5 6803 6803 x 0.08 = 544 6803 + 544 = 7347

  11. (Continued) Example – 3.4 The total amount due at the end of the fifth year, $7347, is the amount that your friend will give you to repay the original loan. Notice that this amount is $347 more than the amount you received for loaning the same amount, for the same period, at simple interest. This, of course, is because of the effect of interest being earned (by you) on top of interest.

  12. Example – 3.5 If $500 were deposited in a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compounded annually. We can draw a diagram of the problem. Note: To have a consistent notation, we will represent receipts by upward arrows (and positive signs), and disbursements (or payments) will have downward arrows (and negative signs).] S OLUTION: From the view point of the person depositing the $500,the diagram for "today" (Time=0) through Year3 is as follows: F=? Reciept(+$) 0 1 2 3 Disbursements(-$) P=500 n=3 i=0.06

  13. (continued) Example – 3.5 We need to identify the various elements of the equation. The present sum P is $500 . The interest rate per interest period is 6%, and in 3 years there are three interest periods. The future sum F is to be computed from the formula F = P(1 + i)^n = 500(1 + 0.06)^3 = $595.50 where P = $500,i = 0.06, n = 3, and F is unknown. Thus if we deposit $500 in the bank now at 6% interest, there will be $595.50 in the account in three years. ALTERNATE SOLUTION: The equation F =P (I + i)^n need not be solved with a hand calculator. Instead, the single payment compound amount factor, (1 + i)^n, is readily determined from computed tables . The factor is written in convenient notation as (1 + i)^n = (F /P, i, n) and in functional notation as (F/P, 6%, 3) Knowing n = 3, locate the proper row in the 6% table. To find F given P, look in the first column, which is headed "Single Payment, Compound Amount Factor": for n =3,we find 1.191.

  14. (Continued) Example – 3.5 Thus, F = 500(F / P, 6%, 3) = 500(1.191) = $595.50 Before leaving this problem, let's draw another-diagram of it, this time from the bank's point of view. Receipts(+) P=500 i=0.06 n=3 0 1 2 3 Disbursements (-) This indicates the bank receives $500 now and must make a disbursement of F at the end of 3 years. The computation, from the bank's point of view, is F =500(F/P, 6%,3) =500(1.191) =$595.50 This is exactly the same as what was computed from the depositor's viewpoint, since this is just the other side of the same transaction. The bank's future disbursement equals the depositor's future receipt.

  15. Example – 3.6 If you wish to have $800 in a savings account at the end of 4 years, and 5% interest was paid annually, how much should you put into the savings account now? SOLUTION: F =$800 ; i = 0.05 ; n = 4 ; P =unknown P = F(1 + i)^-n= 800(1 + 0.05)^-4 P= 800(0.8227) =$658.16 Thus to have $800 in the savings account at the end of 4 years, we must deposit $658.16 now.

  16. (Continued) Example – 3.6 ALTERNATE SOLUTION: F=800 RECIEPTS (+) 0 1 2 3 P=? n=4 DISBURSEMENTS(-) i=0.05 P =F(P/F,i,n)=$800(P/F, 5%,4) From Compound interest tables: (P/F,5%,4)=0.8227 Hence, P =F(P/F,i,n)=$800(P/F, 5%,4) = $800(0.8227)=$658.16

  17. Example – 3.7 Suppose the bank changed their interest policy in Example 3-5 to "6% interest, compounded quarterly.“ For this situation, how much money would be in the account at the end of 3 years, assuming a $500 deposit now? SOLUTION: First, we must be certain to understand the meaning of ( 6% interest, compounded quarterly. There are two elements: 6% interest: Unless otherwise described , it is customary to assume that stated interest is for one year period, if the stated interest is for other than one year period, the time frame must be clearly stated. Compounded quarterly: This indicates there are four interest period per year; that is, an interest is 3 months long. We know that 6% interest is an annual rate because if it were for a different period, it would have been stated. Since , we are dealing with four interest periods per year, it follows that the interest rate per interest period is 1½ % . For the total 3 year duration, there are 12 interest periods.

  18. (Continued) Example – 3.7 Thus P = $500 ; i = 0.015 n=(4 x 3) = 12 ; F = unknown F = P(1 + i)^n= P(F / P, i, n) = $500(1 + 0.015)^12= $500(F / P,11/2%, 12) = $500(1.196) = $598 A $500 deposit now would yield $598 in 3 years.

  19. Example – 3.8 Consider the following situation: Year Cash Flow O +P 1 0 2 0 3 -400 4 0 5 -600 Solve for P assuming a 12% interest rate and using the compound interest tables. Recall that receipts have a plus sign and disbursements or payments have a negative sign. Thus, the diagram is: Reciept(+) P 0 1 2 3 4 5 Disbursements(-) SOLUTION: 400 P = 400(P/ F, 12%,3) + 600(P / F, 12%,5) 600 = 400(0.7118) + 600(0.5674) = $625.16 It is important to understand just what the solution, $625.16,represents.We can say that $625.16 is the amount of money that would need to be invested at 12% annual interest to allow for the withdrawalof $400 at the end of 3 years and $600 at the end of 5 years.

  20. (continued) Example – 3.8 There is another way to see what the $625.16 value of P represents. Suppose at Year0 you were offered a piece of paper that guaranteed you would be paid $400 at the end of 3 years and $600 at the end of 5 years. How much would you be willing to pay for this piece of paper if you wanted your money to produce a 12% interest rate? This alternate statement of the problem changes the signs in the cash flow and the diagram : Year Cash Flow O -P 1 0 2 0 3 +400 4 0 5 +600

  21. (Continued) Example – 3.8 Reciept(+) 400 600 0 1 2 3 4 5 P Disbursement(-) Since, the goal is to recover our initial investment P together with 12% interest per year. We can see that the P must be less than the total amount to be received in the future ($400+600=$1000) . We must calculate the present sum P that is equivalent at 12% interest, to an aggregate on $400 in 3years and $600 in 5years.

  22. (Continued) Example – 3.8 Since we have already derived the relationship p =F (1 + i) ^(-n ) we write p =400(1 + 0.12)^-3 + 600(1 + 0.12)^-5 =$625.17 This is virtually the same amount computed from the first statement of this example. [The slight difference is due to the rounding in the compound interest tables. For example, (1 + 0.12)-5 = 0.567427 ,but the compound interest table for 12% shows 0.5674.]

  23. Example – 3.9 The second set of cash flows in Example 3-8 was: Year Cash Flows 0 -P 1 0 2 0 3 +400 4 0 5 +600 At a 12% interest rate, P was computed to be $625.17. Suppose the interest rate is increased to 15%. Will the value of P be larger or smaller? SOLUTION: One can consider P as a sum of money invested at 15% from which one is to obtain $400 at the end of 3 years and $600 at the end of 5 years. At 12%, the required P is $625.17. At 15%, P will earn more interest each year., indicating that we can begin with a smaller P & still accumulate enough money for subsequent cash flows.

  24. (Continued) Example – 3.9 P =400 (P / F, 15'%, 3) + 600 (P/ F, 15%, 5) = 400 (0.6575) + 600 (0.4972) = $561.32 The value of P is smaller at 15% than at 12% interest

  25. EQUIVALENCE Repaying a Debt: • Plan 1 (Constant Principal) repays 1/n th of the principal each year. • Plan 2 (Interest only) only the interest due is paid each year, with no principal payment. • Plan 3 (Constant Payment) calls for five equal end-of-year payments. • Plan 4 (All at Maturity) no payment is made until the end of year 5, when the loan is completely repaid.

  26. EQUIVALENCE Plan – 1 Constant principal payment plus interest due

  27. EQUIVALENCE Plan – 2 Annual interest payment and principal payment at end of 5 years

  28. EQUIVALENCE Plan – 3 Constant annual payment * Calculations in Example 4-3, 1252.28 rounded off to even amount.

  29. EQUIVALENCE Plan – 4 All payment at end of 5 years

  30. Plan 1 Constant Principal At end of each year pay $1000 principal plus interest due

  31. Plan 2 Interest Only Pay interest due at end of each year and principal at end of 5 years

  32. Plan 3 Constant Payment Pay in five equal end-of-year payments

  33. Plan 4 All at maturity Pay principal and interest in one payment at end of 5 years

  34. EQUIVALENCE Pay principal and interest in one payment at end of 5 years

  35. EQUIVALENCE Plan 1 Plan 2 Plan 3 Plan 4

  36. EQUIVALENCE Plan 1 Plan 2

  37. Difference in Repayment Plans Shaded area = (Money owed in Year 1) (1 Year) Shaded area = (Money owed in Year 2) (1 Year) Shaded area = (Money owed in Year 3) (1 Year) Shaded area = (Money owed in Year 4) (1 Year) Shaded area = (Money owed in Year 5) (1 Year) Or Shaded area [(Money owed)(Time)] = dollar-years

  38. Difference in Repayment Plans The dollar-years for the four plans would be as follows:

  39. Equivalence is Dependent on Interest Rate • Interest rate increased to 9% • Each plan repay a sum less than $5,000 • Equivalent present sum at 9%

  40. Equivalence is Dependent on Interest Rate

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