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Time value of money

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Time value of money

  1. 1. Contents • Interest: Cost of Money • Simple Interest & Compound Interest • Discounting & Compounding • Comparison of Different Money series by Economic Equivalence
  2. 2. 1. Interest: Cost of Money The Interest Rate reflects the market rate, which takes into account the earning power as well as the effect of inflation perceived in the market place. The rate at which money earn interest should be higher than the inflation rate to make any economic sense of the delayed purchase The economic value of money depends on when it is received; because money has both earning and purchasing power
  3. 3. Elements of Transactions Initial Amount of Money (Principal) Time Period (Interest Period) Cost of Money (Interest Rate) Number of Interest Periods Plan (Cash Flow Pattern) Future Amount of Money
  4. 4. Elements of Transactions An = A discrete amount occurring at the end of some interest period i = The interest rate for a period N = Total number of interest periods P = Initial amount at time zero, referred as Present Value (Present Worth) F = Future some of money at the end of period A = An end-of-period payment in a uniform series that continues for N periods; where A1 = A2 = ----- = AN Vn = An equivalent amount of money at the end of a specific period ‘n’ that considers the effect of time value of money; that is, V0 = P and VN = F
  5. 5. Cash Flow Diagram • Cash Flow Diagrams give a convenient summary of all the important elements of a problem in a graphical form to determine whether the statement of the problem has been converted into its appropriate parameters. • Cash flow diagram represents time by a horizontal line marked off with the number of interest periods. • Upward arrows denote positive flows (receipts) and downward arrows negative flows (payments). Arrows represents net cash flows. • End of period convention is the practice of placing all cash flow transactions at the end of an interest period.
  6. 6. Cash Flow Diagram Period-1 Period-2 Period-3 Period-4 Period-5 Outflows Inflow Period-0
  7. 7. 2. Methods of Calculating Interest 1. Simple Interest Total amount available in N periods F = P+I = P(1+iN) For a deposit of P at a simple interest rate i for N periods; Total Interest I = (iP)N Interest earned only on the principal amount during each interest period 2. Compound Interest Total amount at the end of N periods For a deposit of P at interest rate i, total amount at end of period-1 P+iP = P(1+i) Interest earned in each period is calculated based on total amount at end of previous period. Total amount = principal amount + accumulated interest
  8. 8. Illustration Question If you deposit SR 1000 in a bank savings account that pays interest rate at a rate of 10% compounded annually. Assume that you don’t withdraw the interest earned at the end of each period (one year), but let it accumulate. How much money would you have at the end of year 3? Compare your answer with the simple interest.
  9. 9. Answer P = 1000 N = 3 i = 10% = 0.10 Simple Interest F = P(1+iN) = 1000 [1+(0.10)3] = 1000 x 1.3= 1300 Compound Interest  N iPF )1( 3 )10.01(1000  3 )1.1(1000 )331.1(1000 1331 Comparison of Interest = 1331-1300 = SR31 more under Compound Interest
  10. 10. 3. Economic Equivalence  N iPF )1( 5 )12.01(1000  34.1762SR At 12% interest, SR1000 received now is equivalent to SR1762.34 received in 5 years
  11. 11. Principles of Equivalence Calculation 1. Equivalence calculations made to compare alternatives require a Common Time basis (Present worth of future worth) 2. Equivalence depends on Interest Rate. 3. Equivalence calculations may require the conversion of multiple payment cash flows to a Single Cash flow 4. Equivalence is maintained regardless of point of view (borrower or lender)
  12. 12. Five Types of Cash Flows Years 0 1 2 3 4 5 Years 0 1 2 3 4 5 Years 0 1 2 3 4 5 Years 0 1 2 3 4 5 Years 0 1 2 3 4 5 Single Cash Flow Series Equal (Uniform) Payment Series Linear-gradient Series Geometric- gradient Series Irregular Payment Series
  13. 13. Types of Cash Flows Cash flow Series Payment Characteristics 1. Single Cash Flow Series Deal with only Single amount: Present amount P & its future worth F 2. Equal (Uniform) Payment Series Equal cash flows at regular intervals 3. Linear-gradient Series Cash flows increases (decreases) at uniform fixed amount 4. Geometric-gradient Series Cash flow increases (decreases) at uniform fixed rate expressed as percentages 5. Irregular Payment Series Cash flows are irregular
  14. 14. 4.Equivalence in Single Cash Flow 0 N 0 N P P F F Compounding Process Discounting Process
  15. 15. 4.1 Compounding Process The process of finding F is called Compounding Process N i)1(  is called Compounding Factor (Table Values)Where Functional Notation for Single payment Compound-amount Factor ),,/( NiPF which is read as ‘Find F, given P, i, and N’
  16. 16. 4.2 Discounting Process Finding the Present Worth of a future sum is simply the reverse of compounding and is known as the Discounting Process N N iFP i FP           )1( )1( 1 N iWhere   )1(, is known as Present worth factor Or Discounting factor (Table Value) Functional Notation for Single payment Present Worth Factor (Discounting Factor) ),,/( NiFP which is read as ‘Find P, given F, i, and N’
  17. 17. Illustration Question: 1 If you had SR 2000 now and invest it at 10%, how much would it be worth in eight years?
  18. 18. Answer to Question: 1 P = SR 2000 i = 10% = 0.10 N = 8 years 18.4287 )10.01(2000 )1( 8 SRF F iPF N   
  19. 19. Illustration Question: 2 Suppose that SR 1000 is to be received in five years. At an annual interest rate of 12%, what is the Present Worth of this amount?
  20. 20. Answer to Question: 2 F = SR 1000 i = 12% = 0.12 N = 5 years 43.567 )56743.0(1000)12.01(1000 )1( 5 SRP P iFP N     
  21. 21. Illustration Question: 3 • You have just purchased 100 shares of SABIC at SR 60 per share. You will sell the stock when its market price has doubled. If you expect the stock price to increase 20% per year, how long do you anticipate waiting before selling the stock?
  22. 22. Answer to Question: 3 Share price = SR60 Number of shares = 100 P = 60 x 100 = SR 6000 F = SR 12000 (Doubled) i = 20% = 0.20 yearsN N N iPF N N N N 480.3 20.1log 2log 20.1log2log 20.12 20.1 6000 12000 )20.01(600012000 )1(       
  23. 23. 5. Equivalence in Uneven payment (Irregular) Series We can find the Present Worth of any uneven stream of payments by calculating the present value of each individual payments and summing the results Once the Present Worth if found, we can calculate other Equivalence Calculations
  24. 24. Illustration Question: 1 If your business wishes to set aside money now to invest over the next 4 years to use to automate its customer service department. The business can earn 10% on a lump sum deposited now, and it wishes to withdraw the money in the following increments. Year-1: SR 25,000 to purchase a computer and database software designed for customer service use. Year-2: SR 3,000 to purchase additional hardware to accommodate anticipated growth in use of the system. Year-3: No expenses Year-4: SR 5,000 to purchase software upgrades How much money must be deposited now to cover the anticipated payments over the next 4 years? (Assume each transaction occur at the end of each year)
  25. 25. Answer to Question: 1 (Cash Flow Diagram) 0 1 2 3 4 SR 25,000 SR 3,000 SR 5,000 P P1 P2 P4 P = P1 + P2 + P3 + P4
  26. 26. Answer (Continues) F1 = SR 25,000 F2 = SR 3,000 F3 = SR 5,000 P = P1 + P2 + P3 + P4 622,28415,30479,2727,224321 415,3)6830.0(5000)10.1(5000)10.01(50004 03 479,2)8264.0(3000)10.1(3000)10.01(30002 727,22)9091.0(25000)10.1(25000)10.01(250001 )1( 44 22 11 SRPPPPP SRP P SRP SRP iFP N          
  27. 27. Illustration Question: 2 Suppose that you have a savings account in a Bank. By looking at the history of the account, you learned the interest rate in each period during the last five years as shown in the cash flow diagram below. Show how the bank calculated your balance in the fifth year? 0 1 2 3 4 5 SR 500 SR 300 SR 400 F = ? 5% 6% 6% 4% 4%
  28. 28. Answer to Question: 2 P1 = SR 300; i = 5% (N=1)+ 6% (N=2)+4% (N= 2) P2 = SR 500; i = 6%(N=1) + 4% (N=2) P3 = SR 400; i = 4% (N=1) 82.382)0816.1(934.353)04.1(934.353)04.01(934.353 )934.353&2%(4 934.353)1236.1(315)06.1(315)06.01(315 )315&2%(6 315)05.01(300 )1%(5 300 )1( 22 22 1         PNiWhen PNiWhen NiWhen SRofonContributi iPF N
  29. 29. Answer Continues 25.573)0816.1(530)04.1(530)04.01(530 )530&2%(4, 530)06.1(500)06.01(500 )500&1%(6 500 )1( 22 1 SR PNiWhen PNiWhen SRofonContributi iPF N       416)04.1(400)04.01(400 )400&1%(4, 400 )1( 11 SR PNiWhen SRofonContributi iPF N     Total Balance after 5 years = 382.82+573.25+416 =SR1,372.06
  30. 30. 6. Equivalence in Equal Payment Series If an amount A is invested at the end of each period for N periods, the total amount F that can be withdrawn at the end of the N periods will be the sum of the compound amounts of the individual deposits. This means there exists a series of the following form. :& )4()1()1( :lim32 )3()1(..............)1()1()1( :)1(1 )2()1(.............)1()1(( exp1 )1()1(...........)1()1( 2 12 21 givewillAFforSolving iAAFiF gettotermscommoninateetoEquationfromEquationgSubtractin iAiAiAFi ibyEquationgMultiplyin iAiAiAAF followsaselyalternativressedisEquation AiAiAiAF N N N NN          
  31. 31. Compound-amount Factor & Sinking- fund Factor in Equal Payment Series )5( 1)1(         i i AF N The term in the bracket is known as Compound-amount Factor in Equal Payment Series )6( 1)1(         N i i FA The term in the bracket is known as Sinking –fund Factor in Equal Payment Series
  32. 32. Capital-recovery Factor (Annuity factor) & Present-worth factor The term in the bracket is knows as Capital-recovery Factor & A is Annuity Factor )8( 1)1( )1( )7( 1)1( )1( :,)1(6Re                   N N N N N i ii PA OR i i iPA getweiPbyEquationinFforplacing )9( )1( 1)1( :8           N N ii i AP givesPforsolvingEquationIn The term in the bracket is knows as Present-worth Factor
  33. 33. Illustration Question: 1 Suppose you make an annual contribution of SR 3000 to your savings account at the end of each year for 10 years. If the account earns 7% interest annually, how much can be withdrawn at the end of 10 years? 0 1 2 3 4 5 6 7 8 9 10 SR 3000 i=7% Cash Flow Diagram F=?
  34. 34. Answer to Question: 1 A = SR 3000 N = 10 years i = 7% = 0.07 20.449,41)8164.13(3000 07.0 1)07.1( 3000 07.0 1)07.01( 3000 1)1( 10 10 SRF F i i AF N                         
  35. 35. Illustration Question: 2 A Biotechnology Company has borrowed SR 250,000 to purchase laboratory equipments. The loan carries an interest rate of 8% per year and is to be repaid in equal installments over the next 6 years. Compute the amount of annual installment 0 1 2 3 4 5 6 SR 250,000 i= 8% A= ?
  36. 36. Answer to Question: 2 P = SR 250,000 i = 8% = 0.08 N = 6 years   54075 2163.0000,250 1)08.1( )08.1(08.0 000,250 1)08.01( )08.01(08.0 000,250 1)1( )1( 6 6 6 6 SRA i ii PA N N                           
  37. 37. Illustration Question: 3 Suppose you are joining in an insurance scheme, where you have to deposit an amount now, so that you will receive an annual amount of SR 10,576,923 for next 26 years at an interest rate of 5%. Estimate the initial deposit amount? 0 1 2 ----------------- 25 26 A = SR 10,576,923 i= 5% P= ?
  38. 38. Answer to Question: 3 i = 5% = 0.05 A = SR 10,576,923 N = 26 years 383,045,152 )3752.14(923,576,10 )05.1(05.0 1)05.1( 923,576,10 )05.01(05.0 1)05.01( 923,576,10 )1( 1)1( 26 26 26 26 SRP P P ii i AP N N                           
  39. 39. 7. Linear-Gradient Series A linear gradient includes an initial amount A during period-1, which increases/ decreases by G during the interest periods. Each Cash Flow consists of a Uniform series of N payments of amount A1 and Gradient increment/ decrement of G amount 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 + -- = = Increasing Gradient Decreasing Gradient A1 A1 G 2G 3G 4G G 2G 3G 4G A1 A1 A1 -- G A1 + G A1 + 4G A1 -- 4G
  40. 40. Present worth Factor (Discounting) Derivation To find P, the single payment Present worth Factor is applied to each term :,&Re )3( )1( )1(1 2 )1( )1(1 )1(20 })1(,.......,2,,0{ )2()1(20[ )1(20 )1/(1 )1()1()1( )1( )1( )1( 2 )1( 0 2 1 2 2 1 12 12 12 32 1 32 obtainwexAforvaluesoriginalplacing x xNNx axP asrewrittenisEquation x xNNx xxNXx sumfinitethehasxNxxseriesgeometricarithmetictheSince xNxxax axNaxaxP yieldsxiandaGLetting iGnP OR i GN i G i G P NN NN N N N N N n n N                                        
  41. 41. Present Worth Factor The factor in the bracket is called as Present Worth factor in the gradient series.          N N ii iNi GP )1( 1)1( 2
  42. 42. Annuity Factor (A) The figure in the bracket is known as Gradient uniform series factor             ]1)1[( 1)1( : N N ii iNi GA obtaintoseriespayment equalinfactorAnnuity invaluesPngsubstitutiBy
  43. 43. Illustration Question: 1 A textile mill has just purchased a lift truck that has a useful life of 5 years. The engineer estimates that maintenance cost of for the truck during the first year will be SR 1000. As the truck ages, maintenance costs are expected to increase at a rate of SR 250 per year over the remaining life. Assuming that the maintenance costs occur at the end of each year. The firm wants to set up a maintenance account that earns 12% annual interest. All future maintenance expenses will be paid out of this account. How much does the firm have to deposit in the account now?
  44. 44. Answer 0 1 2 3 4 5 1500 1750 2000 1250 1000 P=? i=12% 1000 1000 1000 1000 250 500 750 1000
  45. 45. Answer Continues A1 = 1000 G = 250 i = 12% = 0.12 N = 5 P = P1 (Equal payment series) + P2 (Gradient Series) 8.36041 )6048.3(10001 )12.1(12.0 1)12.1( 1000 )12.01(12.0 1)12.01( 10001 )1( 1)1( 1 5 5 5 5 SRP P P ii i AP N N                            520425.15998.360421 25.1599)397.6(2502 )12.01(12.0 15*12.0)12.01( 250 )1( 1)1( 2 52 5 2                    PPP P ii iNi GP N N
  46. 46. Illustration Question: 2 Mr. X and Mr. Y have just opened two savings accounts in a Bank. The accounts earn 10% annual interest. Mr. X wants to deposit SR 1000 in his account at the end of the first year and increase this amount by SR 300 for each of the next 5 years. Mr. Y wants to deposit an equal amount each year for the next 6 years. What would be the size of Mr. Y’s annual deposit so that the two accounts will have equal balances at the end of 6 years. 0 1 2 3 4 5 6 1000 1000 1000 1000 1000 1000 300 600 900 1200 1500 1300 1600 1900 2200 2500 Mr. X’s Deposit Plan 1000
  47. 47. Answer A1 = 1000 G = 300 i = 10% = 0.10 N = 6 A = A1 (Equal payment ) + A2 (Gradient payment) A1 = 1000 08.166708.667100021 08.667)2236.2(3002 ]1)10.1[(10.0 16*10.0)10.1( 300 ]1)10.01[(10.0 16*10.0)10.01( 3002 ]1)1[( 1)1( 6 6 6 6                             AAA A A ii iNi GA N N
  48. 48. 8. Geometric Gradient Series If ‘g’ is the percentage change in cash flow, the magnitude of nth payment ‘An’ is related to the first payment A1 by the following formula Cash flows that increase or decrease over time by a constant percentage (Compound Growth) decreasewillseriesthegincreasewillseriesthegIf NngAA n n    ,0&,0 )1(.,,.........3,2,1,)1( 1 1 
  49. 49. A1 P P A1 (1+g) A1 A1 (1+g) Geometric Gradient Series 0 1 2 3 4 5 0 1 2 3 4 5 2 1 )1( gA  1 1 )1(   N gAincreasewillseriesthegWhen  ,0 decreasewillseriesthegIf  ,0 2 1 )1( gA  1 1 )1(   N gA
  50. 50. Estimation of P : )6()1( 1 )( )()1( )( :45 )5()( :4 )4()( :3. 1 1 1 )3( 1 1 )1( :)1( )2()1()1( 1 1 1 1432 32 1 1 1 1 1 1 1 1 xandaforvaluesoriginalreplacingBy x x xxa P xxaxP xxaxPP equationfromequationgSubstratin xxxxaxP gettoxbyequationMultiply xxxxaP asequationrewriteThen i g xand g A aLet i g g A P yieldssummationtheoutsidegABringing igAP N N N N N nN n N n nn                                       
  51. 51. Formulae of P                     N i g gi A P giIf 1 1 11 i NA P giIf    1 1 The factor in bracket is known as Geometric-gradient series present-worth factor
  52. 52. Illustration Question Ansell, Inc., a medical device manufacturer uses compressed air in solenoids and pressure switches in its machines to control various mechanical movements. Over the years, the manufacturing floor has changed layouts numerous times. With each new layout, more piping was added to the compressed-air delivery system to accommodate new locations of manufacturing machines. None of the extra, unused old pipe was capped or removed; thus the current compressed-air delivery system is inefficient and fraught with leaks. Because of the leaks, the compressor is expected to run 70% of the time that the plant will be in operation during the upcoming year. This will require 260kWh of electricity at a rate of SR 0.05/kWh for 6000 hours. The Plant runs 250 days a year, 24 hours per day. If Ansell continues to operate the current air delivery system, the compressor run time will increase by 7% per year for the next 4 years because of ever-worsening leaks (After 5 years, the current system will not be able to meet the plant’s compressed air requirements, so it will have to be replaced). If Ansell decides to replace all of the old piping now, it will cost SR 28,570. The compressor will still run the same number of days; however, it will run 23% (or will have 70% (1-0.23) = 53.9% usage during the day) less because of the reduced air pressure loss. If Ansell’s interest rate is 12%, is the machine worth fixing now? (Calculate A1 and P)
  53. 53. Answer Current Power Consumption (g) = 7% i = 12% = 0.12 N = 5 Step: 1 Calculate the cost of power consumption of current piping for 1st year Power Cost = (% of operating days) (Days operating per year) (Hours per day) (kWh) ($/KWh) = (70%) (250) (24) (260) (0.05) = SR 54,600
  54. 54. 54,600 Step: 2 Each year, the annual power cost increase @ 7%. The anticipated power cost over 5 years is as shown in the diagram below. The equivalent present lump-sum cost @ 12% for this geometric gradient series is: 0 1 2 3 4 5 58,422 62,512 66,887 71,569 g=7% 222937 12.1 07.1 1 05.0 54600 )( 12.01 07.01 1 07.012.0 54600 1 1 1)( 5 5 1 SRoldP i g gi A oldP N                                                      
  55. 55. Step: 3 If Ansell replaces the current compressed air system with the new one; the annual power cost will be 23% less during the 1st year and will remain at that level over the next 5 years. The equivalent Present lump-sum cost at 12% is: A = 54600 (1- 23%) = 54600 (1 - 0.23) = 54600 (0.77) = SR42042 552,151 )6048.3(42042 )12.01(12.0 1)12.01( 42042 )1( 1)1( 5 5 SR ii i AP SeriesPaymentEqualIn N N                    
  56. 56. Step: 4 The net cost for not replacing the old system = = 222,937 – 151,552 = SR 71,385 Since the new system costs only SR28,570, the replacement should be made now

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