Chap2 Compound Interest and Compound Discount

Chap 2 Compound Interest and Compound Discount § 2.1 Fundamental Compound Interest Formula ① P ② I ③ Original principal, present value of S, discounted value of S Compound Interest: interest added to principal at the end of each interest period thereafter earns interest S Compound amount accumulated value: the compound amount of P, or the accumulated value of P ④ interest period: time between two successive interest computation Conversion period: time unit, need not be a year Payable quarterly, compound semi-annually, convertible monthly Example1 Find the compound interest earned on $100 for 2 years at 10% compounded semi-annually and compare with the simple interest earned on $100. Solution conversion period: 6 months , rate=10% , 2 years = 4  6 = 24 months,  there are 4 interest periods in 2 years. At the end of 1st interest period 2nd …………………. Compound Interest 1000  0.05=50 1050  0.05=52.50 Accumulated value 1000+50=1050 1050+52.5=1102.5 3rd ………………… 1102.5  0.05=55.13 1102.5+55.13=1157.63 th 4 ………………….. 1157.631102.5=57.88 1157.63+57.88=1215.51  Compound interest on $100 for 2 years at 10% compounded semi-annually is $215.51 if simple interest rate r=10% ,then I=1000  0.1  2=$200 since 215.51 > 200, compound is better. Notation P: principal, present value, discounted value of S S; Compound amount, accumulated value, future value of P n: number of interest (or conversion) periods involved m: number of interest periods per year ,frequency of compounding j m : nominal (yearly) interest rate that is compounded (payable, convertible) m times per year, i m also i: interest rate per conversion(compound period) Note 1. m=1,compound annually; m=2, semi-annually; m=4, quarterly; m=12,monthly; m=52, weakly; m=365 daily ( leap year or not) 2. j i  m , used in the compound interest calculation m  9% (compound monthly), i  12 j j 12 9% 3   %  0.075  0.75% 12 12 4 p. 20 i.e. i=0.75% is the interest rate per month . e.g. let P: Original principal , i: interest rate per conversion period n: # of periods Find S for n periods interest due accumulated value S end of 1st period … … 2nd …….. 3rd… I  Pi P(1  i)i P(1  i ) 2 i P(1  i ) n 1 i P  Pi P(1  i )  P(1  i )i  P(1  i) 2 P(1  i )2  P(1  i) 2 i  P(1  i)3 ………nth….. P (1  i ) n  fundamental compound interest formula S  P (1  i ) n Note When P=$1, S=1. (1  i )  (1  i ) n n accumulation factor /accumulated value of $1, Example 2 Accumulate $100 at jn  12% for (a) t= 5 years (b) t=25 years Solution:( a) P=100, i= n j12 12%   1%  0.01 , n=5.12=60. 12 12 60 S= P(1  i )  100(1  0.01) (b) P=100,i=  $181.67 j12 12%   1%  0.01 ,n=25.12=300 12 12  S= P(1  i ) n  100(1  0.01)300  $1978.85 Compound with simple interest: I = Prt =100(0.12)(25)= $300 Using the table to find Growth of $100 at compound interest rate j12 Example 3 A person deposits $100 into a savings account that earns interest at 4.25% compound daily How much interest will be earned: (a) during the first year (b)2 nd year Solution: ( a) P=100, j365  4.25% , i = j365 4.25%  , n = 365 365 365  S= P(1  i)n  100(1  ( 0.0425 365 ))  $1043.41 365 The compound interest in first year = $43.41 (b) P=100, j365  4.25% , i= j365 4.25%  , n = 365  2 = 730 365 365 0.0425 730 ))  $1088.71  S = P(1  i)n  100(1  ( 365 Compound interest earned during the second=$1088.71-$1043.41=$45.30 45.30 > 43.41  第二次利息更多 p. 21 Note using all digits ,don’t round on the value of I § 2.2 Equivalent Compound Interest Rates Nominal interest rate: jm must have m-frequency of conversion m = 1, 2, 4, 12, 52, 365 yearly , semi-annually, quarterly, monthly, weekly, daily. let P=$10000 will accumulate in 10 years Frequency of conversion Rate per period number of periods m=1 m=2 m=4 m=12 m=365 Note Amount $21,589.25 $21,911.23 $22,080.40 $22,196.40 $22,253.41 i  j1  8% 10  1 =10 10  2 =20 10  4 =40 10  12=120 10  365=3650 j1 8%   4% 2 2 j 8% i 1   2% 4 4 j 8% i 1  12 12 j 8% i 1  365 365 i at the same j1  8% , the accumulated value S depends on the m-frequency of conversion increasing n S = S (m) since S  P(1  i )  P(1  ( jm mt )) ,where P , jm ,t is fixed (here t=10) m jm ,define a rate j,if compound annually, will produce m S is function of m, m 上升，S 上升。是增函数。 Annual effective rate j: for given jm , i  the same amount of interest per year. Let P=$1,compare the accumulated value of $1 at the end of 1 year . $1 will accumulate to 1+j. At rate i  jm ,$1 will accumulate at the end of 1 year will accumulate to m S  P(1  i ) n  1(1  (i)) m1  (1  i) m . Thus,1+j= (1  i) ,j= (1  i) -1(annually effective rate) m m EX1. Find the annual effective rate j corresponding to j= (1  (a) j2  10% (b) j12  6% (c) j365  13 % jm m ) -1 m 1 4 j2 2 0.1 ) 1  (1  ) 2 1  (1.05) 2 1 10.25% 2 2 jm m j12 12 0.06 12 ) -1= (1  )  1  (1  )  1  (1.005) 12  1  6.17% (b) j= (1  m 12 12 Solution: (a) j= (1  i) -1= (1  m p. 22 (c) j= (1  jm m j ) -1= (1  365 )365  1  14.17% m 365 same 2.Two nominal (yearly) compound interest rates are equivalent if they yield the accumulated values at the end of one year and hence at the end of any number of years. EX2. Find the rate j4 equivalent to (a) j12  12% (b) j2  10% Solution:(a) (1  i ) = (1  4 0.12 12 ) 12 1+i= (1.01) 12 4  (1.01)3 i= (1.01)3 -1  0.030301 i j4  j4  4i  4  (0.030301)  0.121204  12.12% 4 4 (b) j2  10% $1 at rate j4 will accumulate to (1  i )  (1  i )4 = (1  2 4 0.1 2 )  (1.005)2 2 1 2 1+i= (1  i ) , i  (1  i )  1  0.02469508  j4  4i  4  (0.02469508)  9.88%  j4  9.88% equivalent to j2  10% EX3. What simple interest is equivalent to j2  9% if money is invested for 3 years. Solution: Let P=$1,r-simple interest rate t=3,m=2,n=6,S=P(1+rt),s= P(1  i ) 1+3.r= (1  Note n 0.09 32 ) ,1  3r  (1.045)6 , r  0.10075338  10.08% 2 round of yearly rates of interest to the nearest hundredth of a percent. 3, 5, 6 课堂练习 p.32 Part A , 8 (a), 9, 再讲 Part B § 2.3 Discounted Value at Compound interest S= P (1  i)  P  S (1  i) n n P: discounted /present value Compound discount D=S-P Discount factor (1  i ) n discounted value of $1,using j * EX1 Find discounted value of $100000 due in (a) 10 years (b)25 years ,if money is worth j12  6% Solution:(a)s=100,000, i j12 0.06   0.005, n  10 12  120.P  100, 000(1  0.005) 120  $54,963.27 12 12 p. 23 (b) EX2. P  1 0 0 , 0 0 0 1 ( 0 .2 5015 ) 0 20  0 1 0 0 , 0 0 03(01 . 0 0 5 )  $22, 396.57 A lot for $84,000 cash or for payments of $50,000 now, $20,000 in 1 year and $20,000 in 2years If money is worth jn  9% ,which option is better for you . Solution: ① Option ② ① Discounted value of option 1 2.50,000+20,000 (1  85,000.39>$84,000 0.09 12 12 )  20, 000(1  0.05 ) 12  $85, 000.39 12  option 1 is better savings=$1001.39 ② If j12  12% ,the discounted value of option 2 50,000+20,000 (1.01) 12  20, 000(1.01) 24  $83,500.30 , savings=$499.70.option 2 is better. EX.3 A note for $2,000 dated Sept 1,01,is due with compound interest at j12  12% ,3 years who charges after the date .On Dec 1,/02, the holder of mote has it discounted by a j4  13 1 % .Find the proceeds and the compound discount 4 Solution: S=2000 (1  0.12 36 12 )  $2861.54 P=2861.54 (1  0.1325 7 )  $2277.87 4 Compound discount D=2861.54-2277.87=$583.67 习题选讲 p.35 11, p.36 Part B 3 § 2.4 Accumulated and Discounted value for a Fractional period of Time From formula: P  S  S (1  i )  n ,S  P(1  i ) n ,nint eger n (1  i ) Now n can be a fraction –- a fraction part of interest conversion period ① Exact / theoretical method Example1 Find the accumulated and discounted value of $1500 for 16 months at j4  12% , using the exact method. ① P=1500, j4  12% , i  j4 0.12 16 1   0.03, n   5 ( periods) 4 4 3 3 5 1 3 S  P(1  i ) n  1500(1  0.03)  $1756.13 p. 24 ② S=1500, P  S (1  i ) n  1500(1  0.03) 5 1 3 =$1281.23 ② Approximate /practical method Compounding interest for the full of conversion period. Example 2 Find the accumulated and discounted value of $1500 for 16 months at j4  12% , using the approximate methods ,and compare the results with those of Example 1. Solution 1. accumulated value: P=1500, j4  12% , n  16 1  5 ( periods) 3 3 S  (1500(1.03)5 .[1  (0.12) Note 1  $1756.30 12 the simple interest accumulation using S = P(1+rt), r = j4 , t in years ③ S=1500, n  16 1  5 ( periods) , at j4 =12%. Find P 3 3 First, discount $1500 for the smallest number of whole periods containing the given time and then accumulate this value for ② months At a simple interest rate of 12% Solution P= 1500[1  90.03)] [1  0.12( Note 6 2 )]  $1281.35 > $1281.23 12 using practical method is great than theoretical method. We use practical method . Example 3 A note of $3000 without interest is due on Aug ,18,2003. On June 11,2002, the holder of the mote has it discounted by a lender who changs j12  12% .what are the proceeds? Solution P= 3000[1  ( 习题选讲 p.39. Part B 3, 0.12 15 24 )] [1  (0.12) )]  $2604.44 12 365 HW: P.37, Part A 3, 8,9 §2.5 Find the Rate and the Time Find the Rate: S, P, n given, S  P (1  i ) , (1  i )  n n S S 1 ,1  i  ( ) n P p Example1 At what nominal rate jn will money triple itself in 12 years? let P =x, hen s = 3x, and n = (12) (12)=144 1 1 Solution: 144 144 3x = x(1  i ) , (1  i )  3,1  i  3144 , i  3144  1  0.007658429 j12  12i  0.091901147  9.19% Find the time s, p, i given, S  P (1  i ) n p. 25 Method 1 (1  i )  n S S ，取对数，得 n = log(1+i) = log P P Method 2 Example 2 Interpolation How long will it take $500 to accumulate to $850 at j12  12% , assume that (a) the theoretical method of accumulation is in effect. Solution let n be the number of months then P=500, s=850 ,i= 0.12  0.01 12 500(1.01)n  850,(1.01) n  4 years 5 months 10 days. Hw p.43 850 log1.7  1.7,n log(1.01)  log(1.7),n   53.3277 months)  500 log1.01 Part A 3, 1, 6, 11, 14. PartB.5 § 2.6 Equation of value Equivalence X is equivalent to Y at i due n periods. If Y=X (1  i ) orX  Y (1  i ) n n Note 1. 2. moving money forward, accumulate, multiply (1  i ) moving money backward, discount, …… n accumulate factor. discount factor. (1  i )  n Property of transitivity is held in compound interest rate: if X  Y , Y  Z , then X  Z proof: if X  Y then Y  X (1  i ) n2  n1 and Y  Z ,then Z  y (1  i ) n3  n2  Z  X (1  i ) n3  n1 , X  Z ① equivalence of dated values at compound interest is an equivalence relation; while at simple interest is not. ② at compound interest do not depend on focal date Example 1 An obligation of $500 falls due at the end of 3 years .Find an equivalence debt at the end of (a) 3 month (b)3 years 9 months j4  12% Solution: X= 500(1  0.12 11 )  500(1.03)11  $361.21 4 3 Y= 500(1.03)  $546.36 Verifying: Y  X (1.03) 151  361.21(1.03)14  546.36 Dated value of the set .sum of all equivalent dated valued due on the same date. p. 26 Property various dated values of the same set are equivalent. Example 2. A person owes $200 due in 6 months and $300 due in 15 months, what single payment (a) now (b) in 12 months. Will liquidate these obligations if money is worth j12  15% Solution X  200[1  ( 0.12 6 0.15 15 )]  300[1  ( )]  $434.63 12 12 0.12 6 0.15 3 Y  200[1  ( )]  300[1  ( )]  $504.51 12 12 n 12 therefore X  Y Verify: Y  X [1  i ]  434.63(1.0125)  $584.50 X  Y [1  i ] n  504.51(1.0125) 12  $434.64 Example 3 (compound interest not depended on the focal date) A debt of $1000 at j4  10% will be replaced by $200 at the end of 3 months and 3 equal payments at the end of 6,9,12 months. What will these payments take? Solution ① At the end of 12 months 0.1 3 )]  x(1.025) 2  x(1.025)  x  1000(1.025) 4 , x = $288.86 4 0.1 1 )]  x(1.025)2  x(1.025) 3  x(1.025) 4  1000 ② At the present time: 200[1  ( 4 200[1  ( Note X = $288.86 you can select any date as focal date-the simplest one. Example 4 A man leaves an estate of 50,000, that is invested at jn  9% .At the time of his death ,he has two children aged 13 and 18.Each child is to receive an equal amount from he estate who they reach 21.How much does each child get? Solution ① the older child will get $x in 3 years =36 months ② the younger child will get $x in 8 years =96 months Select today as focal date: x(1  0.09 36 0.09 96 )  x(1  )  50, 000 12 12 36 X=$39,929.38. Answer: each child will get $39,929.38. Checking: Amount in focal at the end of 3 years= 5000(1.0075) Payment to the older child =39,929.38 Balance is focal =25,502.89 Amount in focal after 8-3=5 years =25,502.89 (1.0075) 512  65, 432.27  $39,929.39 (Assume two children have the same birthday and the same interest day) Hw p.48. 2.6 Part A 3,6,11, Part B 2, 3. p. 27 § 2.8 Other application of Compound Interest Theory Inflation and the real rate of interest Simple interest: S=P(1+rt),compound : S  P (1  i ) :Geometric growth n Example 1 (Geometric growth) A tree measured in 1998, contains 150 m ,(cubic meter) of wood. If the tree grows at rate of 3% per annual, how much wood would it produce in 2008? Solution: S  P (1  i ) = 150(1  0.03) n 10 3  202(m3 ) Example 2 Population of Canada in July 1987 was 26.5 million. In July ,1997,it was 30.3 million ,(a) what was the nominal growth rate 1987-1997 (b)At this rate, when will the population reach to 40 million Solution (a) 30.3  26.5(1  i ) , i  1.35% 10 (b) 40  30.3(1  0.0135) , (1.0135)  t t 40 , 30.3 t log(1.0135)  log( 40 ), t  20.71 years = 20 years 9 months 30.3 From July 1997 add 20 years 9 months reaches March, 2018. Inflation and the real rate of interest An annual effective rate of change is the Consumer Price Index (CPI) Example 3 In June 1992, the Canada CPI was set at 100, In June 1999 it was 110.5.That means that if goods cost an average of $100 in June 1992,they cost $110.5 in June 1999. Over that 7-year period, what was the average annual compound percentage rate of change? If the rate of inflation were to continue, how long would it take before the purchasing power of a June 1992 Canada dollar was only 804? Solution 100(1  i )7  110.5, i  1.44% Find t years 0.8(1  i)t  1, (1.0144)t  1 0.80 t = log1.0144 / log0.80 =15.61 years= 15 years 7 months 10 days. Note Inflation rates vary from country to country, time to time Deflation prices drop and CPI decreases Real Rate of Interest: real relationship between interest rate and inflation rate Let i: annual interest rate , r: annual inflation rate then: ireal  1 i ir 1  1 r 1 r If inflation rate r is low 1+r r  1 , the ireal  i  r Example 4 (a) Jack invested $1000 for 1 year at j1  8% ,annual inflation rate r=2% What was the real annual rate of return on Jack’s investment? p. 28 (b) (c) What was Jack’s annual real after-tax rate of return, if he paid tax at 40% marginal rate? Repeat (b) for 50% marginal tax rate Solution (a) ireal  (b) (c) i  r 0.08  0.02   5.89% 1 r 1  0.02 0, 08(0.6)  0.02 after 40% tax: $1.0.6=$0.6  ireal   2.75% 1  0.02 0, 08(0.5)  0.02 after 50% tax:  ireal   1.96% 1  0.02 Example 5 (a) Suppose that the forecast for next year’s annual inflation rate is r=0.05, and i=0.04, what is the real interest rate ? (a) Using r, i in (a),suppose borrow $10,000 for a year at i=0.04 and buy 5000 units of a certain item tha has a current cost of $2 per unit. If the price of this item is tied to a rate of inflation r=0.05 and you sell the items one year from now at the inflation rate, what will be your net gain on this transaction? Solution: (b) Net (a) gain r=0.05, i=0.04,  ireal  = Increasing in i  r 0,04  0.05   0.95%  0 1 r 1  0.05 value of 5000 units-cost of borrowing = 5,000.2(0.05)  1000(0.04)  500  400  $100 Note where r =0.05 > 0.04 = i, i.e. inflation rate >interest rate, there is an incentive to borrow at negative real interest rate  ireal  0.95% Similarly, at low real interest rate, there is an incentive to consume rate than same 同样是 $1000 , 借它只需要 4%利率，卖$1000 货物 5%利润多赚 100 元. 即： ireal  1 i ir 1  is :r>i, then i-r<0, ireal  0 , 此时买东西好于存款 1 r 1 r Hw p.57 Part A 2, 3, 5, 7(a), Part B 1, 2 习题选讲 p.57. Part B 3 § 2.9 Continuous Compounding m x lim(1  m )m  e x , e  2.718 ① base of natural logarithm if nominal rate of interest compounding continuously, then use e Let jm  12% m 1 2 4 12 x Annual Accumulation Factor (1.12)1  1.12 (1  0.12 ) 2  1.1236 2 (1  0.12 ) 4  1.12550881 4 (1  0.12 )12  1.12682503 12 p. 29 52 365 When m   ,i.e.   j  12% ,we have (1  0.12 )52  1.127340987 52 (1  0.12 )365  1.12747614 365  利息变数为 m lim(1  0.12 m )  e0.12  1.127496852 m Find the rate j12 equivalent to j  15% 0.15 Example 1 Solution: (1  i)12  e0.15 , i  e 12 1  0.012578452  j12  12i  15.09% jm : S  P(1  i ) n  P(1  jm mt ) m m If j interest compounded continuously S  lim P[(1  Similarly, P  Se  j t jm m t ) ]  Pe jt m , j   : force of interest Example 2 Find the accumulated and discounted value of $5,000 over 15 months at a nominal rate 8% or 18% compound continuously? Solution: j  8% ,t= 15  1.25, P  5000 12 S  Pe jt  5000e(0.081.25)  $5525.85 P  Se jt  5000e0.081.25  $4524.19 Similarly, j  S S ln( P ) ln( P ) ,t  . t j 若设为 18%， j  18% ,t= 15  1.25, P  5000 S  Pe jt  5000e(0.181.25)  $6261.61 12 P  Se jt  5000e0.181.25  $3992.58 Examples 3 A natural fund deposit of $1000 increased in value by $1560 over 30 months .Find:(a) the continuous interest rate (b)the annual effective interest rate Solution (a) P=1000, S=1560, t = 30  2.5 (years) 12 1000e j (2.5)  1560, e10 j  1560,10 j  ln1.560 j  ln1.560  0.17784329  17.79% 2.5 j (b) 1  j  e  e0.17784329 , j  e0.17784329  1  19.47%  17.79% Note j1  j ( 19.47%  17.79% compound 次数越多, 利息越高, 利率越低) p. 30 Example 4 How long will it take to triple your investment at j  15% be P ,P=X, 3 X  Xe 0.15t Solution let X t= , e0.15t  3, 0.15t  ln 3, t  ln 3 , 0.15 ln 3  7.324081924 years = 17 years 118 days 0.15 Part B 1.2.6 讲 part B 3 Hw p.61.4.6 § 2.10 Compound Discount at Discount Rate and Equivalent Discount Rate D 1 Recall simple discount rate: d= , P  S (1  dt ), S  p (1  dt ) st Now: compound discount rate d (m) : nominal discount rate compound m times per year * Discount rate per conversion period: d  d (m) m d (m) n P  S (1  )  P  S (1  dt ) m d (m) n S  P (1  )  S  P (1  dt ) 1 m Compound discount at a discount rate is an important topic in developing theoretical models in actuarial science. Example 1 Find the discounted value of $1000 due in 2 years at (a) d (12)  12% (b) d (365)  7% Solution (a) S=100, n=24, P= S [1  d (m) n 0.12 24 ]  1000(1  )  $785.68 m 12 (b) Example 2 (a) n = 2  365 = 730, P= 1000(1  0.07 730 )  $869.35 365 (2) Find the accumulated value of $500 at the end of 3 years (b) at d at 8% simple discount  8% compound semi-annually 1 Solution (a) d=8%, P=500, t=3, S  P(1  dt )  500(1  0.08  3) 1  $657.89 (b) d (2)  8% , P=500,t=2  3=6, S  P(1  (2) d (m)  n 0.08 6 )  500(1  )  $638.77 m 2 Note 当数值 d =8% = d  8% ,P=500, compound discount rate smallest value of S, annual effective discount rate d, discounted value of $1. 1  d  (1  d (m) m d (m) m ) , d  1  (1  ) m m p. 31 Example 3 Find the annual effective discount rate at to d 12  9% Solution d  1  (1  d (12) 12 0.09 12 ) = 1  (1  )  8.64% 12 12 Two nominal compound discount rates are equivalent if they yield same discounted values over the same period. Example 4 Find d (12) equivalent to (a) d (2)  6% (b) d (365)  18% d d (12) 12 0.08 2 )  (1  2 ) 2  (1  ) Solution:(a) (1  12 2 2  d (12)  6.08% (b) (1  d d (12) 12 0.1 365 )  (1  365 )365  (1  ) 12 365 365  d (12)  9.96% Example 5 Find d (4) equivalent to an annual effective compound interest rate of 9% j1  9% Solution let S=$1,compare discounted values P of $1 due in 1 year P  S (1  d d ( m) m )  1.(1  4 ) 4 , P  S (1  j1 ) 1  1.(1  0.09) 1 m 4  (1  d4 4 d )  (1.09)1 ,1  4  (1.09) 4 , d 4  4[1  1.094 ], d 4  8.53% 4 4 T-Bills (Treasury Bills) short-term, low-risk issued by Canadian Government every Thursday 26/13 week (1821/91 days later) Active secondary market reported in terms of nominal interest rate (0.01% convertible every 91/182 days ,and quoted per 100 of face values to the nearest $0.001 Example 6. A 91-day T-bill with face value of $100,000 is purchased for $97,250.Find the 91-day effective interest and discount rates, the nominal interest rate convertible every 91 days ,and the quoted price per 100 of face value. Solution Interest earned for 91-day :100,000-97250=$2,750. 91-day effective interest rate i = I 2750   0.028277635 Pt 97250 D 2750   0.0275 St 100, 000 91-day effective discount rate d = Nominal interest convertible every 91-days jm  0.113421283  11.34% Quoted price per 100 of face value : $97250 = 97250 100  97.25 /100 100, 000 p. 32 d = 0.0275 in 91 days  100  (1-0.0275)=$97.250 ┐ Example 7 Show that 91-day effective interest and discount rates of Example 6 are equivalent . Solution. let P = $1 find S at end of 91-days at i =9.7250. S = P(1+i) n = 1+i 1+2750/97250 = 1.028277635 find S at end of 91-days at d = 2750/100000. S = P (1-d)-n = (1-d) -1 (1-2750/100000)-1 = 1.028277635 满足:1+i = (1-d) -1 尽管从数字上看 i > d.但此处 i 与 d 等价，因 S 相同。 讲 p.65 Part A 7, part B. 4.(a).(b). HW. p.65 part A .1(a), 2(a), 3(a), 4(a), (c),(d), 6, 8, Part B.1,3,5. p. 33