mathematics of finance Faculty of commerce – English Section Second year Review Problems Derive the formula Sn (for) = R[(1+i) n+p – (1+i) ]/ i p Sn (for) 1 … . …. n-2 n-1 n 1 2 Payment interval of n periods 3 P-1 P P interest periods after last payment Sn (for) = R sn+ p┐- Rsp┐ = R (sn+ p┐- sp┐) = R [((1+i)n+p-1)/i - ((1+i)p-1)/i] = R [((1+i)n+p - (1+i)p)/i] = R ((1+i)n – 1 /i) (1+i)p = R sn┐(1+i)p Derive the formula An (def) = R an┐(1+i)-m …. An 1 2 …. m 1 Deferment interval of m periods 2 3 n-1 n Payment interval of n periods An (def) = R am+n┐- Ram┐ = R (am+n┐- am) = R [(1-(1+i)-(m+n)/ i) – (1-(1+i)-m/ i)] = R [( (1+i)-m- (1+i)-(m+n)/ i)] = R [1 - (1+i)-n / i] (1+i)-m = R an┐(1+i)-m A man aged 25 is considering 2 types of life insurance policies. The first is an ordinary life with an annual premium of $200. The second is a 20-year endowment with an annual premium of $510. He decides to take the ordinary life and to deposit the difference in premiums each year into an investment paying 51/2% compounded annually. He dies just before making his 17th deposit but after receiving interest for the first 16 deposits. How much was in the account at that time? Sn (due) = R (1+i)n-1 (1+i) i = 310 [(1.055)16 – 1 / .055] (1.055) = $8058.88 (a) You need a loan. Which is the best deal? Bank A: 7.15% compounded monthly, Bank B: 7% comp. annually, or Bank C: 6.85% comp. quarterly (b) Convert 9% compounded monthly to an equivalent nominal rate converted semi-annually Bank A: 7.15% comp. monthly i = 7.15%/12 r = (1+7.15%/12)12-1 r = 7.39% Bank B: 7% comp. annually r = 7% Bank C: 6.85% comp. quarterly i = 6.85%/4 r = (1+6.85%/4)4-1 r = 7.03% (c) (1+.09/12)12 = (1+J/2)2 (1.0075)6 = (1+J/2) J = 9.17% 1 Faculty of commerce – English Section mathematics of finance Second year Review Problems On February 18, 2002, a person deposited $4500 in an account earning 3.5% compounded quarterly. If simple interest is allowed for part of an interest conversion period, find the amount that will be in the account on August 28, 2010. Also, state the total amount of interest earned 18/2/2002 $4500 18/8/2002 Compound 28/8/2010 Simple Year 2010 2002 Month 8 2 Day 28 18 8 Years 6 months 10 days n = (8 × 4)+6/3 = 32+2=34 Simple interest S = P(1+i)n= 4500 (1.00875)34 = $6,051.36 I = Prt = 6,051.36 × 3.5/100 × 10/360 = $5.88 Final amount = 6,051.36 + 5.88 = $6057.24 OR: Final amount = P(1+i)n + (1+rt) = 4500 (1.00875)34 (1+3.5%×10/360) = $6057.24 A person deposits $250 monthly in a saving account earning 6% compounded monthly. The first deposit was made on April 1, 1989 and the last deposit was made on February 1, 2004 a. Find the amount in the account just after the last deposit on February 1, 2004. b. If no more deposits are made after February 1, 2004 but the money is allowed to continue earning interest, find the amount in the account on October 1, 2010 Year 2004 1989 Month 2 4 14 Years 10 months Day 1 1 0 days n = 14 × 12 + 10 + 1 = 179 Sn = R (1+i)n-1 i = 250 [(1.005)179 – 1 / .005] = $72,094.21 Year 2010 2004 Month 10 2 Day 1 1 6 Years 8 months 0 days P = 6 × 12 + 8 = 80 Sn (for) = R (1+i)n-1 (1+i)p i 179 = 250 [(1.005) – 1 / .005] (1.005)80 = $107,444.78 2 mathematics of finance Faculty of commerce – English Section Second year Review Problems A person buys a TV that costs $1200. The payment plan requires $150 paid down and the balance financed at 18% compounded monthly for 2 years. Find the size of the payments made at the end of each month. Also state the total interest paid on this loan 1200 150 1050 R R R R 2Y An = R an┐ = R (1-(1+i)-n/ i) 1050= R ×(1-(1+.015)-24/.015) R = $52.42 Total interest = (52.42 × 24) – 1050 = $208.08 Determine the amount of money you need to deposit in an account earning 7% compounded monthly on the day your child is born so that on the child's 18th birthday, he/she can begin withdrawing $500 per month for 4 years to help cover college expenses. Also state the total amount of interest earned on this account. 18 Y R R P.v. comp. R R R 22 Y 4Y An(due) = R an┐(1+i) = R (1-(1+i)-n/ i) (1+i) = 500 × (1-(1+0.07/12)-48/0.07/12) (1+0.07/12) = $21,001.9 P = S(1+i)-n = 21,001.9 (1+0.07/12) -216 = $5,979.12 Interest = (500 × 48) - 5,979.12 = $18,020.88 Your uncle deposited $10,000 in an account for you that earns 8% compounded monthly. How much can you withdraw monthly for 5 years if the first withdrawal made immediately? An(due) = R an┐(1+i) 10,000= R (1-(1+.08/12)-60/ .08/12i) (1+.08/12) R = $201.42 When fred turned 30 he deposited $45,000 into his savings account which earns 7% interest compounded annually. How much can be withdrawn each year for 15 years to empty his account, if his first withdrawal is made on his 65th birthday Y.30 45,000 Y.65 R Y.80 ….. R An (def) = R an┐(1+i)-m R = 45,000 / [1-(1+.07)-15/.07)] (1.07)-34 = 45,000/ 0.913 = $49,299.45 Phil was loaned $50,000 at 4% compounded monthly. He makes regular monthly payments of $900 3 Faculty of commerce – English Section mathematics of finance Second year Review Problems (a) How many full payments must he make? (b) What is the size of the smaller concluding payment? An = R an┐ 50,000 = 900 an┐ an┐= 55.556 1-(1+.04/12)-n/.04/12 = 55.556 (1+.04/12)-n = 0.814813333 -n log (1+.04/12) = log 0.814813333 n = 61.54 This means 61 full payment + smaller concluding payment 50,000 (1+.04/12) 62 = 61,457.51 Sn = R (1+i)n-1 (1+i) i = 900 [(1+.04/12) 61 – 1 / .04/12] (1+.04/12) = $60,970.54 Size of concluding payment = 61,457.51 - 60,970.54 = $486.97 Lex decides to give the Kents a break. He loans them $50,000 and asks them to pay him back just $53,880 at the end of 3 years. What interest rate converted quarterly did lex charge the kents? S = P(1+i)n 53,880 = 50,000 (1+i) 12 (1+i) 12 = 1.0776 i = 0.62 J4 = 0.62 × 4 = 2.50% You just deposited $50,000 into an account which earns 5% compounded annually. What is the most you can withdraw each year so that the account is never depleted, if your first withdrawal is made 20 years from now? 50,000 n 20 R R R 19 S = P (1+i) = 50,000 (1.05) = 126,347.51 1 Period = 1 Year, so, we can use simple interest formula I = Prt = 126,347.51 × .05 × 1 = $6,317.38 Mr. Bigshot wants to set up an annual scholarship of $5,000 in his honor at NCSU. The university can earn 7% interest compounded monthly. If the first scholarship is awarded at the end of 1 year, how much does Mr. Bigshot need to donate? First, we convert 7% compounded monthly to % compounded annually r = (1+i)m-1 = (1+.07/12) 12-1= 7.229% OR (1+.07/12) 12 = (1+j) j = 7.229% We need to deposit enough money so that the amount of interest earned each year is $5,000. But since 1 year is 1 conversion period, simple and compound interest are the same. So, I = Prt 5,000 = P × .07229 × 1 P = $69,165.86 Simon's Construction Company (SCC) borrows $80,000 at 10% simple interest from the first bank of America. The bank requires the SCC to make quarterly interest-only payments and pay the full $80,000 at the end of 5 4 Faculty of commerce – English Section mathematics of finance Second year Review Problems years. In order to meet the 5 year obligation of $80,000, SCC makes equal deposits at the end of each quarter into a sinking fund which earns 7% compounded quarterly? (a) Find SCC's total quarterly obligation? (b) Find the equivalent amortization rate compounded quarterly for this loan? The quarterly interest payments must be: I = Prt = 80,000 × 10% × 3/12 = $2,000 Sn = R (1+i)n-1 i 80,000 = R [(1+.07/4) 20 – 1 / .07/4] R = $3,375.30 SCC's total quarterly obligation is : 2,000 + 3,375.3 = $5,375.3 An = R 1-(1+i)n i 80,000 = 5,375.3 sn┐ a20┐= 14.883................................... The equivalent amortization rate = 11.984% Converting compound interest rates. You don't need to draw a time diagram for this problem. (a) 5.75% compounded quarterly is equivalent to what rate compounded annually? (b) 10% compounded daily (use Banker's Rule) is equivalent to what rate compounded semiannually? (a) i = .0575/4 = 0.014375 r = (1.014375)4 – 1 = 5.88% OR (1.014375) 4 = (1+J) J2 = 5.88% (b) i = 0.1/360 r = (1+0.1/360)360 – 1 = 10.52% J = (1+10.52/100)1/2 – 1 = 5.128% J2 = 10.25% OR (1+0.1/360) 360 = (1+ J/2)2 (1+0.1/360) 180 = (1+ J/2) J2 = 10.25% In June, 1986, A person owes $8,000 due in 3 years, $15,000 due in 8 years and $12500 due in 15 years. The debtor want replace these, with 2 payments and twenty annuities, The first of all $9000 due in 4 years, the second $7000 due in 13 years, and he has to pay annuities will be start in June 1, 1992 and at the end of each 6 months, If money is worth 9% converted annually. What is the ordinary annuity? [Putting the focal date on Jen 3Y . 1, 1996] 1/6/1986 4Y . 8000 r = (1+i)2 – 1 8Y . 10Y. 13Y. 15Y. 15000 9000 r = 9% 6Y . RRR R 12500 R 7000 Focal date i = 4.403% ….R . 8000(1.04403)14+15000(1.04403)4 + 12500(1.04403)-10 = 9000(1.04403)12 + 7000(1.04403)-6+[ R a9┐+ R s11┐] [ R a9┐+ R s11┐] = 20,071.62 R [(1-(1.04403)-9/.04403) + ((1.04403)11-1/.04403)] R = $952.48 On June 1, 1995, a woman owes $8000, which she is unable to repay. Her creditor charges her 12% compounded quarterly from that date. If the debtor pays $3000 on June 1, 1997, and she have to pay annuity at 1/6/1995 5 1/6/1997 1/6/2001 Faculty of commerce – English Section mathematics of finance Second year Review Problems the end of each 3 months for 4 years. What is the ordinary annuity at the end of each 3 months? (Putting the focal date on June 1, 2001) 8000 3000 R ……… ……… R 8000(1.03)24 = 3000(1.03)16 + R s16┐ R R s16┐ = 11448.23 R =$567.96 A company can put $9500 in an investment that is expected to result in cash inflows of $3000 in 3 months and $5000 in 8 months and $2000 in 1 year. Find the internal rate of return. 3m. 9500 3000 8m. 1 Y. 5000 2000 At 10%: P.V. = 3000/(1+0.1×3/12) + 5000/(1+0.1×8/12) + 2000/(1+0.1×1) = 9432.51 At 8%: P.V. = 3000/(1+0.08×3/12) + 5000/(1+0.08×8/12) + 2000/(1+0.08×1) = 9539.86 Rate 8% 2% d r 10% Present value of inflows 9539.86 9500 39.86 107.35 9432.51 d/2 = 39.86/107.35 d = 2× 39.86/107.35 =.74 r = 8+.74 = 8.74% An investment of $8000 is made for 15 years and 2 months. During the first 7 years the interest rate is 9% converted semiannually. Then the rate drops to 7% converted semiannually for the remainder of the time. What is the final amount? 7 Y. 9% semi 15 Y. 15 Y.+2m. 7% semi 8000 S1 = P (1+i)n = 8000 (1.045)14 = $14,815.56 Using 9% semiannually S2 = P (1+i)n = 14,815.56 (1.035)16 = $25,689.97 Using 7% semiannually I = Prt = 25,689.97 × .07 × 2/12 = $299.72 Final amount = 25,689.97 + 299.72 = $ 25,989.69 Calculate the present value of LE 1000 due in 3 years (a) at a simple interest rate of 10% P.a. P = S/(1+rt) = 1000/(1+.1×3) = $769.23 (b) at a simple discount rate of 9% P.a. P = S(1-dt) = 1000(1-.09×3) = $730 Example 3: A couple gets an $80,000, 30-year, 12% loan. The monthly payment is $822.9. How much of the first two payments goes to interest and how much to principal? 6 Faculty of commerce – English Section mathematics of finance Second year Review Problems As to first payment : I = Prt = 80,000 × .12 × 1/12 = $800 Payment to principal = $822.9 – 800 = $22.9 As to second payment : New P = 80,000 – 22.9 = $ 79, 977.1 I = Prt = 79, 977.1× .12 × 1/12 = $799.77 Payment to principal = $822.9 – 799.77 = $23.13 Payments of LE 10 are due for the next 6 years. What is the present value of these payments at 1% per month? What is the amount value if the LE 10 payments are due at : (a) The end of each month (b) The beginning of each month Ordinary annuity Sn= Rn+ n / 2 × R × r × t (n-1) 1 1 Sn= 10 × 72 + 72 / 2 × 10 × .12 × 1/12 (72-1) = 720 + 36 × 1.2 × 1/12 × 71 = 720 + 255.6 = 975.6 annuity due Sn= Rn+ n / 2 × R × r × t (n+1) 1 1 Sn= 10 × 72 + 72 / 2 × 10 × .12 × 1/12 (72+1) = 720 + 36 × .1 × 73 = 720 + 262.8 = 982.8 d= r/(1+rt)=.12/(1+.12×1)= 10.71% An= Rn- n / 2 × R × d (t1+ tn) 1 Present value Present value of ordinary =10×72–72/2×10×.1071(1/12+72/12) An= Rn- n / 2 × R × d (t1+ tn) Present value Present value due = 10×72–72/2 ×10×.1071 (0 + 71/12) 1 Find the amount of LE 100 accumulated for 20 years at the following rate of interest:- 7 Faculty of commerce – English Section mathematics of finance Second year Review Problems (a) Effective rate of interest corresponding to a nominal rate of 4% P.a. convertible half yearly? (b) 6% p.a. convertible three times p.a. for 12 years, thereafter 3.5% p.a. convertible every second year (c) the rate of interest p.a. at which a sum of money will treble itself in 21 years (a) r = (1+ .04/2)2 -1 = 4.04% S = P (1+i) n = 100(1.0404)20 = $220.80 (b) S1 = P (1+i) n = 100(1+.06/3)36 = $203.99 S2 = P (1+i) n = 203.99(1+.035/360×24×60×60)31104000 = $211.22 (c) S = P (1+i) n 3 P = P (1+i)21 (1+i) 21 = 3 i = 5.37% S = P (1+i) n S = 100 (1+.0537)20 = $284.67 8 mathematics of finance Faculty of commerce – English Section Second year Review Problems An automobile is purchased for $12,000. The down payment, including the trade-in value of the old car, was worth $3,000. The interest is 10% on an add-on basis. The loan is to be repaid in 3 years. Find the nominal interest rate the buyer is paying. Amount owed = 12,000 – 3,000 = $9,000 Add-on = 9,000 × 3 × 10% = 2,700 Total amount owed = 11,700 Monthly payments = 11,700 / 36 = $325 An = R an┐ 9,000 = 325 a36┐ a36┐= 27.6923 j 15% 3% d Present value of inflows 28.8473 i 27.6923 18% 28.8473 1.155 1.1866 d / 3 = 1.155/1.1866 d = 2.92 j = 15+2.92 = 17.92 % The parents of four children, a 13 year-old, a 12-year-old, and 10-year-old twins, die suddenly. Their wills call for a trust fund of $200,000 to be held, with equal amounts of the money to be given to each of the children at age 18. If the fund earns 10% converted quarterly, find the amount each child will get Y.10 8 Y. Y.12 Y.13 5 Y. 200000 6 Y. X X 2X X (1.025)-20 + X (1.025)-24 + 2X (1.025)-32 = 200,000 X [(1.025)-20 + (1.025)-24 + 2 (1.025)-32] = 200,000 X = $96,586.28 Sixty days after borrowing, a person pays back exactly LE 200. How much was borrowed if the LE 200 payment includes the principal and simple interest at 9% P.a. S = P (1+rt) 200 = P (1+.09×60/360) P =200/1.015 = $197.04 9 Faculty of commerce – English Section mathematics of finance Second year Review Problems Deposits of LE 400 are made at the beginning of each half year for 5 years into an account paying 16% compounded half yearly. How much is in the account: a. at the end of 5 years b. Just before the sixth deposit (a) Sn (due) = R (1+i)n-1 (1+i) i = 400 [(1+.08) 10 – 1 / .08] (1+.08) = $6,258.19 (b) Sn (due) = R (1+i)n-1 (1+i) i = 400 [(1+.08) 5 – 1 / .08] (1+.08) = $2,534.37 Eight hundred pounds is due at the end of 2 years and LE700 at the end of 8 years at rate 12% compounded monthly, find an equivalent single amount at the end of: (a) 2 Years (b) 6 Years (c) 10 Years 2 Y. (a) 8 Y. 800 700 Focal date X = 800 + 700(1.01)-72 = $1141.95 What amount should be invested at outset to give an accumulated amount of L.E. 5,000 at the end of 15 years, if the interest rate is 9% p.a. for the first 8 years, 7% p.a. for the next 4 years, and 5% p.a. for the last 3 years? -n 8Y. 4Y. 3Y. 9% 7% 5% 5000 -3 P1 = S (1+i) = 5000 (1+.05) = $4319.19 P2 = S (1+i)-n = 4319.19 (1+.07)-4 = $3295.09 P3 = S (1+i)-n = 3295.09 (1+.09)-8 = $1653.69 A person borrows $9000 on 3 years and $11000 on 5 years at 19% interest. The debt will be repaid with three payments, X, Y, and Z, that Y equal 3X and Z equal 2Y, A one on 7 years, the second on 9 years, and the other on 12 years. Put the focal date on 7years. And find the size of the payments (using time diagram) Y = 3X Z = 2Y=6X 3 Y. 5 Y. 7 Y. X 9 Y. Y 12 Y. Z 9000 11000 9000(1+0.19×4)+11000(1+0.19×2) = X+3X/(1+0.19×2)+6X/(1+0.19×5) 15840 + 15180 = X + 3/1.38X + 6/1.95X 31020 = 6.25 X X = 4963.2 Y = 3 × 4963.2 = 14889.6 Z = 6 × 4963.2 = 29779.2 10