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# Solved Examples: Bonds Formulas Used: Mathematics of Finance
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(1.) Divine Prosperity Inc. issued a thirty-year semiannual coupon bonds with a face value of one thousand dollars.
If the annual coupon rate is six percent and the current yield to maturity is seven percent, what is the firm's current price per bond?

$t = 30\:years \\[3ex] Semiannual\:\:coupon\:\:bonds \rightarrow m = 2 \\[3ex] FV = \$1000 \\[3ex] BCR = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] YTM = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] BP = ? \\[3ex] BP = \dfrac{FV * BCR}{YTM} * \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] + \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[10ex] Too\:\:long...let\:\:us\:\:solve\:\:in\:\:parts \\[3ex] \underline{First\:\:Part} \\[3ex] \dfrac{FV * BCR}{YTM} \\[5ex] = \dfrac{1000 * 0.06}{0.07} \\[5ex] = \dfrac{60}{0.07} \\[5ex] = 857.142857 \\[3ex] \underline{Second\:\:Part} \\[3ex] \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] \\[10ex] mt = 2(30) = 60 \\[3ex] 1 + \dfrac{YTM}{m} = 1 + \dfrac{0.07}{2} = 1 + 0.035 = 1.035 \\[5ex] \left(1 + \dfrac{YTM}{m}\right)^{mt} = (1.035)^{60} = 7.8780909 \\[7ex] \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = \dfrac{1}{7.8780909} = 0.126934306 \\[10ex] 1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = 1 - 0.126934306 = 0.873065694 \\[10ex] \underline{Third\:\:Part} \\[3ex] \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[7ex] = \dfrac{1000}{7.8780909} \\[5ex] = 126.934306 \\[3ex] \underline{Bond\:\:Price} \\[3ex] BP = 857.142857 * 0.873065694 + 126.934306 \\[3ex] BP = 748.342023 + 126.934306 \\[3ex] BP = 875.276329 \\[3ex] BP \approx \$875.28$
(2.) Favor Inc. has outstanding one thousand dollars face value four percent coupon bonds that make semiannual payments, and have ten years remaining to maturity.
If the current price for these bonds is nine hundred and thirty eight and fifty seven cents, what is the annualized yield to maturity?

$FV = \$1000 \\[3ex] BCR = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] Semiannual\:\:coupon\:\:bonds \rightarrow m = 2 \\[3ex] t = 10\:years \\[3ex] BP = \$938.57 \\[3ex] Annualized\:\:YTM = ? \\[3ex] Annualized\:\:YTM \approx \dfrac{2(t * FV * BCR + FV - BP)}{t(FV + BP)} \\[5ex] Annualized\:\:YTM \approx \dfrac{2(10 * 1000 * 0.04 + 1000 - 938.57)}{10(1000 + 938.57)} \\[5ex] \approx \dfrac{2(400 + 1000 - 938.57)}{10(1938.57)} \\[5ex] \approx \dfrac{2(461.43)}{19385.7} \\[5ex] \approx \dfrac{922.86}{19385.7} \\[5ex] \approx 0.0476051935 \\[3ex] to\:\:percent = 0.0476051935(100) \\[3ex] \approx 4.76\%$
(3.) Future and Hope Inc. issued a twenty-year one thousand dollars face value, eight percent annual coupon bonds, with a yield to maturity of ten percent.
Calculate the current price of the bond.

$t = 20\:years \\[3ex] Annual\:\:coupon\:\:bonds \rightarrow m = 1 \\[3ex] FV = \$1000 \\[3ex] BCR = 8\% = \dfrac{8}{100} = 0.08 \\[5ex] YTM = 10\% = \dfrac{10}{100} = 0.1 \\[5ex] BP = ? \\[3ex] BP = \dfrac{FV * BCR}{YTM} * \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] + \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[10ex] Too\:\:long...let\:\:us\:\:solve\:\:in\:\:parts \\[3ex] \underline{First\:\:Part} \\[3ex] \dfrac{FV * BCR}{YTM} \\[5ex] = \dfrac{1000 * 0.08}{0.1} \\[5ex] = \dfrac{80}{0.1} \\[5ex] = 800 \\[3ex] \underline{Second\:\:Part} \\[3ex] \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] \\[10ex] mt = 1(20) = 20 \\[3ex] 1 + \dfrac{YTM}{m} = 1 + \dfrac{0.1}{1} = 1 + 0.1 = 1.1 \\[5ex] \left(1 + \dfrac{YTM}{m}\right)^{mt} = (1.1)^{20} = 6.72749995 \\[7ex] \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = \dfrac{1}{6.72749995} = 0.148643628 \\[10ex] 1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = 1 - 0.148643628 = 0.851356372 \\[10ex] \underline{Third\:\:Part} \\[3ex] \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[7ex] = \dfrac{1000}{6.72749995} \\[5ex] = 148.643628 \\[3ex] \underline{Bond\:\:Price} \\[3ex] BP = 800 * 0.851356372 + 148.643628 \\[3ex] BP = 681.085098 + 148.643628 \\[3ex] BP = 829.728726 \\[3ex] BP \approx \$829.73$
(4.) Peace Inc. has outstanding one thousand dollars face value twelve percent coupon bonds that make semiannual payments, and have eight years remaining to maturity.
If the current price for these bonds is one thousand, two hundred and seventy four dollars and thirty five cents, what is the annualized yield to maturity?

$FV = \$1000 \\[3ex] BCR = 12\% = \dfrac{12}{100} = 0.12 \\[5ex] Semiannual\:\:coupon\:\:bonds \rightarrow m = 2 \\[3ex] t = 8\:years \\[3ex] BP = \$1274.35 \\[3ex] Annualized\:\:YTM = ? \\[3ex] Annualized\:\:YTM \approx \dfrac{2(t * FV * BCR + FV - BP)}{t(FV + BP)} \\[5ex] Annualized\:\:YTM \approx \dfrac{2(8 * 1000 * 0.12 + 1000 - 1274.35)}{8(1000 + 1274.35)} \\[5ex] \approx \dfrac{2(960 + 1000 - 1274.35)}{8(2274.35)} \\[5ex] \approx \dfrac{2(685.65)}{18194.8} \\[5ex] \approx \dfrac{1371.3}{18194.8} \\[5ex] \approx 0.0753676875 \\[3ex] to\:\:percent = 0.0753676875(100) \\[3ex] \approx 7.54\%$
(5.)

$t = 4\:years \\[3ex] Semiannual\:\:bond \rightarrow m = 2 \\[3ex] FV = \$1000 \\[3ex] BCR = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] YTM = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] BP = ? \\[3ex] BP = \dfrac{FV * BCR}{YTM} * \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] + \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[10ex] Too\:\:long...let\:\:us\:\:solve\:\:in\:\:parts \\[3ex] \underline{First\:\:Part} \\[3ex] \dfrac{FV * BCR}{YTM} \\[5ex] = \dfrac{1000 * 0.06}{0.07} \\[5ex] = \dfrac{60}{0.07} \\[5ex] = 857.142857 \\[3ex] \underline{Second\:\:Part} \\[3ex] \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] \\[10ex] mt = 2(30) = 60 \\[3ex] 1 + \dfrac{YTM}{m} = 1 + \dfrac{0.07}{2} = 1 + 0.035 = 1.035 \\[5ex] \left(1 + \dfrac{YTM}{m}\right)^{mt} = (1.035)^{60} = 7.8780909 \\[7ex] \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = \dfrac{1}{7.8780909} = 0.126934306 \\[10ex] 1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = 1 - 0.126934306 = 0.873065694 \\[10ex] \underline{Third\:\:Part} \\[3ex] \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[7ex] = \dfrac{1000}{7.8780909} \\[5ex] = 126.934306 \\[3ex] \underline{Bond\:\:Price} \\[3ex] BP = 857.142857 * 0.873065694 + 126.934306 \\[3ex] BP = 748.342023 + 126.934306 \\[3ex] BP = 875.276329 \\[3ex] BP \approx \$875.28$
(6.) Joy Inc. issued ten-year zero-coupon bonds with a thousand dollars face value.
If the bonds are currently selling for five hundred and fourteen dollars and eighty seven cents, what is the yield to maturity?

$t = 10\:years \\[3ex] FV = \$1000 \\[3ex] BP = \$514.87 \\[3ex] YTM = ? \\[3ex] YTM = \left(\dfrac{FV}{BP}\right)^{\dfrac{1}{t}} - 1 \\[7ex] YTM = \left(\dfrac{1000}{514.87}\right)^{\dfrac{1}{10}} - 1 \\[7ex] = \left(1.94223785\right)^{0.1} - 1 \\[5ex] = 1.06863708 - 1 \\[3ex] = 0.06863708 \\[3ex] to\:\:percent = 0.06863708(100) \\[3ex] = 6.863708\% \\[3ex] \approx 6.86\%$
(7.)

If the question does not state that the interest is simple interest, then assume compound interest.

First: This is a case of Compound Interest because their retirement savings of $\$25,000$earns interest compounded annually/yearly at the interest rate of$6\%$. Second: This is also a case of Annuity Due because the couple would have to deposit$\$1,000$ at the beginning of each year.

$\underline{Compound\:\:Interest\:\:and\:\:Annuity\:\:Due} \\[3ex] P = \$25000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] PMT = \$1000 \\[3ex] CFV = \$80000 \\[3ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] t = ? \\[3ex] t = \dfrac{\log\left[\dfrac{rCFV + PMT(m + r)}{rP + PMT(m + r)}\right]}{m\log\left(1 + \dfrac{r}{m}\right)} \\[10ex] = \dfrac{\log\left[\dfrac{0.06(80000) + 1000(1 + 0.06)}{0.06(25000) + 1000(1 + 0.06)}\right]}{1 * \log\left(1 + \dfrac{0.06}{1}\right)} \\[10ex] = \dfrac{\log\left[\dfrac{4800 + 1000(1.06)}{1500 + 1000(1.06)}\right]}{1 * \log\left(1 + 0.06\right)} \\[10ex] = \dfrac{\log\left[\dfrac{4800 + 1060}{1500 + 1060}\right]}{\log\left(1.06\right)} \\[7ex] = \dfrac{\log\left[\dfrac{5860}{2560}\right]}{\log\left(1.06\right)} \\[7ex] = \dfrac{\log(2.2890625)}{0.0253058653} \\[5ex] = \dfrac{0.359657651}{0.0253058653} \\[5ex] = 14.2124226\:years \\[3ex] Let\:\:us\:\:check\:\:to\:\:confirm \\[3ex] \underline{Check} \\[3ex] \underline{Compound\:\:Interest} \\[5ex] P = \$25000 \\[3ex] r = 0.06 \\[3ex] m = 1 \\[3ex] t = 14.2124226\:years \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] = 25000\left(1 + \dfrac{0.06}{1}\right)^{1(14.2124226)} \\[7ex] = 25000\left(1 + 0.06\right)^{14.2124226} \\[7ex] = 25000(1.06)^{14.2124226} \\[5ex] = 25000(2.2890625) \\[3ex] = \$57226.5625 \\[3ex] \underline{Annuity\:\:Due} \\[3ex] PMT = \$1000 \\[3ex] r = 0.06 \\[3ex] m = 1 \\[3ex] t = 14.2124226\:years \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] * \left(1 + \dfrac{r}{m}\right) \\[10ex] = 1 * 1000 * \left[\dfrac{\left(1 + \dfrac{0.06}{1}\right)^{1(14.2124226)} - 1}{0.06}\right] * \left(1 + \dfrac{0.06}{1}\right) \\[10ex] = 1000 * \left[\dfrac{\left(1 + 0.06\right)^{14.2124226} - 1}{0.06}\right] * \left(1 + 0.06\right) \\[10ex] = 1000 * \left[\dfrac{\left(1.06\right)^{14.2124226} - 1}{0.06}\right] * \left(1.06\right) \\[10ex] = 1000 * 1.06 * \left[\dfrac{2.2890625 - 1}{0.06}\right] \\[7ex] = 1060 * \left[\dfrac{1.2890625}{0.06}\right] \\[7ex] = \dfrac{1060 * 1.2890625}{0.06} \\[5ex] = \dfrac{1366.40625}{0.06} \\[5ex] = \$22773.4375 \\[5ex] \underline{Combined\:\:Amount} \\[3ex] CFV = 57226.5625 + 22773.4375 = \$80000$
(8.)

$\underline{First\:\:Semester} \\[3ex] \Sigma CH = 12 \\[3ex] GPA = 3.583333333 \\[3ex] \Sigma GP = 3.583333333(12) = 43 \\[3ex] \underline{Second\:\:Semester} \\[3ex] \Sigma CH = 16 \\[3ex] \Sigma GP = 55 \\[3ex] \underline{First\:\:and\:\:Second\:\:Semesters} \\[3ex] \Sigma CH = 12 + 16 = 28 \\[3ex] \Sigma GP = 43 + 55 = 98 \\[3ex] CGPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] CGPA = \dfrac{98}{28} \\[5ex] CGPA = 3.5 \\[3ex] CGPA = 3.50$
(9.)

$\underline{Amortization} \\[3ex] PV = R1500000 \\[3ex] t = 20\:years \\[3ex] r = 10.5\% = \dfrac{10.5}{100} = 0.105 \\[5ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] PMT = ? \\[3ex] (9.1) \\[3ex] PMT = \dfrac{PV}{m} * \left[\dfrac{r}{1 - \left(1 + \dfrac{r}{m}\right)^{-mt}}\right] \\[10ex] PMT = \dfrac{1500000}{12} * \left[\dfrac{0.105}{1 - \left(1 + \dfrac{0.105}{12}\right)^{-1 * 12 * 20}}\right] \\[10ex] = 125000 * \left[\dfrac{0.105}{1 - \left(1 + 0.00875\right)^{-240}}\right] \\[7ex] = 125000 * \left[\dfrac{0.105}{1 - \left(1.00875\right)^{-240}}\right] \\[7ex] = 125000 * \left[\dfrac{0.105}{1 - 0.123580101}\right] \\[5ex] = 125000 * \left[\dfrac{0.105}{0.876419899}\right] \\[5ex] = \dfrac{125000 * 0.105}{0.876419899} \\[5ex] = \dfrac{13125}{0.876419899} \\[5ex] = 14975.6983 \\[3ex] \approx R14,975.70 \\[3ex] (9.2) \\[3ex] 12\:months = 1\:year \\[3ex] 144th\:\:payment \rightarrow \dfrac{144}{12} = 12th\:year\:\:payment \\[5ex] Remaining = 20 - 12 = 8\:years \\[3ex]$ Ask your students how they will calculate the outstanding balance after the $144^{th}$ payment.
Note their responses.
Some of the responses may include:
Find the total payment...for 20 years ...which is 240 months
Find the payment for 144 months
Subtract that payment from the total payment
...among other responses.

Remind them that these are Mortgage payments...Fixed Rate Mortgage
Show them how it works...using the example on my website
You can also use the Loan Amortization Calculator to confirm the calculations
Any monthly payment made is used to offset the interest earned during that period...first step
Then, any remaining balance is applied toward the principal.
The same principle applies to car payments and other amortized payments.

We can find the outstanding balance at the end of the $144th$ payemnt (at the end of $12$ years) in at least two ways.
Use any method you prefer.

First Method: Find the present value of the ordinary annuity for the remaining $8$ years.
OR
Second Method:
Compare to the previous response of "probably" one of your students.
But, explain the concept well for your students to understand.

Use the Compound Interest formula to calculate the future value of the loan amount for $12$ years.

Remember that the monthly payment of $14975.70$ was found using the Amortization formula for the entire term of the loan.

So, use the Future Value of an Ordinary Annuity formula and calculate the future value of that payment for $12$ years.
Then, subtract the future value of that payment from the future value of the loan amount.

$\underline{First\:\:Method} \\[3ex] \underline{Present\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] t = 8\:years \\[3ex] PMT = R14975.6983 \\[3ex] r = 0.105 \\[3ex] m = 12 \\[3ex] PV = m * PMT * \left[\dfrac{1 - \left(1 + \dfrac{r}{m}\right)^{-mt}}{r}\right] \\[10ex] PV = 12 * 14975.6983 * \left[\dfrac{1 - \left(1 + \dfrac{0.105}{12}\right)^{-1 * 12 * 8}}{0.105}\right] \\[10ex] = 179708.38 * \left[\dfrac{1 - \left(1 + 0.00875\right)^{-96}}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{1 - \left(1.00875\right)^{-96}}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{1 - 0.43329075}{0.105}\right] \\[5ex] = 179708.38 * \left[\dfrac{0.56670925}{0.105}\right] \\[5ex] = \dfrac{179708.38 * 0.56670925}{0.105} \\[5ex] = \dfrac{101842.401}{0.105} \\[5ex] = 969927.629 \\[3ex] \approx R969,927.63 \\[3ex] \underline{Second\:\:Method} \\[3ex] \underline{Compound\:\:Interest\:\:and\:\:Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] \underline{Compound\:\:Interest} \\[3ex] P = R1500000 \\[3ex] r = 0.105 \\[3ex] t = 12\:years \\[3ex] m = 12 \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] A = 1500000\left(1 + \dfrac{0.105}{12}\right)^{12 * 12} \\[7ex] = 1500000\left(1 + 0.00875\right)^{144} \\[7ex] = 1500000\left(1.00875\right)^{144} \\[7ex] = 1500000(3.50615308) \\[3ex] = R5259229.62 \\[3ex] \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] m = 12 \\[3ex] PMT = R14975.6983 \\[3ex] r = 0.105 \\[3ex] t = 12\:years \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 12 * 14975.6983 * \left[\dfrac{\left(1 + \dfrac{0.105}{12}\right)^{12 * 12} - 1}{0.105}\right] \\[10ex] = 179708.38 * \left[\dfrac{\left(1 + 0.00875\right)^{144} - 1}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{\left(1.00875\right)^{144} - 1}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{3.50615308 - 1}{0.105}\right] \\[5ex] = 179708.38 * \left[\dfrac{2.50615308}{0.105}\right] \\[5ex] = \dfrac{179708.38 * 2.50615308}{0.105} \\[5ex] = \dfrac{450376.71}{0.105} \\[5ex] = R4289302 \\[3ex] \underline{Balance\:\:after\:\:144th\:\:payment} \\[3ex] Balance = 5259229.62 - 4289302 = R969,927.62 \\[3ex]$  (10.)

$\underline{First\:\:Semester} \\[3ex] \Sigma CH = 12 \\[3ex] GPA = 3.583333333 \\[3ex] \Sigma GP = 3.583333333(12) = 43 \\[3ex] \underline{Second\:\:Semester} \\[3ex] \Sigma CH = 16 \\[3ex] \Sigma GP = 55 \\[3ex] \underline{First\:\:and\:\:Second\:\:Semesters} \\[3ex] \Sigma CH = 12 + 16 = 28 \\[3ex] \Sigma GP = 43 + 55 = 98 \\[3ex] CGPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] CGPA = \dfrac{98}{28} \\[5ex] CGPA = 3.5 \\[3ex] CGPA = 3.50$