Prepared by Khanimamba Ntshana
install.packages(“tinytex”) tinytex::install_tinytex()
\(PMT_{29} = 7000 + (25 - 1)\times(-250) = 70000 - 24\times250 = 1000\)
from the fact that the payments follow a linear sequence with the general formula for the n-th term being
\[T_{n} = a + {(n-1)}\times{d}\]
Hence we have the value of the original loan as:
\[PV = 7000v + 6750v^2+ 6500v^3 + ... + 1000v^{25}\]
\[PV = 7250\times{(v+v^2+v^3 + ... + v^{25})} - 250\times{(v+2v^2+3v^3+...+25v^{25})}\] and this simplifies to: \[PV = 7250a_{\overline{25}|} - 250(Ia)_{\overline{25}|}\] \[a_{\overline{25}|} = \frac{1-1.085^{-25}}{0.085} = 10.23419078\]
\[(Ia)_{\overline{25}|} = \frac{\frac{1-1.085^{-25}}{1-1.085^{-1}} - 25\times1.085^{-25}} {.085} = 92.37355765\] \(\text{And so,}\)
\[\text{Loan Amount} = PV = R 51104.49\]
b)Before constructing the schedule, we first need to caluclate the loan outstanding just after the seventh repayment (L7, which is the loan outstanding at the beginning of year 8) and subsequently the loan outstanding immediately after the eighth repayment (L8, which is the loan outstanding at the beginning of year 9) and the respective interest and capital repayments for that year.
For L7, as at the beginning of the 8th year we have 18 (25 - 7 = 18) repayments left.The value of the next repayment (repayment at the end of year 8) is :
\[PMT_ 8 = 7000 - (8-1)\times 250 = 7000 - 7\times250 = 5250\] \[\text{And so :}\]
\[L_7 = 5250v + 5000v^2 + 4750v^3 + ... + 1000v^18\] \[L_7 = 5500(v+v^2+v^3+...+v^{18}) - 250(v+2v^2+3V^3+ ... + 18v^{18})\]
\[L_7 = 5500a_{\overline{18}|} - 250(Ia)_{\overline{18}|}\] \[a_{\overline{18}|} = \frac{1-1.085^{-18}}{.085} = 9.055476438\] \[(Ia)_{\overline{18}|} = \frac{\frac{1-1.085^{-18}}{1-1.085^{-1}} - 18\times1.085^{-18}}{.085 }= 66.82436335\] \[\implies L_7 = R 33099.03\]
After this, we can easily get \(X_8\),\(L_8\), \(X_9\) and the interest and capital elements of \(X_8\) AND \(X_9\):
\[X_8 = R5250\]
The interest paid on the 8th payment (\(b_8\)) is:
\[b_8 = 0.085\times{L_7} = 0.085\times33099.03 = R2813.42\] Which means the capital element (\(f_8\)) is;
\[f_8 = 5250 - 2813.42 = R2436.58\] and, \[L_8 = 33099.03-2436.58 = 30622.45\] \[X_9 = 5000 , b_9 = 0.085\times30622.45 = 2606.31, f_9 = 5000-2606.31 = 2393.69\]
And we can put this in a pretty table as below:
| Year | Loan o/s at start of year | Payment | Interest | Capital |
|---|---|---|---|---|
| 8 | 33099.03 | 5250 | 2813.42 | 2436.58 |
| 9 | 30662.45 | 5000 | 2606.31 | 2393.69 |
\[PV = 4200a_{\overline15|} - 200(Ia)_{\overline15|}\] \[ and , a_{\overline15|} = \frac{1-(1+1.04)^{-15}}{.04} = 11.11838743\]
\[(Ia)_{\overline15|} = \frac{\frac{1-(1.04)^{-15}}{1-(1.04)^{-1}}-15(1.04)^{-15}}{.04} = 80.85388472\]
So the loan amount is, \[PV = 4200\times11.11838743 - 200\times80.858388472\] \[PV = R30526.45\]
\[X_8 = 4000 + (8-1)\times-200 = 2600\] and so, \[L_7 = 2800(Ia)_{\overline8|} - 200(Ia)_{\overline8|}\] \[a_{\overline8|} = \frac{1-(1.04)^{-8}}{.04} =6.732744875\] \[(Ia)_{\overline8|} = \frac{\frac{1-(1.04)^{-8}}{1-(1.04)^{-1}} - 8(1.04)^{-8}}{.04} = 28.91332575\] So,\[PV = 3200\times6.732744875 - 200\times28.91332575 = R13069.0205\] So now that we have the loan outstanding right after the 7th payment , we can easily get the 8th installment, interest and capital repayments as:
\[X_8 = R2400, b_8 = 13069.0205\times0.04 = R522.76082,f_8 = 2600 - 522.76082 = R2077.24\]
\[L10 = 2200a_{\overline5|} - 200(Ia)_{\overline5|}\] \[a_{\overline5|} = \frac{1-(1.04)^{-10}}{.04} = 4.451822331\] \[(Ia)_{\overline5|} =\frac{\frac{1-(1.04)^{-10}}{1-(1.04)^{-1}} - 10(1.04)^{-10}}{.04} = 13.00649226\]
\[\implies L_{10} = 2200\times 4.451822331 - 200\times13.00649226=R7192.710676\] This amount was cut in half, so the value of the “new” loan is: \[\text{New loan amount} = \frac{1}{2}\times 7192.710676 = R3596.355338\] And the repayment structure is as follows: \[PV = Xv + (X+20)v^2 + (X+2\times20)v^3 + ...+(X+9\times20)v^10\] \[\implies 3596.355338 = (X-20)a_{\overline10|} + 20(Ia)_{\overline10|}\] \[\implies X = \frac{3596.355338 -20(Ia)_{\overline10|}+20a_{\overline10|}}{a_{\overline10|}}\] Since, \[a_{\overline10|} = \frac{1-(1.1)^{-10}}{.1} = 6.144567106\] and, \[(Ia)_{\overline10|} =\frac{\frac{1-(1.1)^{-10}}{1-(1.1)^{-1}} - 10(1.1)^{-10}}{.1} = 29.03590922\] \[\implies X = R510.78\]
\[PV = 4Xa^{(4)}_{\overline5|} + 4(X+500)Xa^{(4)}_{\overline5|}v^5 +...+4(X+1500)a^{(4)}_{\overline5|}v^{15}\] \[PV = 4a^{(4)}_{\overline5|}(X+(X+500)v^5+...+(X+1500)v^{15})\] \[PV =4a^{(4)}_{\overline5|}(X+(X+500)v_j+..+(X+1500)v^3_j))\] Where j is the 5-yearly effective interest rate
\[PV = 4a^{(4)}_{\overline5|i}((X-500)\ddot{a}_{\overline4|j} +500(I\ddot{a})_{\overline4|j)})\] \[\implies X = \frac{\frac{160000}{4a^{(4)}_{\overline5|i}}-500(I\ddot{a})_{\overline4|j)}+500{\ddot{a}_{\overline4|j}}}{\ddot{a}_{\overline4|j}}\]
Since \[PV = R160000\] Now,\[i = (1+\frac{8%}{2})^2 - 1 = 0.0816\] So \[j = (1.0816)^5 - 1 = 0.4802442849\] And,\[a^{(4)}_{\overline5|i} = \frac{1-(1.0816)^{-5}}{4((1.0816)^{\frac{1}{4}} -1)} = 4.095604737\]
\[\ddot{a}_{\overline4|j} = \frac{1-(1.4802442849)^{-4}}{1-1.4802442849^{-1}} = 2.440269783\] \[(I\ddot{a})_{\overline4|j)} =\frac{\frac{1-1.4802442849^{-4}}{1-1.4802442849^{-1}} - 4\times1.4802442849)^{-4}}{1-1.4802442849^{-4}} = 4.953563848\]
\[\implies X = R3487.29\]
1 January 1999 is the end of the second quarter, so only one quarterly payment has been made before that, which was \(X_1 = 3487.29\) and \(f_1 = 160000\times(1.0816^{\frac{1}{4}} - 1) = 3168.624435\),\(b_1 = 3487.29-3168.624435 =R318.665565\).
\(L_2 = 160000-318.665565 = R159681.3344\) \(b_2 = 3487.29-R159681.3344\times(1.0816^{\frac{1}{4}} - 1)= R324.98\)
Because we’re smart, and we ALL drew our timelines, we know that 1 July 2011 is the end of year 13 (quarter \(13\times4 = 52\)).So we want the loan outstanding as at year 13 (quarter 52).
Hence,\[L_{52} = 4\times4487.29a^{(4)}_{\overline2|i}+4\times4987.49a^{(4)}_{\overline5|i}v^{2}\] \[a^{(4)}_{\overline2|i} =\frac{1-(1.0816)^{-2}}{4((1.0816)^\frac{1}{4} -1)} = 1.832919135\] and \(a^{(4)}_{\overline5|i} = 4.095604737\) from before.So \(L_{52} = R102740.17\)
And if you don’t see why this is, its because you did not draw your timeline.DRAW YOUR TIMELINE!
\[PV = 4a^{(4)}_{\overline5|}(X+(X+40)v_j+(X+80)v^2_j)\] Where j is the 5-yearly effeective interest rate.
\[\implies PV = 4a^{(4)}_{\overline5|i}((X-40)\ddot{a}_{\overline3|j} +40(I\ddot{a})_{\overline3|j})\] \[ \implies X = \frac{\dfrac{11820}{4a^{(4)}_{\overline{5|}}} - 40(I\ddot{a})_{\overline{3|}j} + 40\ddot{a}_{\overline{3|}j}} {\ddot{a}_{\overline{3|}j}}\]
\[a^{(4)}_{\overline5|i} = \frac{1-(1.12)^{-5}}{4((1.12)^{\frac{1}{4}} -1)} = 3.763162085\]
\[j = (1.12)^{5} - 1 = .7623416823\]
\[\ddot{a}_{\overline3|j} = \frac{1-(1.7623416823)^{-3}}{1-(1.7623416823)^{-1}} = 1.889400092\]
\[(I\ddot{a})_{\overline3|j}) = \frac{\frac{1-(1.7623416823)^{-3}}{1-(1.7623416823)^{-1}} - 3(1.7623416823)^{-3}}{1-(1.7623416823)^{-1}} = 3.100773421\]
So, \[\implies X = 389.96\]
bi) We’ll use the retrospective method to get the loan outstanding after the second year(fourth quarter) to switch it up.
\[L_4 =11820(1.12)^2 - 4\times 389.96a^{(4)}_{\overline2|}(1.12)^2\] \[a^{(4)}_{\overline2|} = \frac{1-(1.12)^2}{4((1.12)^{\frac{1}{4}}-1)}=1.764308119\] \[\implies L_4 = R11374.85106\]
\(X_5 = 389.96\), and the total principle amount paid during that year is the The loan outstanding at the beginning of that year minus all the quarterly interest payments,i.e:
\[b_5 = (389.96-11374.85106\times((1.12)^{\frac{1}{4}} - 1) = 63.07698393\] \[L_5 = 11374.85106-63.07698393 = 11311.77408\] \[b_6 = (389.96-11311.77408\times((1.12)^{\frac{1}{4}} - 1) = 64.88964896\] \[L_6 = 11311.77408-64.8896489 = 11246.88443\] \[b_7 = (389.96-11246.88443\times((1.12)^{\frac{1}{4}} - 1) = 66.75440506\]
\[L_7 = 11246.88443 - 66.75440506 = 11180.13002\] \[b_8 = (389.96-11180.13002\times((1.12)^{\frac{1}{4}} - 1) = 68.67274944\]
And , finally, \[\text{Total capital repaid} = b_5+b_6+b_7+b_8 = R263.39\]
bii) Same idea as in bi.
\[L_{33} = 4\times (389.96 + 40)a^{(4)}_{\overline1.75|} + 4\times(389.96 + 80)a^{(4)}_{\overline5|})v^{1.75}\]
And,\[a^{(4)}_{\overline1.25|} = \frac{1-(1.12)^{-1.75}}{4(1.12)^{1/4} - 1)} = 1.56500965\] \[a^{(4)}_{\overline5|} = 3.763162085\].
hence, \[L_{33} = R8493.092574\] since the remaining term is \(15\times4 - 33 = \text{27 quarters} = \text{6.75 years}\),we have
\[8493.092575 = 4Xa^{(4)}_{\overline6.75|}\] at the new level repayments X.
\[\implies X = \frac{8493.092575}{4a^{(4)}_{\overline6.75|}}\]
\[a^{(4)}_{\overline6.75|} = \frac{1-(1.12)^{-6.75}}{4((1.12)^{1/4} - 1)} = 4.651190589\] \[\implies X = R456.50\]