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Math Help - Annuities certain (arrear and due)

  1. #1
    Senior Member
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    Annuities certain (arrear and due)

    Hey guys, I have a few questions that I've been stuck on for a few days, these questions have no solutions provided so I was wondering if anyone can check them for me. The first 3 questions I have completed with full working however I'm not entirely sure if they are correct, the last 4 I have no clue how to solve :3 Any help would be appreciated!



    So first we need to work out the amount of each installment, R, thus 1000000=Ra_{25:0.08} \implies R = 93678.779

    Now X purchases the annuity on 1 Jan 1986, so we need to work out the price that X bought the annuity for.

    Since he purchased it on 1 Jan 1986 and the annuity expires at the end of 2005, the annuity has a maturity left of 19 years.

    But since X pays a tax rate of 0.45 on the interest portion of each payment of R, then we need to split up R each period to find the interest payment and capital payment so we can account for the tax.

    To derive an expression for the interest payment and capital payment of R each period we let L be the lump sum borrowed at the start of an arbitrary period, n be the number of periods and i be the effective interest rate per period. Then for the first payment period, the loan outstanding at the start of the period is L, the interest payment for the first period is iL = i(Ra_{n:i}) = iR\left(\frac{1-v^n}{i}\right) = R(1-v^n) where v = (1+i)^{-1}. Then clearly the capital component paid at the end of the first period is Rv^n since R(1-v^n) +Rv^n = R, we can show in general that for payment period k, the interest component paid at the end of the period is R(1-v^{n-k+1}) and the capital component paid at the end of the period is Rv^{n-k+1}

    Now back to the question: let u = (1+0.09)^{-1} and v = (1+0.08)^{-1}

    The capital payments of each period starting from 1986 until the end of 2005 when discounted back to 1986 form the cash flow stream:

    u(Rv^{19}) + u^2(Rv^{18}) + \cdots + u^{19}(Rv)

    The interest payments of each period starting from 1986 until the end of 2005 when discounted back to 1986 form the cash flow stream: (Note we take into consideration the tax rate of 45% which is (1-0.45) = 0.55 of the interest payments)

    u(0.55R(1-v^{19})) + u^2(0.55R(1-v^{18})) + \cdots + u^{19}(0.55R(1-v))

    Now we need to combine both of these cash flow streams and form one single expression to evaluate it.

    Let us match together the v^{k} components:

    u(Rv^{19}) + u(0.55R(1-v^{19})) = 0.45uRv^{19} + 0.55uR

    u^2(Rv^{18}) + u^2(0.55R(1-v^{18})) = 0.45u^2Rv^{18} +0.55u^2R

    And so on... so let us now group together the expressions with 0.55 as the coefficient:

    0.55uR + 0.55u^2R + \cdots + 0.55u^{19}R = 0.55R(u+u^2+ \cdots + u^{19}) \cdots [1]

    Now group together the expressions with 0.45 as the coefficient:

    0.45uRv^{19} + 0.45u^2Rv^{18} + \cdots + 0.45u^{19}Rv = 0.45R(uv^{19} + u^2v^{18} + \cdots + u^{19}v) \cdots [2]

    [1] = 0.55Ra_{19:0.09} = 0.55(93678.779)\left(\frac{1-(1.09)^{-19}}{0.09}\right) = 461139.703695

    [2] = 0.45R\left(uv^{19}\left(\frac{1-(uv^{-1})^{19}}{1-uv^{-1}}\right)\right) = 156912.68

    Thus [1]+[2] = 618052.3837

    Now let the lump sum paid on April 1995 be L, L must satisfy the following equation of value: (Let q = 1.1^{-1})

    \left[0.55R(q+q^2+\cdots+q^9)\right] + \left[0.45R(qv^{19}+q^2v^{18} + \cdots + q^9v^{11}\right] + Lq^{9.25} = 618052.3837

    0.55R\left(\frac{1-1.1^{-9}}{0.1}\right) + 0.45R\left[qv^{19}\left(\frac{1-(qv^{-1})^9}{1-qv^{-1}}\right)\right] + Lq^{9.25} = 618052.3837

    Solving the above yields L = 596411.93

    [hr]



    First we need to work out the amount of each installment paid in advance from 1 Jan 1985 to 2005 (20 years).

    Let each installment be R.

    First note (1+0.1) = (1-d)^{-1} where d is the effective annual rate of discount.

    d = \frac{1}{11}

    10000=R\ddot{a}_{20:0.1} \implies 10000=R\left(\frac{1-1.1^{-20}}{d}\right) \implies R = 1067.81477

    Now we need to find on 1 Jan 1990 the total outstanding capital left to be repaid.

    This is given by R\ddot{a}_{15:0.1} = R\left(\frac{1-1.1^{-15}}{d}\right) = 8934.07

    Now from 1 Jan 1990, if B were to increase the yield to 0.12, then we need to find the new installments per period, these new R is given by:

    (Note (1+0.12) = (1-d)^{-1} \implies d = \frac{3}{28})

    8934.07 = R\ddot{a}_{15:0.12}  \implies 8934.07 = R\left(\frac{1-1.12^{-15}}{\frac{3}{28}}\right) \implies R = 1171.19499

    Now let A's final payment be L, L must satisfy the following equation of value:

    R\ddot{a}_{5:0.12} + L(1.12)^{-5}= 8934.07

    \implies R\left(\frac{1-1.12^{-5}}{\frac{3}{28}}\right) + L(1.12)^{-5} = 8934.07

    \implies L = 7411.61448

    [hr]



    Let C_n be the capital component paid in arrear for the nth period and I_n be the interest component paid in arrear for the nth period. Let L_0 be the loan outstanding.

    From the question C_1+(1-0.3)I_1 = 5000

    But I_1 = 0.1L_0 \implies C_1 +0.07L_0 = 5000

    Likewise, C_2+0.7I_2=5000

    But I_2 =0.1L_1 = 0.1(L_0 - C_1) [This is obvious as L_1 represent the loan outstanding at the end of the first period, which is equal to the loan outstanding at the beginning of the period minus the capital paid in the first period]

    So C_2+0.07(L_0-C_1) = 5000

    Furthering this pattern, C_3+0.7I_3 = 5000

    But I_3 = 0.1L_2 = 0.1(L_1-C_2) = 0.1(L_0-C_1-C_2)

    Hence C_3+0.07(L_0-C_1-C_2)=5000

    So we have:

    C_1 +0.07L_0 = 5000 \cdots [1]

    C_2+0.07(L_0-C_1) = 5000 \cdots [2]

    C_3+0.07(L_0-C_1-C_2)=5000 \cdots [3]

    From [1] we have C_1 = 5000 - 0.07L_0

    So [2] becomes C_2+0.07(L_0 - 5000 +0.07L_0) = 5000 \implies C_2 + 0.07L_0 - 0.07(5000) + 0.07^2L_0 = 5000 \implies C_2 = 5000+0.07(5000) - (0.07+0.07^2)L_0 \implies C_2 = 1.07(5000-0.07L_0) = 1.07C_1

    In [3] we have:

    C_3 = 5000- 0.07(L_0-C_1-C_2) \implies C_3 = 0.07C_1+0.07(1.07)C_1+C_1 \implies C_3 = 1.07C_1 + 0.07(1.07)C_1 = 1.07^2C_1

    This forms a recurrence relation and we can show that C_n = 1.07^{n-1}C_1

    Now we need to find what L_0 is. Note that if we assume the person who bought this decreasing annuity does not pay tax, then we should have:

    \sum_{i=1}^{10} \left(C_i+I_i\right)v^i = L_0 where v = 1.1^{-1}

    \implies \sum_{i=1}^{10} \left(C_i + \frac{1}{0.7}(5000-C_i)\right)v^i=L_0 \implies \sum_{i=1}^{10} \left(1.07^{i-1}C_1 + \frac{1}{0.7}(5000-1.07^{i-1}C_1)\right)v^i=L_0 \implies \sum_{i=1}^{10} \left(1.07^{i-1}(5000-0.07L_0) + \frac{1}{0.7}(5000-1.07^{i-1}(5000-0.07L_0))\right)v^i=L_0

    Now we can solve for L_0, L_0 = 35117.0977

    The price that someone would pay for this decreasing annuity if they do not pay tax is given by:

    \sum_{i=1}^{10} \left(1.07^{i-1}(5000-0.07L_0) + \frac{1}{0.7}(5000-1.07^{i-1}(5000-0.07L_0))\right)v^i where v^i = 1.08^{-i}

    The price is 38252.909

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  2. #2
    MHF Contributor
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    Re: Annuities certain (arrear and due)

    Gave 1st one a quick read; quite confused:

    Mr.X steps in Jan.1/86 and buys annuity (19 annual payments of $93,678.78)
    at 9%, thus for $838,435.82.
    So whatever happened before is now history; really, the whole problem could
    start with the above; no need to bring in anything prior to Jan.1/86.

    Then you've calculated that if Mr.X receives $596,411.93 on Apr.1/95 (so after
    receiving 9 of the annual payments), then Mr.X has realised 10% on:
    "THE ENTIRE TRANSACTION" : what d'heck does that mean?

    Also confusing is your v = 1.08^(-1); how is it possible that the 8% (which was
    the rate BEFORE Mr.X stepped in) be used in ANY calculation afterwards?

    Can you clarify? If I'm dumb for having missed something: tell me
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