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Math Help - load and inflation

  1. #1
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    load and inflation

    hello

    let say you get a loan of 1,000 with bank rate of interest 5% for 10 years

    an online calculator says that you will totally have paid at the end of the 10 year period 1272.79, each month 10,61 (the 272.79 are the total interest)

    how are these calculated? I mean, what is the procedure?

    at the same time there is inflation of x,y,z% etc each year (or for making it more easy, let's say standard inflation 3% each year)

    I think that at the end, you will ofcourse have paid a total of 1272.79, but given the fact that inflation made the doses become more and more "cheaper", the total amount will "seem" less, do I make any mistake here?

    if not, how can that less amount be calculated? any equation?
    (for standard inflation or variable per year)

    thanks
    Last edited by mr fantastic; June 22nd 2009 at 12:54 AM. Reason: Merged posts
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  2. #2
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    Hi mathos,

    First we must know that the bank rate of interest is compound and is annual. A certain amount is paid monthly, we can presume that this is fixed per month. To start we must define some terms;

    \mu_n~ : ~\mbox{The total amount needed to be paid back after n months.}

    \lambda~ : ~\mbox{The value of the loan taken.}

    \imath~:~\mbox{The rate of interest, monthly, expressed as a decimal.}

    \varphi~:~\mbox{The amount paid per month.}

    At the start the total amount to be paid is the total amount borrowed since no interest has yet to be accumulated. In other words after 0 months the amount to be paid back is what was borrowed, hence

    \mu_0=\lambda

    Now after the first month, interest has been added. The value of the interest for the first month is the monthly rate of interest multiplied by the initial value of the loan. Also some of the loan is paid back after the first month and so from our definitions

    \mu_1=\mu_0+\lambda\imath-\varphi

    =\lambda+\lambda\imath-\varphi

    =\lambda(1+\imath)-\varphi~~~(*)

    After the second month, more interest has been added. The value of the interest added for the second month is the monthly rate of interest multiplied by the amount needed to be paid back after the first month. Again some of the loan is paid back, hence

    \mu_2=\mu_1+\mu_1\imath-\varphi

    =\mu_1(1+\imath)-\varphi

    From (*) it becomes clear that

    \mu_2=\left[\lambda(1+\imath)-\varphi\right](1+\imath)-\varphi

    =\lambda(1+\imath)^2-\varphi(1+\imath)-\varphi

    =\lambda(1+\imath)^2-\varphi\left[1+(1+\imath)\right]~~~(**)

    The third month follows the same procedure and so

    \mu_3=\mu_2+\mu_2\imath-\varphi

    Using (**) and rearranging some we arrive at

    \mu_3=\lambda(1+\imath)^3-\varphi\left[1+(1+\imath)+(1+\imath)^2\right]

    So we can conjecture that

    \mu_n=\lambda(1+\imath)^n-\varphi\left[1+(1+\imath)+(1+\imath)^2+ ... +(1+\imath)^{n-1}\right]

    =\lambda(1+\imath)^n-\varphi\cdot\sum_{k=0}^{n-1}(1+\imath)^k

    We can prove this by induction, but for this purpose we will assume it to be true. Now

    \sum_{k=0}^{n-1}(1+\imath)^k

    is a Geometric Progression with a=1 and r=(1+\imath) so

    \sum_{k=1}^{n-1}(1+\imath)^k=\frac{(1+\imath)^n-1}{1+\imath-1}=\frac{1}{\imath}\cdot\left[(1+\imath)^n-1\right]

    hence

    \mu_n=\lambda(1+\imath)^n-\frac{\varphi}{\imath}\cdot\left[(1+\imath)^n-1\right]

    So this is the formula for the amount needed to be paid back after n months. Clearly after a certain number of months, say N, all will need to be paid back and so \mu_N=0 and thus

    \lambda(1+\imath)^N-\frac{\varphi}{\imath}\cdot\left[(1+\imath)^N-1\right]=0

    \implies \lambda\imath(1+\imath)^N=\varphi\left[(1+\imath)^N-1\right]

    \implies \varphi=\frac{\lambda\imath(1+\imath)^N}{(1+\imath  )^N-1}=\frac{\lambda\imath}{1-(1+\imath)^{-N}}

    We know that \lambda=1000. We also know that after 10 years or 120 months all of the loan plus the interest must be paid back and so N=120. Also the rate of interest is 5% annually which is (5/12)% monthly and so

    \imath=\frac{5}{1200}=\frac{1}{240}

    hence the amount needed to be paid monthly is

    \varphi=\frac{1000(1/240)}{1-(241/240)^{-120}}\approx10.61

    The total amount to be paid is therefore the amount needed to be paid monthly multiplied by the number of months which is approximately equal to

     <br />
10.61\times120 = 1272.79<br />

    Inflation would not affect the amount of interest paid directly. Note that the this interest is fixed at an annual rate of 5%. However, that being said inflation may have an indirect impact as inflation tends to lead to wage rate increases which will lead to higher nominal incomes and thus meaning that consumers may decide to pay more back monthly since as a proportion of total income, the amount paid back on this loan will decrease.

    Hope this helps.
    Last edited by Sean12345; June 23rd 2009 at 03:12 AM.
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  3. #3
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    Quote Originally Posted by Sean12345 View Post
    Inflation would not affect the amount of interest paid directly. Note that the this interest is fixed at an annual rate of 5%. However, that being said inflation may have an indirect impact as inflation tends to lead to wage rate increases which will lead to higher nominal incomes and thus meaning that consumers may decide to pay more back monthly since as a proportion of total income, the amount paid back on this loan will decrease.
    Hope this helps.
    thanks for your reply

    as for the second question, about how inflation alters the total paid amount:

    your opinion of how inflation affects the total paid amount by increasing salaries and thus lowering the total period of full payment of the load (which is the choice and will of the own that got the loan) is very useful

    however, I was thinking of another indirect way that inflation affects the total amount paid, without altering, by choice of the person that got the loan, the total period of payment of the loan:

    in the same example, the total amount paid at the end of the 20 year period, is 1272.79

    ofcourse, the first doses of the loan and the last ones as well, all they are 10.61 in absolut amount

    however, 10.61 in 20 years ago are much money, than 10.61 now, if the inflation is increasing each year

    so I wonder if eventually, we will not have paid 1272.79, but less, since the later doses will have less value, due to inflation increase

    that less amount ofcourse, will not be the absolute total amount of the payment, but it will be an amount that will be ABLE to be compared with other amounts of the present year (now)

    this is called deflation or disinflation (not sure) of amounts, that enables us to compare amounts of past years with amounts of the present year

    maybe this is a mistake, and the total absolute amount paid has nothing to do with inflation

    but, is this safe to say so? in order to compare the total amount paid in 20 year loan with the amount that we ended up with at the present year? to check how much are the gains?

    thanks
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  4. #4
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    Quote Originally Posted by mathos View Post
    however, 10.61 in 20 years ago are much money, than 10.61 now, if the inflation is increasing each year
    Inflation does not directly mean that things are cheaper. In fact inflation means that prices across the economy are rising. However with increases in prices often comes increases in REAL incomes (i.e. income adjusted for the trends in inflation), especially during boom periods. It's this increase in nominal and real income which means money is less value than it was years ago. It is the increase in real income which means goods account for a small proportion of income than years back.

    Take the following example;

    Let's denote some terms;

    P_y~:~\mbox{Price of Good x in y years after 2000 expressed in sterling.}
    L_y~:~\mbox{Level of disposable income in y years after 2000 expressed in sterling.}
    D_y~:~\mbox{Proportion of disposable income taken by good x in y years after 2000.}

    Now let's take a fictional scenario. Say in the year 2000 the price of a certain good is 10 and the level of disposable income is 1000 (remember this is fictional) and so

    P_0=10

    L_0=1000

    Note that the proportion of disposable income taken by good x in this year is

    D_0=\frac{P_0}{L_0}\cdot 100 = 1

    Now lets say that in 2001 due to inflation of 10% good x costs 11. People tend to bargain for real wage increases. So if they spot that inflation is going to be 10% they will bargain for 15% increases in wages and so wages in 2001 will be 1150 therefore

    P_1=11

    L_1=1150

    hence

    D_1=\frac{P_1}{L_1}\cdot 100 \approx 0.957 < 1

    thus the proportion is lower in 2001 and so goods appear to be cheaper.

    Hope this helps.
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