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Thread: Force of interest problem

  1. #1
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    Question Force of interest problem

    Hi,
    i'm looking for help with a force of interest problem that i'm having great trouble with. Any help would be greatly appreciated. Here's the question:

    An investor deposited 1,000,000 in an account with a variable interest rate. The deposit had accumulated to 1,036,915.04 after 6 months and 1,056,870.83 after 9 months. Under the supposition that the force of interest is a linear function of time, find the amount accumulated after 1 year.

    please help with this i don't even no where to start

  2. #2
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    This would be my guess, but I haven't reviewed force of interest problems lately, so maybe compare my idea with the answer of course.

    Since the force of interest is variable, the force of interest you use to accumulate the 1 million to time t=.5 is not the same force of interest you use to accumulate the 1 million to time t=.75, both starting at time t=0.

    Find the force of interest that gives you the said accumulated values. You now have two points on a line, and can use this to estimate the force of interest used to get from t=0 to t=1.

    I get a force of interest from t=0 to t=1 as ~.075 so the accumulated value at t=1 should be 1000000e^.075=1077884.141

    Just a stab

  3. #3
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    Force of interest is a linear function of time, so:
    delta(t) = mt+b
    a(t) = e^ [ int_0^t ms+b ds ]
    a(t) = e^ [1/2 mt^2+bt ]
    A(t) = (1,000,000) e^ [1/2 mt^2+bt ]
    Now use the two other values you have for A(t) to solve for m and b.
    Last edited by nonlnear; October 3rd 2010 at 05:02 PM.

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    Quote Originally Posted by Ronan1990 View Post
    Hi,
    i'm looking for help with a force of interest problem that i'm having great trouble with. Any help would be greatly appreciated. Here's the question:

    An investor deposited 1,000,000 in an account with a variable interest rate. The deposit had accumulated to 1,036,915.04 after 6 months and 1,056,870.83 after 9 months. Under the supposition that the force of interest is a linear function of time, find the amount accumulated after 1 year.

    please help with this i don't even no where to start
    Force of Interest means that interest is accumulated every moment. A force of interest of mu would mean that an investment of 1 would accumulate to e^(mu*t) after t years. When the force of interest isn't constant, you have to integrate the exponent. For a constant of mu: integrate[mu, dt] = mu*t. But you were told that for this problem, the force of interest is a linear function of time so the equation that's going to rule this problem is integrate[m t + b, dt] = .5 m t^2 + b t. I'll call this f(t).
    e^f(.5) = 1.03691504
    e^f(.75) = 1.05687083
    take the natural log of both sides:
    ln(1.03691504) = .125 m + .5 b
    ln(1.05687083) = .28125 m + .75 b

    Solve for m and b using your favorite methods.

  5. #5
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    I get the same answer either way, its the same concept.

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    Quote Originally Posted by brandond View Post
    I get the same answer either way, its the same concept.
    Yes, you're right. Sorry about my first sentence (now redacted). It works out the same way because the problem asks about a linear force of interest. Your procedure takes the average force of interest from 0 to 0.5, and the average from 0 to 0.75, constructs a linear function for the average force of interest, and then uses the average force of interest from 0 to 1. Fortunately it works out for this problem that the two calculations give identical solutions. However if the problem posed almost any nonlinear force of interest function (and they do show up on Exam P), then the direct method is the only reliable way to get a solution.

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    Quote Originally Posted by Kenneth View Post
    Force of Interest means that interest is accumulated every moment. A force of interest of mu would mean that an investment of 1 would accumulate to e^(mu*t) after t years. When the force of interest isn't constant, you have to integrate the exponent. For a constant of mu: integrate[mu, dt] = mu*t. But you were told that for this problem, the force of interest is a linear function of time so the equation that's going to rule this problem is integrate[m t + b, dt] = .5 m t^2 + b t. I'll call this f(t).
    e^f(.5) = 1.03691504
    e^f(.75) = 1.05687083
    take the natural log of both sides:
    ln(1.03691504) = .125 m + .5 b
    ln(1.05687083) = .28125 m + .75 b

    Solve for m and b using your favorite methods.
    Thanks a lot for that I understand the problem now good explanation

  8. #8
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    Quote Originally Posted by nonlnear View Post
    Yes, you're right. Sorry about my first sentence (now redacted). It works out the same way because the problem asks about a linear force of interest. Your procedure takes the average force of interest from 0 to 0.5, and the average from 0 to 0.75, constructs a linear function for the average force of interest, and then uses the average force of interest from 0 to 1. Fortunately it works out for this problem that the two calculations give identical solutions. However if the problem posed almost any nonlinear force of interest function (and they do show up on Exam P), then the direct method is the only reliable way to get a solution.
    Agreed, I got lucky with my guess.

  9. #9
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    Quote Originally Posted by Kenneth View Post
    Force of Interest means that interest is accumulated every moment. A force of interest of mu would mean that an investment of 1 would accumulate to e^(mu*t) after t years. When the force of interest isn't constant, you have to integrate the exponent. For a constant of mu: integrate[mu, dt] = mu*t. But you were told that for this problem, the force of interest is a linear function of time so the equation that's going to rule this problem is integrate[m t + b, dt] = .5 m t^2 + b t. I'll call this f(t).
    e^f(.5) = 1.03691504
    e^f(.75) = 1.05687083
    take the natural log of both sides:
    ln(1.03691504) = .125 m + .5 b
    ln(1.05687083) = .28125 m + .75 b

    Solve for m and b using your favorite methods.
    i solved for m and b and got a total accumulated value of 1,333,961.91 but i don't know if that's correct. For m i got -0.02445535201 and for b i got 0.1379452925

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    Quote Originally Posted by Ronan1990 View Post
    i solved for m and b and got a total accumulated value of 1,333,961.91 but i don't know if that's correct. For m i got -0.02445535201 and for b i got 0.1379452925
    Nope. Did you take the natural log first? The interest rate is increasing over time, not decreasing.

    I get m = .01 and b = .07

    It's easy to drop a step in force of interest problems. You'll often get an answer that isn't obviously wrong. That answer is also sometimes one of the other multiple-choice answers.

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