Banner Ad 1



No announcement yet.

Variable force of interest question

  • Filter
  • Time
  • Show
Clear All
new posts

  • Variable force of interest question

    I'm looking at Cherry section 1g #13

    I know the answer to this kind of problem, but I'm not clear on the why...

    given that a fund accumulates at a force of interest:

    kt for 0<t<=5

    .04kt^2 for 5<t<=10

    and given that a deposit of 1 will accumulate to 2.7183 in 10 years, find k.

    The problem I have is that I want to take the AV of the $1 after the first 5 years, then use that figure (here e^(25k/2)) as the deposit that accumulates for the next 5 years. So, I end up with e^(25k/2) times e^(875k/75). This is wrong, as it is necessary to add the two integrals together instead of mulitply.

    Instinctually I know the correct way to proceed, but I get tripped up because I'm uncertain as to why I need to add instead of multiply.

    Can anyone clear this up for me?

  • #2
    Product rule: (a^m)*(a^n) = a^(m+n)
    Power of a power rule: (a^m)^n = a^(mn)

    Does this help? I am sure you already know this, maybe you are just looking at the problem wrong?


    • #3
      I think your trouble lies with the variable force of interest over time. Remember section 1a(iii):

      AV = (1+i1)(1+i2)...every period 1 and 2 has it's own rate and gets multiplied with the next period. In your case

      the rate lasts for more than one year so you get;

      AV = (1+i)^5*(1+j)^5 apply that to growth in terms of Force of interest

      you get;

      AV = e^int(0,5) of ktdt*e^int(5,10) of 1/25kt^2dt

      The rest is the Algebra bmathew22 described. Ask more question if you don't follow our explanations.


      • #4
        2 mistakes for the price of 1

        Thank you, humans.

        It's always easier to find 1 mistake instead of 2.

        My instincts were correct in the first place. When I went to solve for k, I incorrectly calculated the exponent in the first term. (I wanted to add (25k/2) + (875k/75), and I multiplied 25k by 150 instead of 75). In any case, I calculated wrong.

        Then, when I went to the solution to find out what I did wrong, what I thought I saw in the first line was e^int(0,5) ktdt + e^int(5,10) (1/25)kt^2. So I thought I was supposed to add both the integrals together, and as per my previous post, my instincts told me not to. In truth, the first line of the solution is of the form e^(integral + integral).

        I get it now.