Discounting of payment stream using force of interest

Discussion in 'CT1' started by acoovadia, Sep 6, 2007.

  1. acoovadia

    acoovadia Member

    Hi

    I am totally confused by the example given in the CT1 file in chapter 5 on page 16 - the Present value of a payment stream using the force of interest. I can follow the first 3 lines of the solution but after that I am totally lost - can anyone interpret into English?

    Thank you

    Anuschka
     
  2. thomasb

    thomasb Member

    this appears to a straight forward enough integration - you have the following

    Int(v' exp(v) dt) where v = 0.005t^2 + 0.04t ....

    this integrates to exp v.

    Hope this helps.

    Thomas
     
  3. John Lee

    John Lee ActEd Tutor Staff Member

    Yup that's absolutely right thomasb

    In general:

    integral of K f'(t) exp {f(t)} is K exp {f(t)}

    This covers the worst case scenario integrals you get in the exam.
     

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    Last edited: Sep 7, 2007
  4. stylz

    stylz Member

    Boy am I glad I found this! Thanks!!:)
     
  5. Question 5.12 Payment Streams problem

    I've looked at the handout and it was helpful for figuring out where the number 30 comes from. However, if the lower limit of the integral is not zero, it seems like something else is going on which is not shown in the handout, but is shown on the answer key. On the second line of the solution to 5.12, you see in parentheses and brackets that the anti-derivative is there but then moving to the third line, you see t has been replaced by 8, the upper limit, but then, you see the anti-derivative again in the parentheses. I don't understand what is going on there. I believe the exam style question at the end of the chapter has this same thing happening in it. Thank you.
     
  6. Mark Mitchell

    Mark Mitchell Member

    Going from the second line to the third line of Solution 5.12 in the Course Notes is just the process for evaluating a definite integral (ie an integral between limits).

    When integrating a function between limits, we integrate the function, then evaluate the integrated function at the upper limit and subtract the integrated function at the lower limit.

    In this case the upper limit is 8 and the lower limit is t.
     

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