Questions

[George2004]

Book: The theory of interest, Second Version,
Chapter I
Question 43: An investment is made for one year in a fund whose accumulation function a second degree polynomial. THe nominal rate of interest earned during the first half of the year is 5% convertible semiannually. THe effective rate of interest earned for the entire year is 7% Find Delta(.5)

Question 44: Find an expression for the fraction of a period at which the excess of accumulated values computed at simple interest over compound interest is a maximum.

[Krieger]

Book: The theory of interest, Second Version,
Chapter I
Question 43: An investment is made for one year in a fund whose accumulation function a second degree polynomial. THe nominal rate of interest earned during the first half of the year is 5% convertible semiannually. THe effective rate of interest earned for the entire year is 7% Find Delta(.5)

Question 44: Find an expression for the fraction of a period at which the excess of accumulated values computed at simple interest over compound interest is a maximum.


43. We are given a(t) = a +bt+ct^2, with a(0)=1.
Over the first half year the effective rate earned is .025, so we have .025 = a(1/2)-1 = 1/2 b + 1/4 c. Over the entire year the effective rate is .07, so we have .07 = a(1) - 1 = b +c. This pair of equations solves for b = .03 and c = .04, so we have a(t) = 1 +.03t + .04t^2. Then delta_.5 = a'(t)/a(t) at t =.5 is 14/205.

44. Define such excess at time t as E_t = (1+ti)-(1+i)^t. The value of t which maximizes this excess is the value for which d/dt E_t = 0. You will find such t = 1/delta * (ln i - ln delta).

[George2004]

Hi Krieger:
Thank you very much!
Arround 44, although you showed me the idea, I am still confused on the force of interest (Delta). Is Delta derived from complex or simple interest here?

George


43. We are given a(t) = a +bt+ct^2, with a(0)=1.
Over the first half year the effective rate earned is .025, so we have .025 = a(1/2)-1 = 1/2 b + 1/4 c. Over the entire year the effective rate is .07, so we have .07 = a(1) - 1 = b +c. This pair of equations solves for b = .03 and c = .04, so we have a(t) = 1 +.03t + .04t^2. Then delta_.5 = a'(t)/a(t) at t =.5 is 14/205.

44. Define such excess at time t as E_t = (1+ti)-(1+i)^t. The value of t which maximizes this excess is the value for which d/dt E_t = 0. You will find such t = 1/delta * (ln i - ln delta).

[Krieger]

The force of interest delta is just the derivative divided by the function, however it is given.