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+1 vote
in BUS 2016H - Financial Mathematics by (400 points)
edited by

"A bank credits interest on deposits using a variable force of interest. At the start of a given year an investor deposited R10 000 with the bank. The accumulated amount of the investor’s account was R10 511.40 midway through the year and R11 046.18 at the end of the year. Measuring time in years from the start of the given year, and assuming that over the year the force of interest per annum is a linear function of time, derive an expression for the force of interest per annum at time t (0 ≤ t ≤ 1). Hence find the accumulated amount of the account three quarters of the way through the year."

I got to the point where \(\ln(10511.4/10 000)= \int_{0}^{0.5} \delta (s) ds \), for the one equation? What should I do from that point?

1 Answer

+1 vote
by (4.2k points)

"the force of interest per annum is a linear function of time" means you can write it as $$\delta(s) = as + b $$

You've got the right idea with your first equation, but you can do the same thing for the accumulation between \(t = 0.5\) and \(t = 1 \) right? You get $$\int_{0.5}^1 \delta(s) ds = \ln(11046.18/10511.4) $$

Now you have two equations to solve for two variables \(a\) and \(b\), and the rest should be routine. Oh and my answer was off by about 11c but I think that's a rounding error.

by (400 points)

Thank you so much. I understand now