Independent solution

How to solve this Continuous Cash-Flow Accumulation question

Setup

Setup

Under force 1/(8 + t), accumulation from 0 to t is exp of the force integral, equal to (8 + t)/8. Let the continuous deposit rate at time t be k(8 + t).

a(t)=exp⁡(∫0tdr8+r)=8+t8a(t)=\exp\left(\int_0^t\frac{dr}{8+r}\right)=\frac{8+t}{8}

Model

Model

Accumulate each infinitesimal deposit to time 10. The factor a(10)/a(t) cancels its (8 + t) term, leaving the constant integrand 18k.

20000=∫010k(8+t)a(10)a(t) dt20000=\int_0^{10}k(8+t)\frac{a(10)}{a(t)}\,dt

Compute

Compute

The terminal fund is therefore the integral of 18k over ten years, or 180k. Setting 180k = 20,000 gives k = 111.11, hence 111 to the requested precision.

20000=∫01018k dt=180k⟹k=111.1120000=\int_0^{10}18k\,dt=180k\Longrightarrow k=111.11

Answer

Answer

The calculation gives 111 for continuous cash-flow accumulation, matching published choice A.

k=111(A)\boxed{k=111\quad\text{(A)}}