This Exam FM sample reference tests Continuous Cash Flows. Accumulating each infinitesimal payment from its payment time to time 4 gives 656.91, which selects choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 313 is close to integrating the payment rate without accumulating the deposits to time 4. Each payment must earn interest after it enters the account.
BThe value 432 does not result from the required accumulated-value integral. In particular, using only the force factor over the whole interval ignores the payment-time dependence.
CThe value 477 is not produced by either endpoint evaluation of the correct integral. The factor e to the 0.08 times 4 minus t must remain inside the integrand.
DThe value 606 results from an incomplete accumulation treatment. Combining the exponents gives 0.42t plus 0.32 and leads to 656.91.
Original practice · fully worked
Original variant: calibrating a continuous grant stream
A conservation fund receives grants continuously from time 1 through time 3 at rate K exp(0.25t) per year. The fund earns a constant force of interest of 4%. Determine K if the grants must accumulate to 500 at time 4.
A 125.00
B 134.20
C 138.95
D 145.30
E 152.10
Variant answer in brief
The time-4 accumulation factor per unit of K is 3.59839; therefore K = 500/3.59839 = 138.95, choice C.
Setup
Setup
A grant received at time t earns force-of-interest accumulation for the remaining 4 minus t years.
dC(t)=Ke0.25tdt
a(t,4)=e0.04(4−t)
Model
Model
Set the accumulated value of the entire continuous stream equal to the required time-4 target.
500=K∫13e0.25te0.04(4−t)dt
500=Ke0.16∫13e0.21tdt
Compute
Compute
The integral contributes 3.59839 units of accumulated value for each unit of K.
0.21e0.16(e0.63−e0.21)=3.5983896
K=500/3.5983896=138.9510
Answer
Answer
The required scale of the grant rate is 138.95, which is choice C.
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