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Exam FM — Force of Interest & Equations of Value Flashcards

A comprehensive Exam FM deck on the force of interest: the constant force delta and its links to i and d, accumulation and present value under a time-varying force via the exponential of an integral, equations of value with time diagrams, and unknown-time and unknown-rate problems including the method of equated time.

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  1. Force basics
    Define the force of interest $\delta_{t}$ in terms of the accumulation function $A(t)$.
    The force of interest is the instantaneous relative rate of growth of the fund: $\delta_{t}=\frac{A'(t)}{A(t)}=\frac{d}{dt}\ln A(t)$. It measures interest as a continuously-compounded rate at the exact instant $t$.
  2. Constant force
    Under a CONSTANT force of interest $\delta$, write $(1+i)$, $v$, and $A(t)$ in terms of $\delta$.
    A constant force means $\delta_{t}=\delta$ for all $t$. Then $1+i=e^{\delta}$, the discount factor is $v=e^{-\delta}$, and the accumulation of $1$ over $t$ years is $A(t)=e^{\delta t}$.
  3. Constant force
    Given an annual effective rate $i$, express the constant force of interest $\delta$.
    Since $1+i=e^{\delta}$, take logs: $\delta=\ln(1+i)$. Because $\ln(1+i)<i$ for $i>0$, the force is always slightly smaller than the effective rate.
  4. Constant force
    An account earns a constant force of interest $\delta=0.06$. What is the equivalent annual effective rate of interest $i$?
    Use $1+i=e^{\delta}$. So $1+i=e^{0.06}=1.061837$, giving $i=0.061837$, about $6.184\%$. The effective rate slightly exceeds the force because of continuous compounding.
  5. Constant force
    Give the relationship between the constant force $\delta$ and the annual effective rate of discount $d$.
    Since $v=1-d=e^{-\delta}$, we have $d=1-e^{-\delta}$ and $\delta=-\ln(1-d)$. Combined with $i$: the chain $1+i=(1-d)^{-1}=e^{\delta}$ holds, so $\delta=\ln(1+i)=-\ln(1-d)$.
  6. Constant force
    Order these four quantities from smallest to largest for a positive rate: $d$, $\delta$, $i$, and the nominal $i^{(2)}$.
    For $i>0$ the ordering is $d<d^{(2)}<\cdots<\delta<\cdots<i^{(2)}<i$. In particular $d<\delta<i$: the force of interest sits between the effective discount rate and the effective interest rate.
  7. Constant force
    Show that a nominal rate $i^{(m)}$ converging as $m\to\infty$ gives the force of interest.
    The per-period equivalence is $\left(1+\frac{i^{(m)}}{m}\right)^{m}=1+i=e^{\delta}$. Taking $m\to\infty$, $\left(1+\frac{i^{(m)}}{m}\right)^{m}\to e^{\delta}$ requires $i^{(m)}\to\delta$. Thus the force of interest is the limit of the nominal rate as compounding becomes continuous.
  8. Varying force
    A fund grows according to the time-varying force $\delta_{t}$. Write the accumulated value at time $n$ of $1$ invested at time $0$.
    Integrate the force over the period and exponentiate: $A(n)=e^{\int_{0}^{n}\delta_{t}\,dt}$. This generalizes $e^{\delta n}$ to a force that changes with time.
  9. Varying force
    Under a time-varying force $\delta_{t}$, write the present value at time $0$ of $1$ payable at time $n$.
    Discounting uses the NEGATIVE integral: $\text{PV}=e^{-\int_{0}^{n}\delta_{t}\,dt}$. The accumulation and present-value factors are reciprocals.
  10. Varying force
    Write the accumulated value at time $t_{2}$ of $1$ invested at time $t_{1}$ under a force $\delta_{t}$ (with $t_{1}<t_{2}$).
    Integrate the force over only the interval the money is invested: $\text{AV}=e^{\int_{t_{1}}^{t_{2}}\delta_{t}\,dt}$. The lower limit is the deposit time, the upper limit is the valuation time.
  11. Varying force
    The force of interest is $\delta_{t}=0.04+0.01t$ for $t\geq 0$. Find the accumulated value at time $5$ of $1$ invested at time $0$.
    Integrate: $\int_{0}^{5}(0.04+0.01t)\,dt=\left[0.04t+0.005t^{2}\right]_{0}^{5}=0.20+0.125=0.325$. Then $A(5)=e^{0.325}=1.38403$. So $1$ grows to about $1.3840$.
  12. Varying force
    The force of interest is $\delta_{t}=\frac{1}{1+t}$ for $t\geq 0$. Find the accumulated value at time $4$ of an investment of $100$ at time $0$.
    Integrate: $\int_{0}^{4}\frac{1}{1+t}\,dt=\left[\ln(1+t)\right]_{0}^{4}=\ln 5-\ln 1=\ln 5$. Then $A(4)=e^{\ln 5}=5$, so $100$ accumulates to $100\times 5=500$.
  13. Varying force
    The force of interest is $\delta_{t}=0.05+0.002t$. Find the present value at time $0$ of $1000$ payable at time $10$.
    Integrate over $[0,10]$: $\int_{0}^{10}(0.05+0.002t)\,dt=\left[0.05t+0.001t^{2}\right]_{0}^{10}=0.50+0.10=0.60$. The discount factor is $e^{-0.60}=0.548812$, so $\text{PV}=1000\times 0.548812=548.81$.
  14. Varying force
    The force of interest is piecewise: $\delta(t)=0.05$ for $0\leq t<3$ and $\delta(t)=0.04$ for $t\geq 3$. Find the accumulated value at time $5$ of $1$ invested at time $0$.
    Split the integral at the breakpoint $t=3$ and integrate each constant piece separately: $a(5)=\exp\!\big(\int_{0}^{3}0.05\,dt+\int_{3}^{5}0.04\,dt\big)$. The pieces are $\int_{0}^{3}0.05\,dt=0.05\times 3=0.15$ and $\int_{3}^{5}0.04\,dt=0.04\times 2=0.08$, summing to $0.23$. Then $a(5)=e^{0.23}=1.25860$, so $1$ grows to about $1.2586$.
  15. Varying force
    Under a force $\delta_{t}$, what equivalent constant annual effective rate $i$ does an investment over $[0,n]$ earn?
    Equate accumulation factors: $(1+i)^{n}=e^{\int_{0}^{n}\delta_{t}\,dt}$, so $i=e^{\frac{1}{n}\int_{0}^{n}\delta_{t}\,dt}-1$. The equivalent force is the AVERAGE force $\bar{\delta}=\frac{1}{n}\int_{0}^{n}\delta_{t}\,dt$.
  16. Varying force
    Money grows under $\delta_{t}=0.08-0.001t$ over $[0,10]$. What single annual effective rate $i$ is equivalent over those 10 years?
    Average force: $\int_{0}^{10}(0.08-0.001t)\,dt=\left[0.08t-0.0005t^{2}\right]_{0}^{10}=0.80-0.05=0.75$, so $\bar{\delta}=0.075$. Then $i=e^{0.075}-1=0.077884$, about $7.788\%$.
  17. Force basics
    Why must the force of interest stay nonnegative for an ordinary accumulation function to be non-decreasing?
    Because $A'(t)=\delta_{t}\,A(t)$ and $A(t)>0$, the sign of $A'(t)$ equals the sign of $\delta_{t}$. If $\delta_{t}\geq 0$ the fund never shrinks; a negative force over an interval would make the fund decline there.
  18. Equations of value
    What is an equation of value, and what makes one valid?
    An equation of value sets the value of all inflows equal to the value of all outflows, with every cash flow accumulated or discounted to one common COMPARISON DATE at a stated interest rate. A correctly-built equation gives the same solution regardless of which comparison date is chosen.
  19. Equations of value
    Why does the choice of comparison date not change the solution of an equation of value (for a fixed rate)?
    Moving the comparison date multiplies every term by the SAME accumulation/discount factor, which cancels out. So the unknown rate or time that balances the equation at one date balances it at every date. Pick the date that makes the algebra easiest.
  20. Equations of value
    A person deposits $500$ now and $700$ in $3$ years. At $i=0.06$ annual effective, what single deposit at time $5$ has the same value?
    Accumulate both deposits to time $5$. $500(1.06)^{5}+700(1.06)^{2}=500(1.338226)+700(1.1236)=669.113+786.52=1455.63$. The equivalent single deposit at time $5$ is $1455.63$.
  21. Equations of value
    An investor pays $1000$ now to receive $X$ in $4$ years and $X$ in $8$ years. At $i=0.05$, find $X$.
    Equation of value at time $0$: $1000=Xv^{4}+Xv^{8}$ with $v=1/1.05$. $v^{4}=0.822702$, $v^{8}=0.676839$, sum $=1.499542$. So $X=\frac{1000}{1.499542}=666.87$.
  22. Equations of value
    What is a time diagram and why is it the recommended first step in an equation-of-value problem?
    A time diagram is a horizontal time line marking each cash flow's amount above its timing point, with inflows and outflows on opposite signs. It prevents mis-timing errors by making the comparison date and the direction of every accumulation or discount factor explicit before any algebra.
  23. Equations of value
    Deposits of $200$ are made at times $0$, $1$, and $2$. Using $i=0.04$, find the accumulated value just after the time-$2$ deposit.
    Accumulate each to time $2$: $200(1.04)^{2}+200(1.04)+200=200(1.0816)+200(1.04)+200=216.32+208.00+200.00=624.32$. Equivalently $200\,\ddot{s}_{\overline{3|}}$ where $\ddot{s}$ is the accumulated value of an annuity-due, giving the same $624.32$.
  24. Unknown time
    In an unknown-TIME problem, what quantity are you solving for and what is the typical setup?
    You solve for the time $n$ (often non-integer) that makes an equation of value balance at a known rate. The setup isolates a power of $v$ or $(1+i)$, e.g. $(1+i)^{n}=k$, then $n=\frac{\ln k}{\ln(1+i)}$. A non-integer $n$ usually signals a balloon or drop final payment.
  25. Unknown time
    $1000$ is invested at $i=0.07$ annual effective. How long until it accumulates to $1500$?
    Solve $1000(1.07)^{n}=1500$, so $(1.07)^{n}=1.5$. $n=\frac{\ln 1.5}{\ln 1.07}=\frac{0.405465}{0.067659}=5.9928$ years, about $5.99$ years.
  26. Unknown time
    State the Rule of 72 and use it to estimate the doubling time at $i=0.08$.
    The Rule of 72 estimates the doubling time as $n\approx\frac{72}{100\,i}$. At $i=0.08$: $n\approx\frac{72}{8}=9$ years. (Exact: $\frac{\ln 2}{\ln 1.08}=9.006$ years, so the rule is a good approximation.)
  27. Method of equated time
    State the method of equated time for finding the equated time $\bar{t}$ of a set of payments $C_{t_{k}}$ made at times $t_{k}$.
    The method of equated time approximates the single time at which one payment equal to the total $\sum C_{t_{k}}$ has the same value, using a payment-weighted average of the times: $\bar{t}=\frac{\sum_{k} t_{k}\,C_{t_{k}}}{\sum_{k} C_{t_{k}}}$. It is an approximation to the exact equation-of-value solution.
  28. Method of equated time
    Payments of $100$ at $t=1$, $200$ at $t=3$, and $300$ at $t=6$ are made. Find the equated time $\bar{t}$ by the method of equated time.
    $\bar{t}=\frac{100(1)+200(3)+300(6)}{100+200+300}=\frac{100+600+1800}{600}=\frac{2500}{600}=4.1\overline{6}$ years, about $4.17$ years.
  29. Method of equated time
    Does the method of equated time give a time slightly larger or smaller than the EXACT equated time, and why?
    The method of equated time always gives a value at least as LARGE as the exact answer ($\bar{t}\geq t^{*}$). Because the present-value function is convex, the payment-weighted average of times overstates the single equivalent time; the exact time solving $\sum C_{t_{k}} v^{t^{*}}=\sum C_{t_{k}} v^{t_{k}}$ is slightly smaller.
  30. Unknown rate
    In an unknown-RATE problem with multiple cash flows, why is the unknown rate usually found numerically rather than by a closed-form formula?
    The equation of value is a polynomial in $v$ (or $1+i$) of degree equal to the latest payment time, which generally has no algebraic solution. So the yield rate is found by financial-calculator IRR, iteration, or interpolation between trial rates that bracket a zero of the net present value.
  31. Unknown rate
    An investment of $1000$ returns $600$ at the end of year $1$ and $600$ at the end of year $2$. Estimate the annual yield rate $i$ by setting up the equation of value.
    Equation at time $0$: $1000=600v+600v^{2}$ with $v=1/(1+i)$. Let $f(i)=600v+600v^{2}-1000$. Try $i=0.13$: $v=0.884956$, value $=600(0.884956)+600(0.783147)-1000=530.97+469.89-1000=0.86$. Try $i=0.14$: $v=0.877193$, value $=526.32+461.68-1000=-12.00$. Interpolating, $i\approx 0.1307$, about $13.1\%$.