Exam FM · practice
Exam FM — Force of Interest & Equations of Value Practice Flashcards
Thirty exam-realistic multiple-choice problems on the force of interest and equations of value: constant and time-varying force, accumulation and present value via integrated force, rate equivalences, equations of value, and unknown-time, equated-time, and unknown-rate problems.
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- Constant forceAn account is credited interest at a constant force of interest $\delta = 0.055$. Calculate the annual effective rate of interest $i$. (A) $5.35\%$ (B) $5.50\%$ (C) $5.58\%$ (D) $5.65\%$ (E) $5.95\%$**Answer: (D).** Under a constant force, $1+i=e^{\delta}$, so $i=e^{0.055}-1=1.056541-1=0.056541,$ about $5.65\%$. Distractor (A) is $d=1-e^{-0.055}=0.05351$, the discount rate; (B) sets $i=\delta$; (C) treats $\delta$ as a nominal $i^{(2)}$, giving $(1+0.055/2)^{2}-1=0.05576$; (E) overshoots from a term slip.
- Constant forceAn investment earns an annual effective rate of discount $d = 0.07$. Calculate the equivalent constant force of interest $\delta$. (A) $0.0677$ (B) $0.0700$ (C) $0.0718$ (D) $0.0726$ (E) $0.0753$**Answer: (D).** From $v=1-d=e^{-\delta}$ we get $\delta=-\ln(1-d)$: $\delta=-\ln(0.93)=0.072571,$ about $0.0726$. Distractor (A) is $\ln(1.07)=0.06766$, wrongly treating $d$ as $i$; (B) sets $\delta=d$; (E) is the equivalent effective rate $i=\dfrac{d}{1-d}=0.07527$, not the force; (C) is an arithmetic slip.
- Constant forceAn annual nominal rate of interest of $6\%$ compounded monthly is offered. Calculate the equivalent constant force of interest $\delta$. (A) $0.0583$ (B) $0.0598$ (C) $0.0600$ (D) $0.0602$ (E) $0.0617$**Answer: (B).** Match one-year accumulation factors: $e^{\delta}=\left(1+\dfrac{0.06}{12}\right)^{12}=(1.005)^{12}$, so $\delta=12\ln(1.005)=0.059850,$ about $0.0598$. Distractor (C) sets $\delta=i^{(12)}=0.06$; (E) uses the annual effective $i=(1.005)^{12}-1=0.06168$; (A) and (D) are log slips.
- Constant forceFor a positive interest rate, which ordering of the discount rate $d$, the nominal rate $i^{(4)}$, the force of interest $\delta$, and the effective rate $i$ is correct, smallest to largest? (A) $i < i^{(4)} < \delta < d$ (B) $d < \delta < i^{(4)} < i$ (C) $d < i^{(4)} < \delta < i$ (D) $\delta < d < i^{(4)} < i$ (E) $d < i < i^{(4)} < \delta$**Answer: (B).** For $i>0$ the standard chain is $d < d^{(2)} < \cdots < \delta < \cdots < i^{(4)} < i^{(2)} < i.$ The force lies strictly between the effective discount rate $d$ and any nominal interest rate $i^{(m)}$, and nominal interest rates increase toward $i$ as $m$ falls. Hence $d < \delta < i^{(4)} < i$. The other choices reverse the discount/interest relationship or misplace the force relative to $i^{(4)}$.
- Constant forceA constant force of interest is in effect. A deposit of $\$2{,}000$ grows to $\$2{,}915.27$ after 6 years. Calculate the force of interest $\delta$. (A) $0.0600$ (B) $0.0617$ (C) $0.0628$ (D) $0.0648$ (E) $0.0660$**Answer: (C).** Under a constant force, $2000\,e^{6\delta}=2915.27$, so $e^{6\delta}=\frac{2915.27}{2000}=1.457635,\qquad 6\delta=\ln(1.457635)=0.376819.$ Thus $\delta=0.062803$, about $0.0628$. Distractor (D) is the annual effective rate $i=1.457635^{1/6}-1=0.0648$ (the candidate solved $(1+i)^{6}$ instead of $e^{6\delta}$); (A), (B), (E) come from arithmetic slips.
- Constant forceUnder a constant force of interest, $\$1$ accumulates to $\$1.25$ over 4 years. Calculate the force of interest $\delta$. (A) $0.0498$ (B) $0.0558$ (C) $0.0574$ (D) $0.0595$ (E) $0.0625$**Answer: (B).** $e^{4\delta}=1.25$, so $4\delta=\ln 1.25=0.223144$ and $\delta=0.055786,$ about $0.0558$. Distractor (C) is the annual effective $i=1.25^{1/4}-1=0.0574$; (E) is the naive simple-interest guess $0.25/4=0.0625$; (A) and (D) are arithmetic slips.
- Force basicsThe accumulation function for a fund is $A(t)=1+0.03t+0.002t^{2}$ for $t\ge 0$. Calculate the force of interest $\delta_{4}$ at time $t=4$. (A) $0.0388$ (B) $0.0399$ (C) $0.0420$ (D) $0.0460$ (E) $0.0500$**Answer: (B).** The force is $\delta_{t}=\dfrac{A'(t)}{A(t)}$. With $A'(t)=0.03+0.004t$, $A'(4)=0.046,\qquad A(4)=1+0.12+0.032=1.152,$ so $\delta_{4}=\dfrac{0.046}{1.152}=0.039931$, about $0.0399$. Distractor (D) reports $A'(4)=0.046$ but forgets to divide by $A(4)$; (E) divides by $A(0)=1$; (A) and (C) are slips in $A(4)$.
- Force basicsA fund has accumulation function $A(t)$ with force of interest $\delta_{t}$. Which statement is TRUE? (A) $\delta_{t}=A'(t)\cdot A(t)$ (B) $A(t)=\delta_{t}\,e^{t}$ (C) $\delta_{t}=\dfrac{d}{dt}\ln A(t)$, and $A(n)=\exp\!\left(\int_{0}^{n}\delta_{t}\,dt\right)$ (D) The fund can grow even when $\delta_{t}<0$ over an interval (E) $\delta_{t}$ must equal a constant for $A(t)$ to be increasing**Answer: (C).** By definition $\delta_{t}=\dfrac{A'(t)}{A(t)}=\dfrac{d}{dt}\ln A(t)$. Integrating from $0$ to $n$ gives $\ln A(n)-\ln A(0)=\int_{0}^{n}\delta_{t}\,dt$, and with $A(0)=1$, $A(n)=\exp\!\big(\int_{0}^{n}\delta_{t}\,dt\big)$. (A) and (B) misstate the relationship. Since $A'(t)=\delta_{t}A(t)$ and $A(t)>0$, the sign of $A'(t)$ equals the sign of $\delta_{t}$; a negative force makes the fund shrink, so (D) is false. The force need not be constant for growth, only nonnegative, so (E) is false.