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Exam FM · Varying Annuity Payments · Tutorial 208 of 998

Geometric Gradients with Payments at the Beginning

Value geometric payment streams made at the beginning of each period by relating them to otherwise identical immediate annuities.

Intermediate 10 min read

What You'll Learn

  • Define the timing and payment pattern of a geometric annuity-due stream.
  • Shift each payment one period earlier to relate due and immediate present values.
  • Calculate increasing and decreasing annuity-due values, including the special equal-rate case.
  • Accumulate a due stream to the time of its final payment.
  • Identify and correct payment-timing errors in geometric-gradient calculations.

Geometric Gradients Paid at the Beginning of Each Period

A geometric payment pattern changes by a fixed percentage from one payment to the next. If its first payment is made immediately, the stream is an annuity-due. This timing matters: each payment is received one period earlier than the corresponding payment in an otherwise identical annuity-immediate stream, so it has a different value at the same focal date.

The central technique is to keep the payment amounts in the same order, shift each payment one period earlier, and adjust the value of the corresponding immediate annuity. This approach works for increasing and decreasing streams and makes the payment-timing adjustment explicit.

Definition: A geometric annuity-due with \(n\) payments has payment \(Cq^k\) at time \(k\), for \(k=0,1,\ldots,n-1\). Here \(C\) is the first payment and \(q\) is the payment ratio. If \(q=1+g\), payments increase at rate \(g\); if \(q=1-h\), with \(0\le h\le1\), they decrease at rate \(h\). Payments occur at times \(0,1,\ldots,n-1\), not at times \(1,2,\ldots,n\).

Present Value by Shifting the Payments

Assume a constant effective annual interest rate \(i>-1\), with \(v=1/(1+i)\), and annual payments at the stated integer times. The present value at time 0 is the sum of each payment discounted from its actual payment time:

$$ PV_{\text{due}}=\sum_{k=0}^{n-1} Cq^k v^k =C\sum_{k=0}^{n-1}(qv)^k. $$

Now place the same amounts at times \(1,2,\ldots,n\), producing an annuity-immediate stream. Each amount is then discounted one additional period. Its present value is \(vPV_{\text{due}}\), so shifting the payments one period earlier gives:

$$ PV_{\text{due}}=(1+i)PV_{\text{immediate}}, $$

where \(PV_{\text{immediate}}\) refers to the same ordered payment amounts, now paid at times \(1\) through \(n\). This relationship does not require level payments: it follows from the timing shift, and applies to any fixed sequence of payment amounts under the same interest rate.

Theorem: For a geometric annuity-due with first payment \(C\), payment ratio \(q=1+g\), \(n\) payments, and constant effective annual interest rate \(i\), $$ PV_{\text{due}}= \begin{cases} C\dfrac{1-\left(\dfrac{1+g}{1+i}\right)^n}{1-\dfrac{1+g}{1+i}} =C(1+i)\dfrac{1-\left(\dfrac{1+g}{1+i}\right)^n}{i-g}, & i\ne g,\\[1.2em] nC, & i=g. \end{cases} $$

The finite geometric-sum formula gives the first expression because each discounted term is \(C(qv)^k\). The second expression uses \(1-qv=(i-g)/(1+i)\). When \(i=g\), \(qv=1\), so every discounted payment equals \(C\) and the present value is \(nC\). The displayed fraction with denominator \(i-g\) must not be used at \(i=g\); use the limiting value \(nC\) instead.

1
Write the payment dates and amounts.
For a due stream, the first payment is at time 0; list the \(n\) payments as \(C,Cq,\ldots,Cq^{n-1}\).
2
Discount from actual dates.
Use \(v^k\) on the payment at time \(k\), or first value the same amounts as an immediate stream and multiply that value by \(1+i\).
3
Check the special case.
If \(i=g\), each discounted due payment is \(C\); if \(i\ne g\), use the finite geometric-sum formula.

Worked Example: An Increasing Due Stream When Growth Equals Interest

A hypothetical grant pays $700 immediately, then increases each year by 10% for four further payments. The effective annual interest rate is 10%. Find the present value at the grant date.

Method. List the five due payments and discount each from its actual time. Here \(C=700\), \(g=0.10\), \(i=0.10\), and \(n=5\). Since growth equals interest, the general geometric ratio after discounting is:

$$ qv=\frac{1.10}{1.10}=1. $$

The payments are $700, $770, $847, $931.70, and $1,024.87 at times 0 through 4. Their present values are:

$$ 700,\quad 770v=700,\quad 847v^2=700,\quad 931.70v^3=700,\quad 1{,}024.87v^4=700. $$

Thus \(PV_{\text{due}}=5(700)=\$3{,}500.00\). As a timing check, the same five amounts placed at times 1 through 5 have present value \(v(3{,}500)=\$3{,}181.82\); multiplying by \(1.10\) gives $3,500.00. The check confirms the due-versus-immediate adjustment.

Common mistake avoided: Substituting \(i=g\) into the fraction with denominator \(i-g\) would divide by zero. The direct discounted-term calculation gives the valid value.

Increasing and Decreasing Payment Patterns

For an increasing stream, \(g>0\) and the payment at time \(k\) is \(C(1+g)^k\). For a decreasing stream, write the percentage decrease as \(h\), so \(q=1-h\) and the payment at time \(k\) is \(C(1-h)^k\). Both use the same geometric-sum method; only the payment ratio changes.

For a decreasing stream, substituting \(g=-h\) into the present-value formula gives, when \(i+h\ne0\):

$$ PV_{\text{due}} =C(1+i)\frac{1-\left(\dfrac{1-h}{1+i}\right)^n}{i+h}. $$

The general sum \(C\sum_{k=0}^{n-1}(qv)^k\) remains the safest form when a special case is possible. It directly represents the discounted payments and avoids relying on a quotient whose denominator may be zero.

Worked Example: Present Value of an Increasing Due Stream

A hypothetical training fund makes three annual payments. It pays $1,000 immediately, with each later payment 10% larger than the preceding payment. The effective annual interest rate is 20%. Find the value at time 0.

Method. The payments occur at times 0, 1, and 2. The payment ratio is \(q=1.10\), so the discounted ratio is \(qv=1.10/1.20=11/12\). Apply the due-stream sum:

$$ PV_{\text{due}} =1{,}000\left[1+\frac{11}{12}+\left(\frac{11}{12}\right)^2\right] =1{,}000\left(\frac{144+132+121}{144}\right) =1{,}000\left(\frac{397}{144}\right) =\$2{,}756.944\ldots $$

The payments themselves are $1,000, $1,100, and $1,210. Discounting them from their actual dates gives the independent check:

$$ 1{,}000+\frac{1{,}100}{1.20}+\frac{1{,}210}{1.20^2} =1{,}000+916.666\ldots+840.277\ldots =2{,}756.944\ldots $$

To the nearest cent, the present value is $2,756.94. The calculation retains full precision until the final rounding. The common error is to use the annuity-immediate value without multiplying by \(1.20\), which would value these same payments one period too late.

Worked Example: Decreasing Due Payments and Accumulated Value

A hypothetical equipment reserve pays $1,000 immediately, followed by two payments that each decrease by 20%. The effective annual interest rate is 25%. Find both the present value at time 0 and the accumulated value at time 2, when the final payment is made.

Method. The payment ratio is \(q=0.80\), and \(v=1/1.25=0.80\). The discounted ratio is \(qv=0.64\). Therefore:

$$ PV_{\text{due}} =1{,}000(1+0.64+0.64^2) =1{,}000(1+0.64+0.4096) =\$2{,}049.60. $$

The payment amounts are $1,000 at time 0, $800 at time 1, and $640 at time 2. Accumulating each to time 2 gives:

$$ AV_2=1{,}000(1.25)^2+800(1.25)+640 =1{,}562.50+1{,}000+640 =\$3{,}202.50. $$

The present-value result checks against this accumulated value: \(2{,}049.60(1.25)^2=\$3{,}202.50\). As a separate timing observation, moving these same payment amounts to times 1, 2, and 3 makes their accumulated value at time 3 equal to $3,202.50, the value of the due stream at time 2. Each amount has the same number of periods to accumulate in those paired valuations.

Accumulating a Geometric Annuity-Due

For a due stream paid at times 0 through \(n-1\), the accumulated value at time \(n-1\), immediately after the final payment, is:

$$ AV_{n-1} =\sum_{k=0}^{n-1}Cq^k(1+i)^{n-1-k}. $$

This follows by accumulating the payment at time \(k\) for \(n-1-k\) periods. It can also be related to present value: each present-value term \(Cq^kv^k\), when accumulated from time 0 to time \(n-1\), is multiplied by \((1+i)^{n-1}\). Hence \(AV_{n-1}=PV_{\text{due}}(1+i)^{n-1}\). If the desired focal date is time \(n\), one more period of accumulation is required: \(AV_n=AV_{n-1}(1+i)\). Always identify the focal date before accumulating; “at the end” can mean different dates for immediate and due streams.

Common Timing Errors and Practical Checks

The annuity-due adjustment is simple, but a misplaced payment date changes the answer. In the increasing-stream example, the immediate value for the same sequence of payments is \(v(2{,}756.944\ldots)=\$2{,}297.45\), not the due value. The correction is:

$$ (1+i)PV_{\text{immediate}} =1.20(2{,}297.4537\ldots) =2{,}756.944\ldots $$

The incorrect calculation is to use the immediate value directly for payments made at times 0, 1, and 2. It discounts every payment one period too long. Conversely, multiplying an already-due present value by \(1+i\) adds an extra period and overstates the result.

  • Write down dates before applying a factor. A due stream starts at time 0; an immediate stream starts at time 1.
  • Keep the first payment anchored. If the first due payment is \(C\), the payment at time \(k\) is \(C(1+g)^k\), not \(C(1+g)^{k+1}\).
  • Distinguish growth from interest. Growth determines payment amounts; interest determines discounting or accumulation.
  • Check equal rates directly. If \(g=i\), each discounted due payment is \(C\), so the present value is \(nC\).
  • State the accumulated-value date. The final due payment occurs at time \(n-1\), not time \(n\).
Key takeaway: A geometric annuity-due has present value \(C\sum_{k=0}^{n-1}((1+g)v)^k\). It is also \(1+i\) times the present value of the same ordered payments shifted to times 1 through \(n\). Accurate payment dates, rather than a memorized annuity factor, prevent timing errors.

Check Your Understanding

For each problem, identify the payment dates before calculating. Round money to the nearest cent where needed.

  1. A fund pays $1,000 immediately and makes two further annual payments, each 12.5% larger than the preceding payment. The effective annual interest rate is 25%. Find the present value. (Answer: $2,710.00)
  2. A reserve pays $1,200 immediately, with each later payment 25% lower than the preceding one, for four payments total. The effective annual interest rate is 25%. Find the present value. (Answer: $2,611.20)
  3. A six-payment annuity-due starts at $600, and payments grow by 5% each year. The effective annual interest rate is also 5%. Find the present value. (Answer: $3,600.00)
  4. A due stream pays $1,000 at time 0, $900 at time 1, and $810 at time 2. At an effective annual interest rate of 10%, find its accumulated value immediately after the final payment. (Answer: $3,010.00)
  5. A four-payment due stream starts at $500 and increases by 10% per year. The effective annual interest rate is 10%. Find the present value, then state the present value of the same amounts paid at times 1 through 4. (Answer: Due value $2,000.00; immediate value $1,818.18)
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