Annuities and discounted production

Quick refresher

A series of $n$ equal payments $A$ , one at the end of each year, is worth $A\,a_n(i)$ today. Here $a_n(i) = [1 - (1+i)^{-n}]/i$ is the annuity factor: the sum of the discount factors from year 1 to year $n$ .
The factor is always less than $n$ . As the term grows, it approaches $1/i$ , the value of a perpetuity (a series that never ends). At 10%, a 30-year series is worth 9.43 annual payments and a perpetual one 10, so the distant years add little.
Payment in advance, payment in the middle of the year, and a delayed start change the value of a series by the constant factors $(1+i)$ , $(1+i)^{0.5}$ and $(1+i)^{-m}$ . The sum that the payments have grown to by the date of the last payment is given by the factor $s_n(i) = [(1+i)^{n} - 1]/i$ .
A series that changes by the fraction $g$ every year is valued with the growing annuity formula. Declining production is the case $g < 0$ . A perpetual growing series has a finite value only while $g < i$ .
At a constant price, the present value of the revenue equals the price multiplied by the discounted production $Q_\text{disc}$ . Discounting barrels simply means taking the price outside the summation sign (the sign $\Sigma$ that adds up the yearly terms). If the price rises by $e_r$ a year, barrels are discounted at the rate $(1+i)/(1+e_r) - 1$ .
For an exponential decline $q = q_i e^{-Dt}$ with continuous discounting, discounted production is $q_i[1 - e^{-(D+j)T}]/(D+j)$ , where $j = \ln(1+i)$ . In other words, discounting acts like an extra decline.

Key relationships

a_n(i) = \sum_{t=1}^{n} \frac{1}{(1+i)^{t}} = \frac{1-(1+i)^{-n}}{i}

annuity factor: payments at the end of each year

PV_\infty = \frac{A}{i}, \qquad PV_{g,\infty} = \frac{A_1}{i-g}

perpetuity (a series that never ends): constant and growing (for $g < i$ )

PV_R = P\,Q_\text{disc}, \qquad Q_\text{disc} = \sum_t \frac{Q_t}{(1+i)^{t}}

present value of the revenue at a constant price

1 + i_q = \frac{1+i}{1+e_r}

rate for discounting barrels when the price grows in real terms by $e_r$ a year

Q_\text{disc} = \frac{q_i\left[1 - e^{-(D+j)T}\right]}{D+j}, \qquad j = \ln(1+i)

exponential decline in flow rate with continuous discounting

Common pitfalls

Using the factor for end-of-year payments when the payments arrive in the middle of the year or in advance: the error is a factor of $(1+i)^{0.5}$ or $(1+i)$ .
Putting into the continuous formula the yearly decline fraction $d$ and the effective rate $i$ instead of $D = -\ln(1-d)$ and $j = \ln(1+i)$ .
Discounting barrels at the same rate as money when the price is not constant in the money of the calculation. For example, in constant prices the price may grow faster than inflation, or it may stay constant in nominal money and so fall in real terms.
Treating discounted production as the “real” reserves (the volume of oil that can be recovered). Facilities are designed, and reserves are reported, on undiscounted volumes.
Using the formula for a growing perpetuity when $g$ is close to $i$ : the result grows without limit and means nothing.

After this lesson you can

derive the annuity factor, and calculate the present value and the accumulated value (what the payments grow to) of a series of equal payments, for any timing of the payments within the year
find how many years it takes for equal receipts to pay back an investment at a given return, and find the return that they give
value series that grow or decline at a constant rate, and series that never end, and understand when such a valuation stops making sense
calculate discounted production and the present-volume factor, and explain why discounting barrels gives the same result as discounting revenue at a constant price
choose the rate for discounting barrels when the price rises, falls or is given in nominal money (the money of each year)
calculate discounted production for an exponential decline in flow rate (a fall by the same fraction every year) with continuous discounting

The annuity and its factor

In many calculations, money arrives or leaves in equal amounts at equal intervals. Examples are the rent of an installation, a tariff (a fee) for pumping oil through someone else's pipeline, loan repayments, the yearly saving from replacing equipment, and the revenue on the production plateau at a constant price. Such a series is called an annuity. If the payments arrive at the end of each period, it is an ordinary annuity. If they arrive at the beginning, it is an annuity due. Below, unless we say otherwise, the payments are yearly and arrive at the end of the year. The first one comes a year after the time origin (the starting moment).

We can find the present value of this series as for any series, by adding up the discounted payments (present value of a series of payments). But for equal payments the sum collapses into a short formula. The easiest way to see it is through a perpetuity, a series that never ends.

Put the sum $A/i$ into an account at rate $i$ . Every year it earns exactly $A$ in interest. If you withdraw that interest, the deposit stays untouched, and the payments go on forever. So a perpetual series of payments $A$ is worth $A/i$ today. A series of $n$ payments is a perpetual series from which we subtract an identical perpetual series that starts after year $n$ . The subtracted series is worth $A/i$ at the end of year $n$ , and $(A/i)(1+i)^{-n}$ today. The difference is the annuity formula.

Present value of an annuity and the annuity factor

PV = A\,a_n(i), \qquad a_n(i) = \sum_{t=1}^{n} \frac{1}{(1+i)^{t}} = \frac{1 - (1+i)^{-n}}{i}

where:

$PV$ present value of the whole series at the time origin, US$ million
$A$ equal payment at the end of each year, US$ million per year
$a_n(i)$ annuity factor: the present value of one unit received at the end of each of $n$ years, number of annual payments
$i$ discount rate, fraction per year
$n$ number of payments (years), —

Units: any currency; $A$ and $PV$ in the same money units, and the rate is the effective rate for one payment period (the rate that really applies over that period)

Holds when

The payments are equal and come without gaps, once in each period of the rate: a yearly rate means yearly payments.
The first payment is one period after the time origin and the last one $n$ periods after it. Other timings are handled with the factors of the next section.
The rate is constant for the whole term. At $i = 0$ the factor equals $n$ .
The payments and the rate are in the same kind of money: a real rate (inflation taken out) for constant prices, and a nominal rate for nominal money, the money of each year (Inflation: nominal and real money).

The factor $a_n(i)$ shows how many annual payments the whole series is worth today. It is always less than $n$ , because each payment is discounted. As $n$ grows, it approaches $1/i$ , the value of a perpetuity. The higher the rate, the sooner it gets close to this limit: the distant payments add almost nothing.

CalculatedAnnuity factor

a_n(i)

against the number of annual payments at rates of 5, 10 and 20%. The dashed lines show the limits

1/i

that the factor approaches when the term has no end.

Annuity factor

a_n(i)

: how many annual payments the series is worth

Rate $i$: 5%
10 years: 7.722
20 years: 12.462
30 years: 15.372
Forever, $1/i$: 20.000

Rate $i$: 10%
10 years: 6.145
20 years: 8.514
30 years: 9.427
Forever, $1/i$: 10.000

Rate $i$: 15%
10 years: 5.019
20 years: 6.259
30 years: 6.566
Forever, $1/i$: 6.667

Rate $i$: 20%
10 years: 4.192
20 years: 4.870
30 years: 4.979
Forever, $1/i$: 5.000

Rate $i$	10 years	20 years	30 years	Forever, $1/i$
5%	7.722	12.462	15.372	20.000
10%	6.145	8.514	9.427	10.000
15%	5.019	6.259	6.566	6.667
20%	4.192	4.870	4.979	5.000

Payments at the end of each year. The table is useful for rough estimates. It is easier to calculate the factor itself with the formula.

We can also solve the formula for the term. Suppose an investment $PV$ is made now, and receipts $A$ arrive at the end of each year. The present value of the receipts equals the investment when $a_n(i) = PV/A$ . This gives the number of years in which the receipts pay back the investment together with a return $i$ on it. It is the discounted payback period of an equal series, the time needed to get the investment back when future money is discounted (Cumulative cash flow and payback period):

Number of equal payments that pay back an investment

n = -\frac{\ln\left(1 - i\,PV/A\right)}{\ln(1+i)}

where:

$n$ number of annual payments whose present value equals $PV$ ; it need not be a whole number, years
$PV$ investment now (or the amount of a debt), US$ million
$A$ equal receipt (or payment) at the end of each year, US$ million per year
$i$ discount rate, fraction per year
$\ln$ natural logarithm, —

Units: $PV$ and $A$ in the same monetary units, $i$ a fraction per year

Holds when

A solution exists only if $i\,PV < A$ . Otherwise the receipts do not even cover the interest, and the investment is never recovered.
A fractional $n$ is a smooth (continuous) estimate. With end-of-year payments, the investment is recovered with the payment of the year whose number is the next whole number above $n$ .

Without discounting ( $i \to 0$ ), the formula gives the simple payback period $PV/A$ . As the rate grows, the term grows faster and faster. When $i\,PV/A$ approaches 1, the term goes to infinity.

Worked exampleWill the variable speed drives pay back?

Given (assumed data)

Variable speed drives (VSDs, devices that control the speed of the pump motor) for 40 wells with electric submersible pumps (ESPs, pumps driven by an electric motor deep in the well): US$1.9 million now
Electricity saving: US$0.42 million at the end of each year, for 7 years of service, in constant prices
Discount rate: 10% (real)

Solution

Annuity factor: $1.1^{-7} = 0.51316$ , $a_7(10\%) = (1 - 0.51316)/0.1 = 4.8684$ .
Present value of the saving: $0.42\cdot 4.8684 = 2.045$ million dollars. Net present value: $\mathrm{NPV} = 2.045 - 1.9 = 0.145$ million dollars.
Without discounting, the saving over 7 years is $7\cdot0.42 = 2.94$ million dollars, and the simple payback period is $1.9/0.42 = 4.5$ years.
Now allow for the rate: $a_n = 1.9/0.42 = 4.524$ , so $n = -\ln(1 - 0.1\cdot4.524)/\ln 1.1$ $= 6.3$ years.
Suppose the equipment lasts 6 years, not 7. Then $0.42\cdot a_6(10\%) = 0.42\cdot4.3553 = 1.829$ million dollars, which is less than the investment.

Answer: The present value of the saving is 2.045 million dollars, against an investment of US$1.9 million. The NPV is about +0.145 million dollars, and the investment is paid back, with a 10% return, in about 6.3 years.

What it means. The simple calculation gives an illusion of a large margin: US$2.94 million of savings against US$1.9 million of cost, and payback in 4.5 years. Once we allow for the value of money, the margin is small. The project pays back only in the seventh year of seven, and a service life shorter by one year makes the NPV negative. So for decisions like this, the most important thing is how reliable the assumptions are: the service life, and that the wells keep running for all that time. The data in this example are assumed for illustration.

Payment in advance, mid-year payment, delayed start and the accumulated sum

The annuity formula assumes payments at the end of each year. In annual models, the flow of a year is often placed in the middle of that year. Rent is paid in advance, and payments often do not start at once. All these cases are the same series, shifted in time. So its value differs from $A\,a_n(i)$ only by a constant factor (timing conventions).

The same series of

n

payments with different payment timings

When the payments arrive: at the end of each year (ordinary annuity)
Factor: $a_n(i)$
$n = 10$ , $i = 10\%$: 6.145

When the payments arrive: at the beginning of each year (annuity due, payment in advance)
Factor: $(1+i)\,a_n(i)$
$n = 10$ , $i = 10\%$: 6.759

When the payments arrive: in the middle of each year, counted from the start of year 1
Factor: $(1+i)^{0.5}\,a_n(i)$
$n = 10$ , $i = 10\%$: 6.444

When the payments arrive: at the end of each year, but the first payment comes after $m + 1$ years (a delayed start)
Factor: $(1+i)^{-m}\,a_n(i)$
$n = 10$ , $i = 10\%$: 5.078 at $m = 2$

When the payments arrive: accumulated sum at the date of the last payment
Factor: $s_n(i) = (1+i)^{n}\,a_n(i)$
$n = 10$ , $i = 10\%$: 15.937

When the payments arrive	Factor	$n = 10$ , $i = 10\%$
at the end of each year (ordinary annuity)	$a_n(i)$	6.145
at the beginning of each year (annuity due, payment in advance)	$(1+i)\,a_n(i)$	6.759
in the middle of each year, counted from the start of year 1	$(1+i)^{0.5}\,a_n(i)$	6.444
at the end of each year, but the first payment comes after $m + 1$ years (a delayed start)	$(1+i)^{-m}\,a_n(i)$	5.078 at $m = 2$
accumulated sum at the date of the last payment	$s_n(i) = (1+i)^{n}\,a_n(i)$	15.937

The factors come from shifting the whole series in time. Moving it a year earlier means multiplying by $1+i$ . Moving it half a year earlier means multiplying by $(1+i)^{0.5}$ . Moving it $m$ years later means dividing by $(1+i)^{m}$ .

Sometimes we need not the present value but the accumulated sum: for example, how much will build up in a fund if equal sums are paid into it. It is given by the accumulation factor of an annuity. We have already met it when comparing options by terminal value, the value at the end of the term (comparison by terminal value).

Accumulated sum of a series of equal contributions

FV_n = A\,s_n(i), \qquad s_n(i) = \frac{(1+i)^{n} - 1}{i} = (1+i)^{n}\,a_n(i)

where:

$FV_n$ the sum accumulated by the date of the last, $n$ th contribution, US$ million
$A$ equal contribution at the end of each year, US$ million per year
$s_n(i)$ accumulation factor of an annuity, number of annual contributions
$i$ return (growth rate) of the account or fund, fraction per year

Units: any currency; the rate effective per contribution period

Holds when

The contributions are equal and made at equal intervals. The income earned is added to the fund and earns the same rate itself.
The sum is the value at the moment of the last contribution. To get its value at a later date, multiply it by $(1+i)^{k}$ , where $k$ is the number of periods after the last contribution.

The reciprocal $1/s_n(i)$ (one divided by $s_n(i)$ ) shows what equal contribution is needed to accumulate one unit by the due date. It is used to calculate the contributions to a fund for future costs, for example the abandonment of a field.

Worked exampleEqual contributions to an abandonment fund

Given (assumed data)

Abandonment of the field (the project from the library's examples): US$12 million in the middle of year 10
Contributions: equal, in the middle of each year of production from year 3 to year 9, seven in all
Return of the fund: 3% a year above inflation (safe investments)
Discount rate of the project: 10% (real); time origin in the middle of year 1

Solution

By the last contribution, in the middle of year 9, the seven contributions $A$ accumulate to $A\,s_7(3\%)$ . Here $s_7 = (1.03^{7} - 1)/0.03 = 7.66246$ .
The money grows for one more year before abandonment: $A\cdot 7.66246\cdot 1.03 = A\cdot 7.8923 = 12$ . So $A = 1.520$ million dollars a year.
The contributions total $7\cdot 1.520 = 10.64$ million dollars. The income of the fund makes up the rest of the 12 million.
For the project at 10%, the contributions are worth $1.520\cdot\sum_{t=2}^{8} 1.1^{-t}$ $= 1.520\cdot 4.4258 = 6.73$ million dollars. Paying for the abandonment when the time comes is worth $12\cdot 1.1^{-9} = 5.09$ million dollars.

Answer: About 1.520 million dollars must be contributed each year. For the project at a rate of 10%, the fund costs 1.64 million dollars more than paying for the abandonment when it is due.

What it means. The fund is not a saving but a guarantee. The money earns 3% while the project requires 10%, and this difference costs the project 1.64 million dollars in today's money. But the money is certainly there when the abandonment comes, which is why the state or partners often require a fund. If the contributions reduce taxable profit, the fund costs the project less. The lesson on the internal rate of return explains how late costs, and the money set aside for them, affect the return. The data in this example are assumed for illustration.

We can also solve the annuity formula for other quantities. The payment for a known sum: $A = PV/a_n(i)$ . This is how equal loan repayments are calculated. In project appraisal, it gives the equal annual costs that are equivalent to a capital investment. The factor $1/a_n(i)$ is called the capital recovery factor (CRF): the fraction of an investment that must be received each year to recover it with a return. In Russian-language bank practice, where annuity loan payments are calculated, it is called the “annuity coefficient”. Do not confuse it with the annuity factor $a_n(i)$ . The lesson Annual capital charge explains how the capital recovery factor is used to compare options. The rate, when $PV$ , $A$ and $n$ are known, can be found only by trial, that is, numerically. It is the internal rate of return (IRR) of the series (Internal rate of return (IRR)). An example is in the advanced section.

Growing and declining series

Equal payments are a convenient simplification, but a rare one. The flow rate of a well declines, tariffs are indexed (raised each year by an agreed index, such as inflation), and costs grow. Suppose a series changes by the same fraction $g$ each year: it grows for $g > 0$ and declines for $g < 0$ . Its present value also collapses into a formula, the formula of a growing annuity. We derive it in the same way as the annuity formula: take a perpetual growing series and subtract an identical series that starts after year $n$ .

Present value of a series with a constant rate of change

PV = \frac{A_1}{i - g}\left[1 - \left(\frac{1+g}{1+i}\right)^{n}\right], \qquad g \ne i

where:

$A_1$ the first payment, at the end of year 1; the payment of year $t$ is $A_1(1+g)^{t-1}$ , US$ million
$g$ annual rate of change of the payment; for a decline it is negative: $g = -d$ , fraction per year
$i$ discount rate, fraction per year
$n$ number of payments, —

Units: any currency; $g$ and $i$ are fractions for the same period

Holds when

The rate $g$ is constant for the whole term. For a declining series, $d$ is the fraction by which the payment falls in a year.
At $g = i$ the formula is undefined, but the limit is finite: $PV = n A_1/(1+i)$ . Growth exactly cancels discounting.
A perpetual series ( $n \to \infty$ ) is worth $A_1/(i - g)$ only if $g < i$ . For $g \ge i$ the sum is infinite, which shows that the assumption of eternal growth is unrealistic.

We can reduce a growing annuity to an ordinary one: $PV = \dfrac{A_1}{1+g}\,a_n(i^{*})$ , where $1 + i^{*} = (1+i)/(1+g)$ . A decline acts like a surcharge on the rate. With a decline of 18% a year and a rate of 10%, the “net” rate $i^{*}$ is about 34%.

Worked exampleA well after treatment: the income declines every year

Given (assumed data)

Additional net income of the well (gained from the treatment): US$3.2 million for year 1 (at the end of the year), then 18% less each year
Period considered: 8 years
Discount rate: 10% (real)

Solution

The ratio of two neighbouring discounted payments is $(1 - 0.18)/1.1 = 0.74545$ . Over 8 years it is $0.74545^{8} = 0.09536$ .
Use the growing annuity formula with $g = -0.18$ : $PV = \dfrac{3.2}{0.10 + 0.18}\,(1 - 0.09536)$ $= 11.429\cdot 0.90464 = 10.34$ million dollars.
Check with the “net” rate: $i^{*} = 1.10/0.82 - 1 = 34.15$ %, and $a_8(i^{*}) = 2.6493$ . Then $PV = (3.2/0.82)\cdot 2.6493$ $= 3.9024\cdot 2.6493 = 10.34$ million dollars.
For comparison. Without the decline: $3.2\cdot a_8(10\%) = 3.2\cdot 5.3349 = 17.07$ million dollars. Without discounting, the sum of the declining series is 14.14 million dollars. The perpetual declining series is worth $3.2/0.28 = 11.429$ million dollars.

Answer: The present value of the income over 8 years is about 10.34 million dollars, which is 90% of the value of the infinite declining series.

What it means. Decline and discounting add up. Together they act like a rate of about 34% a year. So the value is 39% lower than for an equal income, and everything the well would give after year 8 is worth less than a tenth of the value of the infinite series. One refinement: if the well has fixed costs (costs that do not depend on how much it produces), the net income falls faster than production (economic limit, the lowest production at which the well still pays its costs). A formula that uses the decline rate of production will then overstate the value. The data in this example are assumed for illustration.

Discounted production: why discount barrels

An engineer who sees discounted production for the first time may find it strange. Money can be invested to earn interest, but oil in the reservoir earns no interest. So why multiply barrels by discount factors? The answer is that we do not discount the barrels. We discount the revenue that they will bring. If the price is the same in all years, we can take it outside the summation sign (the sign for adding up the yearly terms), and the discounted barrels remain under it.

Present value of the revenue at a constant price, and discounted production

PV_R = \sum_t \frac{P\,Q_t}{(1+i)^{t}} = P\,Q_\text{disc}, \qquad Q_\text{disc} = \sum_t \frac{Q_t}{(1+i)^{t}}

where:

$PV_R$ present value of the revenue, US$ million
$P$ field price (the price received at the field), the same in all years, $/bbl
$Q_t$ sales of the year that is $t$ years from the time origin, million bbl
$Q_\text{disc}$ discounted production, million bbl
$i$ discount rate, fraction per year

Units: US$ million = $/bbl × million bbl (bbl is the abbreviation for barrel); the rate as a fraction per year

Holds when

The price is constant in the money in which the calculation is made: in constant prices, a constant real price and a real rate.
Revenue and production are placed at the same moments and discounted from one time origin, as in NPV (time origin and conventions).
$Q_t$ are the volumes sold: production minus the oil that the producer uses itself, and minus losses.

Discounted production $Q_\text{disc}$ is the number of barrels that, if sold at the same price at the time origin, would bring the same present value of revenue as the whole profile. The ratio $k_\text{disc} = Q_\text{disc}/\sum_t Q_t$ is called the present-volume factor (uncertainty of flow rates and timing). The same holds for everything that is proportional to production. Variable costs $vQ_t$ (costs that grow with each barrel) and royalty $rPQ_t$ (a payment equal to a share $r$ of the revenue) are discounted like barrels. This is why it makes sense to divide the present value of the costs by $Q_\text{disc}$ (Unit costs per barrel).

Worked exampleDiscounted production of the example project

Given (assumed data)

Production in years 3–9: 1.6, 2.4, 2.4, 2.0, 1.6, 1.2 and 0.8 million bbl, 12 million bbl in all
Price: $50/bbl, constant in real terms
Rate and convention: 10% (real); time origin in the middle of year 1, flows in the middle of each year ( $t = 0 \ldots 9$ ), as in the lesson Net present value (NPV)

Solution

Discount factors for the production years: $1.1^{-2} = 0.8264$ for year 3, …, $1.1^{-8} = 0.4665$ for year 9. The table below gives the discounted production by year. The sum is $Q_\text{disc} = 7.8987$ million bbl.
Present-volume factor: $k_\text{disc} = 7.8987/12 = 0.658$ .
Present value of the revenue: $50\cdot 7.8987 = 394.9$ million dollars. If we discount the revenue itself (80, 120, 120, 100, 80, 60 and 40 million dollars), we get the same result: 394.9 million dollars.
The first three years of production give $1.6 + 2.4 + 2.4 = 6.4$ million bbl. That is 53% of the reserves, but 60% of the discounted production at 10% and 66% at 20%.

Answer:

Q_\text{disc} = 7.8987

million bbl (

k_\text{disc} = 0.658

): at a constant price, the project's 12 million bbl are worth as much as 7.9 million bbl sold at once.

What it means. Dividing the present value of the revenue by the price gives the same result as discounting the barrels. This is what justifies discounted production. The higher the rate, the more the early years weigh. That is why an error in the start-up date, or in how fast production builds up to the plateau, costs more than an error of the same size in the “tail” of production (the last years, when production is low). The data in this example are assumed for illustration.

Discounted production of the project by year at a rate of 10%

Year: 3
$t$: 2
$Q_t$ , million bbl: 1.6
$1.1^{-t}$: 0.8264
$Q_t\,1.1^{-t}$: 1.3223

Year: 4
$t$: 3
$Q_t$ , million bbl: 2.4
$1.1^{-t}$: 0.7513
$Q_t\,1.1^{-t}$: 1.8032

Year: 5
$t$: 4
$Q_t$ , million bbl: 2.4
$1.1^{-t}$: 0.6830
$Q_t\,1.1^{-t}$: 1.6392

Year: 6
$t$: 5
$Q_t$ , million bbl: 2.0
$1.1^{-t}$: 0.6209
$Q_t\,1.1^{-t}$: 1.2418

Year: 7
$t$: 6
$Q_t$ , million bbl: 1.6
$1.1^{-t}$: 0.5645
$Q_t\,1.1^{-t}$: 0.9032

Year: 8
$t$: 7
$Q_t$ , million bbl: 1.2
$1.1^{-t}$: 0.5132
$Q_t\,1.1^{-t}$: 0.6158

Year: 9
$t$: 8
$Q_t$ , million bbl: 0.8
$1.1^{-t}$: 0.4665
$Q_t\,1.1^{-t}$: 0.3732

Year: Total
$t$: —
$Q_t$ , million bbl: 12.0
$1.1^{-t}$: —
$Q_t\,1.1^{-t}$: 7.8987

Year	$t$	$Q_t$ , million bbl	$1.1^{-t}$	$Q_t\,1.1^{-t}$
3	2	1.6	0.8264	1.3223
4	3	2.4	0.7513	1.8032
5	4	2.4	0.6830	1.6392
6	5	2.0	0.6209	1.2418
7	6	1.6	0.5645	0.9032
8	7	1.2	0.5132	0.6158
9	8	0.8	0.4665	0.3732
Total	—	12.0	—	7.8987

The time origin is the middle of year 1, and the flows are in the middle of each year. There is no production in years 1, 2 and 10. The data in this example are assumed for illustration.

CalculatedYearly production of the example project (light bars), and the same production multiplied by the discount factors at 10% (dark bars). The light bars add up to 12 million bbl, the dark ones to 7.90 million bbl.

Discounted production of the project at different rates

Rate $i$: 0%
$Q_\text{disc}$ , million bbl: 12.000
$k_\text{disc}$: 1.000

Rate $i$: 5%
$Q_\text{disc}$ , million bbl: 9.654
$k_\text{disc}$: 0.805

Rate $i$: 10%
$Q_\text{disc}$ , million bbl: 7.899
$k_\text{disc}$: 0.658

Rate $i$: 15%
$Q_\text{disc}$ , million bbl: 6.559
$k_\text{disc}$: 0.547

Rate $i$: 20%
$Q_\text{disc}$ , million bbl: 5.518
$k_\text{disc}$: 0.460

Rate $i$	$Q_\text{disc}$ , million bbl	$k_\text{disc}$
0%	12.000	1.000
5%	9.654	0.805
10%	7.899	0.658
15%	6.559	0.547
20%	5.518	0.460

The time origin is the middle of year 1, and the flows are in the middle of each year. At 10%, every 1000 bbl of the project's reserves are “worth” as much as 658 bbl sold at once. At 20%, they are worth as much as 460 bbl.

We can see why a project's value is concentrated in the first years of production already from the barrels. The distant years are discounted more heavily, and they also give less oil, because production declines. In money terms the effect is even stronger. Fixed costs are spread over an ever smaller volume. Income tax rises towards the end of the project, when the depreciation deductions (the yearly write-off of the capital cost for tax purposes) run out (the project's cash flow). So the project's NPV reacts more strongly to the timing of start-up and the pace of the build-up than to the accuracy of the forecast of late production.

If the price changes: escalation, nominal money and a price scenario

We can take the price outside the summation sign only if it is the same in all years. If the price in the calculation changes at a constant rate, we can still “hide” it, in the rate at which the barrels are discounted.

The rate for barrels at a constant rate of price growth

PV_R = \sum_t \frac{P_0\,(1+e_r)^{t}\,Q_t}{(1+i)^{t}} = P_0\,Q_\text{disc}(i_q), \qquad 1 + i_q = \frac{1+i}{1+e_r}

where:

$P_0$ price at the time origin ( $t = 0$ ), $/bbl
$e_r$ real rate of price growth (growth above general inflation); negative if the price falls, fraction per year
$i_q$ the rate at which barrels are discounted, fraction per year
$Q_\text{disc}(i_q)$ discounted production calculated at the rate $i_q$ , million bbl
$i$ discount rate for money (real), fraction per year

Units: fractions per year; prices in $/bbl, volumes in million bbl

Holds when

The rate of price growth is constant. The real rate is growth above general inflation (escalation).
The price of year $t$ is $P_0(1+e_r)^{t}$ , with the same time origin as for production.

A rising price lowers the rate for barrels. A falling price raises it. A special case is a nominal calculation (in the money of each year). With a nominal rate and a price that is constant in nominal money, barrels are discounted at the nominal rate. Such a price falls in real terms at the rate of inflation. Never mix a constant real price with a nominal rate (consistency of flow and rate).

Worked exampleThe price rises by 2% a year

Given (assumed data)

Production: the example project, 12 million bbl in years 3–9
Price: $50/bbl in year 1 ( $t = 0$ ), then rising by 2% a year in real terms
Rate and convention: 10% (real); time origin in the middle of year 1, flows in the middle of the year

Solution

Rate for barrels: $i_q = 1.10/1.02 - 1 = 7.843$ %.
Discounted production at this rate: $Q_\text{disc}(i_q) = 8.596$ million bbl, instead of 7.899 at 10%.
Present value of the revenue: $50\cdot 8.596 = 429.8$ million dollars. If we discount the revenue year by year, with prices from $52.02 in year 3 to $58.58 in year 9, we get the same result: 429.8 million dollars.
The average price, weighted by discounted production, is $429.8/7.899 = 54.4$ $/bbl. The simple average of the prices of years 3–9 is $55.2/bbl.

Answer: The present value of the revenue is 429.8 million dollars, 8.8% more than at a constant price of $50/bbl.

What it means. Price growth partly cancels the effect of discounting, so barrels are discounted at a lower rate. A simple average of the prices would overstate the revenue, because the highest prices come in the late years, when production is small and has little weight. If the price fell by 2% a year in the same way, barrels would have to be discounted at 12.24%. The data in this example are assumed for illustration.

Suppose the price is given by a scenario (a set of assumptions about the future) year by year, without a constant rate. Then there is only one honest way: discount the revenue. We can still write the result through barrels: $PV_R = \bar P\,Q_\text{disc}$ , where $\bar P$ is the average price weighted by the discounted production of each year. A simple average of prices is no good for this, because it takes account of neither volumes nor timing.

Discounted production is not reserves. Reserves and the production profile are physical volumes. We design facilities, plan sales and report to the state on the basis of these volumes. $Q_\text{disc}$ measures the value of these volumes at the chosen rate.
It is not a forecast of losses. Discounting does not reduce the oil in the reservoir. It shows that late barrels are worth less today than early ones.
It is not a number without conditions. $Q_\text{disc}$ changes with the rate, the time origin, the timing of the flows, and whether the price is constant.

Exponential decline with continuous discounting

An annual table is not the only way. Suppose the flow rate of a well, or of a whole accumulation (an oil pool in the ground), declines exponentially: $q(t) = q_i e^{-Dt}$ (exponential decline; this is the same function as in continuous compounding of interest, continuous compounding). Suppose also that money is discounted continuously by the factor $e^{-jt}$ with the rate $j = \ln(1+i)$ (frequent and continuous compounding). Then discounted production is an integral (a continuous sum) that has an exact formula.

Discounted production under exponential decline

Q_\text{disc} = \int_0^{T} q_i\,e^{-Dt}\,e^{-jt}\dd t = \frac{q_i\left[1 - e^{-(D+j)T}\right]}{D + j}

where:

$q_i$ initial flow rate, expressed as a volume per year (daily rate × 365), bbl per year (m³ per year, t per year)
$D$ nominal decline rate (the exponent in $q = q_i e^{-Dt}$ ); the fraction of decline in a year is $d = 1 - e^{-D}$ , 1/year
$j$ continuous discount rate, $j = \ln(1+i)$ , 1/year
$T$ time until the flow rate falls to the economic limit $q_\text{lim}$ (the lowest flow rate at which the well still pays its way): $T = \ln(q_i/q_\text{lim})/D$ , years
$Q_\text{disc}$ discounted production over the time $T$ , brought to the start of production, bbl (m³, t)

Units: $q_i$ is a volume per year, $D$ and $j$ are in 1/year, $T$ in years; $Q_\text{disc}$ in the same volume units

Holds when

The decline is exponential from the start of production, and the time origin is the start of production ( $t = 0$ ).
Discounting is continuous: $j$ corresponds to the effective annual rate $i$ .
Production stops at the flow rate $q_\text{lim}$ . Without a limit ( $T \to \infty$ ), $Q_\text{disc} = q_i/(D + j)$ .

Without discounting, the production over the same time is $N_p = q_i(1 - e^{-DT})/D$ . Comparing the two formulas shows the main point: discounting acts like an extra decline at the rate $j$ . Without an economic limit, the present-volume factor is $k_\text{disc} = D/(D + j)$ . The faster the decline, the closer it is to one, because the oil is produced sooner.

CalculatedPresent-volume factor

k_\text{disc} = D/(D + j)

for exponential decline without an economic limit, against the discount rate, for three decline rates. The well from the example below is marked, as if it produced for ever.

Worked exampleA new well with an exponential decline in flow rate

Given (assumed data)

Initial flow rate $q_i$: 1000 bbl/d (about 159 m³/d, or 137 t/d at an oil density of 0.86 t/m³)
Nominal decline rate $D$: 0.25 1/year (the flow rate falls by 22.1% in a year)
Economic limit: 50 bbl/d
Rate: 10% a year (effective, real)

Solution

Producing life: $T = \ln(1000/50)/0.25 = 11.983$ years. Continuous rate: $j = \ln 1.1 = 0.09531$ 1/year, and $D + j = 0.34531$ 1/year.
Reserves of the well, that is, the oil it will produce: $q_i = 365{,}000$ bbl per year, $e^{-DT} = 50/1000$ , and $N_p = 365{,}000\cdot(1 - 0.05)/0.25 = 1387$ thousand bbl.
Discounted production: $e^{-(D+j)T} = 0.01596$ . Then $Q_\text{disc} = 365{,}000\cdot(1 - 0.01596)/0.34531$ $= 1040$ thousand bbl.
Present-volume factor: $k_\text{disc} = 1040/1387 = 0.750$ . Without an economic limit it would be $0.25/0.34531 = 0.724$ .
Check with an annual table. If we discount the volume of each year to the middle of its production time, we get 1037.7 thousand bbl (0.2% less). If we discount it to the end of the year, we get 989.4 thousand bbl (4.9% less).

Answer: The well will produce 1387 thousand bbl in about 12 years. The discounted equivalent of this oil is about 1040 thousand bbl (

k_\text{disc} \approx 0.750

What it means. The continuous formula almost agrees with the annual calculation at mid-year, but it differs noticeably from the end-of-year calculation. This is one more argument for the mid-year convention for production that flows steadily. In the first three years the well gives 56% of its reserves but 66% of the discounted production. Here too, the early years decide. The economic limit raises $k_\text{disc}$ slightly, because it cuts off the most heavily discounted “tail”. The data in this example are assumed for illustration.

Level-equivalent production

Sometimes it is convenient to replace a production profile with a level one (the same production every year). For example, we may want to compare projects by capital expenditure per barrel of daily production (capital intensity). A simple average, which is the reserves divided by the production period, understates the value of a profile with an early peak. The level production that is equivalent to a profile must give the same discounted production over the same years. We find it from the sum of the discount factors of those years. For consecutive years, this sum is given by the annuity factor.

Level annual production equivalent to a profile

q_\text{eq} = \frac{Q_\text{disc}}{\sum_{t} (1+i)^{-t}}

where:

$q_\text{eq}$ level annual production equivalent to the profile, million bbl per year
$Q_\text{disc}$ discounted production of the profile, million bbl
$\sum_{t}(1+i)^{-t}$ sum of the discount factors of the production years; for $n$ consecutive years, the first of which is $m + 1$ years away, it is $(1+i)^{-m}\,a_n(i)$ , years

Units: million bbl per year; divide by 365 to get the daily production

Holds when

The production years, the rate and the convention for the timing of flows are the same as for the profile.
Equivalence is understood at a constant price: both productions bring the same present value of revenue.

If the profile is front-loaded (most of the oil comes early), the level production of the same value is higher than the simple average. If production grows towards the end of the term, it is lower.

Worked exampleWhat level production the project's profile is equivalent to

Given (assumed data)

Discounted production of the project: 7.899 million bbl at 10% (middle of year 1, flows in the middle of the year)
Production years: years 3 to 9 ( $t = 2 \ldots 8$ ), 12 million bbl

Solution

Sum of the factors of the production years: $\sum_{t=2}^{8} 1.1^{-t} = a_7(10\%)/1.1$ $= 4.8684/1.1 = 4.4258$ .
Level-equivalent production: $q_\text{eq} = 7.899/4.4258 = 1.785$ million bbl a year, or 4890 bbl/d.
The simple average is $12/7 = 1.714$ million bbl a year (4697 bbl/d). The plateau is 2.4 million bbl a year (6575 bbl/d).

Answer: The project's profile is equivalent to a level production of 1.785 million bbl a year (about 4890 bbl/d), which is 4.1% more than the simple average.

What it means. A level profile with the same total production would be worth less, because it moves some of the oil from the early years to the late ones. Which of the three numbers to use for unit indicators (the plateau, the simple average or the equivalent average) depends on the question you ask. Always state which one you used (costs per barrel of daily production). The data in this example are assumed for illustration.

The same idea works for money too: replace an uneven series with an equal series that has the same present value. The levelised cost per barrel is the constant price at which the present value of the revenue equals the present value of the costs (Unit costs per barrel). The equal annual payments that are equivalent to a capital investment are the subject of the lesson Annual capital charge.

Terms and symbols

Annuity: A series of equal payments at equal intervals. In an ordinary annuity the payments are at the end of the periods, and in an annuity due they are at the beginning.
Annuity factor $a_n(i)$: It equals $[1 - (1+i)^{-n}]/i$ and is the present value of one unit received at the end of each of $n$ periods.
Accumulation factor of an annuity $s_n(i)$: It equals $[(1+i)^{n} - 1]/i$ and shows what $n$ equal contributions grow into by the date of the last one.
Perpetuity: An endless series of payments. It is worth $A/i$ , and one that grows at a rate $g < i$ is worth $A_1/(i - g)$ .
Growing annuity: A series in which each payment is larger than the previous one by the fraction $g$ . Declining production is a growing annuity with $g < 0$ .
Discounted production $Q_\text{disc}$: The production of each year multiplied by the discount factor of that year, and added up. At a constant price, the present value of the revenue equals $P\,Q_\text{disc}$ .
Present-volume factor $k_\text{disc}$: The ratio of discounted production to undiscounted production, that is, the share of the reserves that “reaches” the time origin.
Barrel discount rate $i_q$: The rate at which production is discounted when the price grows at a constant real rate. It equals $(1+i)/(1+e_r) - 1$ .
Nominal decline rate $D$: The exponent in $q = q_i e^{-Dt}$ . The fraction of decline in a year is $d = 1 - e^{-D}$ .
Level-equivalent production: The constant yearly production over the same years that gives the same discounted production as the profile.

Common mistakes and misconceptions

Wrong

“Discounting barrels is meaningless: oil in the reservoir earns no interest.”

In fact

It is not the oil that is discounted but the revenue from it. At a constant price, the present value of the revenue equals the price multiplied by the discounted production. It is the same sum, written in another way. Discounted production measures the value of barrels over time. It does not claim that barrels “grow”.

Wrong

“A 20-year series is worth twice as much as a 10-year series.”

In fact

Only without discounting. At 10%, the annuity factor for 20 years is 8.514 and for 10 years 6.145, so the 20-year series is worth only 39% more. The second ten years are worth less today than the first ten, because each of their payments is discounted for 10 years longer.

Wrong

“The model's flows are assigned to the middle of the year, but the annuity factor does not change because of that.”

In fact

It does change. A series shifted half a year earlier is worth $(1+i)^{0.5}$ times more: 4.9% more at 10%. Multiply the factor for end-of-year payments by this coefficient, or recalculate the series with the convention that you need.

Wrong

“Let us put an annual decline in flow rate of 22% and a rate of 10% into the continuous discounting formula.”

In fact

The continuous formula needs continuous quantities: the nominal decline rate $D = -\ln(1 - 0.22) \approx 0.25$ 1/year and the continuous rate $j = \ln 1.1 \approx 0.095$ 1/year. If we put in 0.22 and 0.10 instead, we understate the decline rate (0.22 instead of 0.25) noticeably more than we overstate the rate (0.10 instead of 0.095). In the end, discounted production comes out too high: by about 8% for the well in the example of this lesson.

Wrong

“The price in the forecast rises, so we calculate discounted production at the usual rate and multiply it by the simple average of prices.”

In fact

A simple average ignores the fact that the high prices come in the late years, when production is small and has little weight. With a constant rate of price growth, barrels are discounted at the rate $(1+i)/(1+e_r) - 1$ . With an arbitrary scenario, the revenue itself is discounted.

In depth: advanced detail

Derivations of the formulas

The annuity factor as a geometric progression. Let $v = 1/(1+i)$ . Then $a_n(i) = v + v^{2} + \ldots + v^{n}$ . This is the sum of $n$ terms of a geometric progression (each term is the previous one times the same ratio), with the first term $v$ and the ratio $v$ . Its sum is $v(1 - v^{n})/(1 - v)$ . Since $1 - v = i/(1+i)$ , the ratio $v/(1 - v)$ equals $1/i$ . So we get $a_n(i) = (1 - v^{n})/i$ , the same formula as the one found through the difference of perpetuities. As $i \to 0$ , the expansion $(1+i)^{-n} \approx 1 - ni$ gives $a_n \to n$ .

The growing annuity. The payment of year $t$ is $A_1(1+g)^{t-1}$ . So $PV = A_1 v\,(1 + q + q^{2} + \ldots + q^{n-1})$ with $q = (1+g)/(1+i)$ . The sum of the progression is $(1 - q^{n})/(1 - q)$ , and $v/(1 - q) = 1/(i - g)$ . This gives the formula of the section on growing and declining series. At $g = i$ the ratio $q = 1$ , each term of the sum is $A_1 v$ , and $PV = n A_1/(1+i)$ . If we take out the factor $1/(1+g)$ , what remains is the sum $q + q^{2} + \ldots + q^{n}$ . This is the annuity factor at the rate $i^{*}$ for which $1/(1+i^{*}) = q$ .

The continuous case. The integral of $q_i e^{-(D+j)t}$ over $t$ from 0 to $T$ is $q_i[1 - e^{-(D+j)T}]/(D+j)$ . At $j = 0$ this is the undiscounted production $N_p$ . The annual calculation comes close to the continuous one if we place the flow of each year at the middle of its production time. Over a short interval the exponential is almost a straight line, so its mean is close to its value at the midpoint. Placing the flow at the end of the year always gives too small a present value, by a factor of about $(1+i)^{0.5}$ .

What return an equal series gives

Suppose we know the investment, the equal receipts and their number. We find the rate at which they are equivalent by trial. At a rate lower than the one we are looking for, the present value of the receipts is larger than the investment. At a higher rate, it is smaller. A convenient way is to halve the interval again and again (bisection; numerical search for the IRR). The rate that we find is the internal rate of return (IRR) of the series. We compare it with the rate that the company requires.

Worked exampleThe return on the variable speed drives

Given (assumed data)

Investment and saving: US$1.9 million now; US$0.42 million at the end of each of 7 years

Solution

The required factor is $a_7(i) = 1.9/0.42 = 4.5238$ .
At 12%, $a_7 = 4.5638$ . This is more than we need, so the rate is above 12%. At 13%, $a_7 = 4.4226$ . This is less than we need, so the rate is below 13%.
Halving the interval again and again gives $i = 12.28$ %. Check: $0.42\cdot a_7(12.28\%) = 1.900$ million dollars.

Answer: The internal rate of return is about 12.28% a year, against the required 10%.

What it means. A margin of 2.3 percentage points agrees with the conclusion of the main example. The project is profitable, but it pays back almost at the end of its service life, and with a service life one year shorter its return would fall below 10%. The data in this example are assumed for illustration.

Connections to other topics

Discounting and present value — discount factors, the present value of a series and conventions for the timing of payments.
Interest and compounding — compounding, continuous compounding and the relation between nominal and effective rates.
Net present value (NPV) — the sum of a project's discounted flows, the time origin and conventions.
Unit costs per barrel — levelised cost per barrel: costs divided by discounted production.
Geological and technical uncertainty of a project — discounted production as a measure of the loss from slow wells and delays.
Sensitivity analysis and scenarios — the slope of NPV against price: discounted production less royalty and tax.
Inflation: nominal and real money — real price escalation and matching the rate to the money of the calculation.
Cumulative cash flow and payback period — payback period and IRR for equal receipts.
Annual capital charge — equal annual payments equivalent to capital investment.

Check yourself

1
Under a contract, the lease of an installation costs US$2.5 million at the end of each year for 6 years. What is the present value of the payments at a nominal rate of 9%?
1. A.US$15.0 million
2. B.US$11.2 million
3. C.US$12.2 million
4. D.US$27.8 million
Show the answer
Correct answer: B. US$11.2 million
$a_6(9\%) = (1 - 1.09^{-6})/0.09 = 4.4859$ , and $PV = 2.5\cdot 4.4859 = 11.2$ million dollars. The answer 15.0 is the sum without discounting. The answer 12.2 would be the value if the payments were made in advance, at the beginning of each year ( $\times 1.09$ ). The answer 27.8 is the value of a perpetuity, $2.5/0.09$ .
2
A buyer is offered the right to US$1 million a year, either for ever or for 30 years. The rate is 12%. How much more is the perpetual right worth, and what exactly does the buyer pay extra for?
Show the answer
The perpetual right is worth $1/0.12 = 8.333$ million dollars, and the 30-year right is worth $a_{30}(12\%) = 8.055$ million dollars. The difference is about 3.5%. The extra is paid for all the payments after year 30. But each of them is discounted by more than 30 years, and together they are worth only 0.28 million dollars today. So at rates of 10–15% a 30-year series is almost equivalent to a perpetual one.
3
In a calculation in constant prices, the oil price rises by 3% a year and the discount rate is 10% (real). At what rate should production be discounted so that the present value of the revenue equals the initial price multiplied by the discounted production?
1. A.7.0%
2. B.6.8%
3. C.10%
4. D.13.3%
Show the answer
Correct answer: B. 6.8%
$1 + i_q = 1.10/1.03$ , so $i_q = 6.80$ %. Subtracting, $10 - 3$ , gives only an approximate 7.0%. The answer 13.3% comes from multiplying, $1.10\cdot1.03$ , instead of dividing. Barrels would then be discounted harder, as if the price were falling. The answer 10% is right only if the price is constant.
4
Two wells start with the same flow rate. The first has a nominal decline rate $D = 0.1$ 1/year, the second $0.4$ 1/year. With no economic limit and at a rate of 10%, compare their reserves and discounted production.
Show the answer
The reserves $q_i/D$ differ fourfold: they are $10\,q_i$ and $2.5\,q_i$ (where $q_i$ is the initial flow rate per year). The discounted production is $q_i/(D + j)$ with $j = \ln 1.1 = 0.0953$ . It is $5.120\,q_i$ and $2.019\,q_i$ , so the difference is already down to 2.54 times. The present-volume factors are 0.51 and 0.81. The slow well loses almost half the value of its reserves to discounting, and the fast one loses a fifth.
5
A transport tariff under a contract grows by 8% a year: US$1.0 million at the end of year 1, then 8% more each year, 5 payments in all. The nominal discount rate is also 8%. What is the present value of the payments?
1. A.Undefined: at $g = i$ the growing annuity formula divides by zero
2. B.US$3.99 million
3. C.US$4.63 million
4. D.US$5.00 million
Show the answer
Correct answer: C. US$4.63 million
When $g = i$ , the growth of each payment exactly cancels its discounting, so all the payments are worth the same today: $1/1.08$ each. So $PV = 5/1.08 = 4.63$ million dollars. The answer 5.00 is $nA_1 = 5 \cdot 1.0$ , which leaves out the discounting of the first payment. The answer 3.99 is the value of the series without growth, $a_5(8\%)$ . The division by zero comes from the way the formula is written, not from the problem itself: the limit is finite.
6
A well gives US$0.6 million of net income at the end of each year. How many years will it take to pay back the US$3.0 million invested in it, at a rate of 12%? And at a rate of 20%?
Show the answer
We need $a_n = 3.0/0.6 = 5$ . At 12%: $n = -\ln(1 - 0.12\cdot5)/\ln 1.12 = -\ln 0.4/\ln 1.12 = 8.09$ years, against 5 years of simple payback. At 20%, the interest on the investment, $0.2\cdot3.0 = 0.6$ million dollars a year, equals the whole income. The debt to the investor does not get smaller, so the investment never pays back ( $1 - 0.2\cdot5 = 0$ , and the logarithm of zero is undefined).

Updated: 28 October 2026