Annual capital charge

Quick refresher

The annual capital charge is $A = C\cdot \mathrm{CRF}_n(i)$ , where $\mathrm{CRF}_n(i) = 1/a_n(i)$ is one divided by the annuity factor. At 10% and a life of 7 years, $\mathrm{CRF} = 0.2054$ : every 100 dollars invested need 20.54 dollars a year.
The factor is made of a return on capital $i$ and a return of capital $1/s_n(i)$ . With a short life, the charge mostly returns the investment. With a long life, it mostly pays for the use of it. For an endless life only $i$ is left, and at a zero rate only $1/n$ .
The investments of several years are brought to one date: the earlier ones are compounded forward (grown with interest; the extra is called interest during construction), and the later ones are discounted. The date must be one period before the first payment. Or the charge is found by dividing the discounted investments by the sum of the discount factors of the payment years.
The charge per barrel (the annual charge divided by the annual production) is unambiguous only when production is level. For an uneven profile, divide by the level-equivalent production, and then it matches the levelised capital cost per barrel.
Equivalent annual cost compares options with different service lives, if each option can be renewed as many times as needed. If the asset is needed for less than its service life, compare the costs over the actual period: the unused life is worth only what the asset can be sold or moved for.
Any depreciation scheme (a way to write off the cost of an asset over its life), together with the return $i$ on the capital not yet returned, has the same present value: the investment. An equal payment goes with a growing (annuity) depreciation. With straight-line depreciation (the same write-off every year), the accounting return of an asset grows with its age, although the true return does not change.

Key relationships

\mathrm{CRF}_n(i) = \frac{1}{a_n(i)} = \frac{i}{1 - (1+i)^{-n}}

capital recovery factor: payments at the end of each year

\mathrm{CRF}_n(i) = i + \frac{1}{s_n(i)}

return on capital plus return of capital

A = \frac{\sum_t C_t\,(1+i)^{-t}}{\sum_{t \in P} (1+i)^{-t}}

an equal payment in the years $P$ for the investments of several years

\mathrm{EAC} = \mathrm{CRF}_n(i)\,C + O - \frac{S}{s_n(i)}

equivalent annual cost of an option with a service life $n$

A_t = i\,B_{t-1} + D_t

the payment of a year is the return on the remaining capital plus depreciation

Common pitfalls

Dividing the investment by the service life: $C/n$ is a return of capital without a return on it. At 10% and 7 years, the charge is understated by almost a third.
Bringing the investments to the date of the first payment and multiplying by the factor for end-of-year payments: the charge is too high by a factor of $1 + i$ .
Dividing the annual charge by the production of the plateau or of one year when the production profile is uneven.
Comparing options by equivalent annual cost over the full service life when the asset is needed for less than that.
Taking the accounting return (profit after straight-line depreciation divided by the book value, which is the value of the asset in the accounts) for the return on the investment.

After this lesson you can

calculate the capital recovery factor and the annual capital charge, and split the charge into a return on capital and a return of capital
bring the investments of several years to one date that fits the moment of the first payment
calculate the capital charge per barrel, and choose which production to divide it by
compare options with different service lives by equivalent annual cost, and recognise when such a comparison is wrong
explain how the capital charge is related to depreciation, income tax and accounting return

What the annual capital charge is

Capital expenditure (capex) is made at the start: a well is drilled, and a unit is built and paid for before it begins to work. But the asset brings benefit for years, in the oil it pumps, the gas it treats and the water it injects. To compare a one-off investment with yearly quantities such as revenue, operating costs and tariffs, we replace it with an equal yearly charge over the whole service life. The charge is chosen so that its present value at the discount rate equals the investment (Discounting and present value). Then, over the service life, it returns the capital together with a return equal to the rate.

It is easiest to picture the asset as bought on credit at the rate $i$ , with the debt repaid by equal yearly payments. Each payment has two parts. The return on capital is the interest on the part of the investment that has not yet been returned. It is largest in the first year. The return of capital is the repayment of the investment itself. It grows as the interest falls. The rate here is not necessarily the interest on a loan. It is the return that the company requires, its discount rate (Choosing the discount rate). By putting money into the asset, the company gives up such a return elsewhere.

Tariffs and prices of services. The owner of a pipeline, a treatment unit or a compressor station (a plant that raises the pressure of gas) must receive for each year of work at least the capital charge plus the operating costs. Divide this sum by the volume, and you get the minimum tariff.
Lease instead of purchase. The rent for a unit, a drilling rig or a floating production system (a ship or platform that produces oil at sea) is in the end the capital charge of the owner plus the owner's costs and profit. Comparing it with your own capital charge shows whether the lease is expensive (buy or lease).
Options with different service lives. A pump that lasts five years and a pump that lasts ten years cannot be compared by their purchase price. They are compared by equivalent annual cost.
Unit costs. The annual charge divided by the annual production is the capital part of the cost per barrel (Unit costs per barrel).

The capital recovery factor

The present value of $n$ equal payments $A$ at the end of each year is $A\,a_n(i)$ , where $a_n(i) = [1 - (1+i)^{-n}]/i$ is the annuity factor (the annuity factor). Set it equal to the investment $C$ , and we get the payment $A = C/a_n(i)$ . The factor $1/a_n(i)$ shows what share of the investment must be received every year. This is the capital recovery factor. In bank calculations it is called the annuity coefficient. In Russian-language valuation practice it is called the instalment to amortise one unit. Do not confuse it with the annuity factor $a_n(i)$ .

Capital recovery factor and the annual capital charge

A = C\cdot \mathrm{CRF}_n(i), \qquad \mathrm{CRF}_n(i) = \frac{1}{a_n(i)} = \frac{i}{1 - (1+i)^{-n}}

where:

$A$ equal capital charge at the end of each of $n$ years, US$ million a year
$C$ the investment, assigned to the moment one year before the first payment, US$ million
$\mathrm{CRF}_n(i)$ capital recovery factor: the share of the investment received each year, 1/year
$a_n(i)$ annuity factor, years
$i$ discount rate: the required return, fraction per year
$n$ number of payments: the service life of the asset, years

Units: any currency; $A$ and $C$ in the same money, and the rate is effective for the period of payment

Holds when

The payments are equal and arrive at the end of each year, the first one a year after the date to which the investment is assigned. For other payment moments, the factor is recalculated in the same way as the annuity factor (payment in advance, mid-year payment, delayed start).
The rate is constant. In constant prices it is a real rate, and the charge is constant in real terms. In a nominal calculation it is a nominal rate, and the charge is constant in nominal money and falls in real terms (Inflation: nominal and real money).
At the end of its life the asset is worth nothing. The residual value is treated separately (see the section on equivalent annual cost).
Income tax is not included: this is a charge before tax (tax is in the advanced section).

$\mathrm{CRF}$ is the share of the investment that must be received each year for the investment to pay back with a return $i$ in exactly $n$ years. Three limits help to check a calculation. At $n = 1$ the factor equals $1 + i$ : the investment and its return come back in one payment. As $n \to \infty$ it tends to $i$ : forever, you can receive only the return, without touching the capital. As $i \to 0$ it tends to $1/n$ : without the cost of money, the charge simply divides the investment equally over the years.

The charge can be split in another way, into two parts, each with its own meaning. The factor always equals the rate plus one divided by the accumulation factor of an annuity, $s_n(i) = [(1+i)^{n} - 1]/i$ . This is the sum to which one dollar paid at the end of each year grows by the end of the term (the derivation is in the advanced section):

Return on capital and return of capital in the factor

\mathrm{CRF}_n(i) = i + \frac{1}{s_n(i)} = i + \frac{i}{(1+i)^{n} - 1}

where:

$i$ return on capital: the interest on the whole investment, 1/year
$1/s_n(i)$ return of capital: the equal yearly contribution that accumulates the whole investment in $n$ years at a return $i$ (in valuation practice, the sinking fund factor), 1/year
$s_n(i)$ accumulation factor of an annuity, years

Units: shares of the investment a year

Holds when

The same as for the capital recovery factor.
The contributions to the fund for the return of capital earn the same return $i$ .

Two ways of returning the investment with a return $i$ cost the same. One is to repay the “debt” by equal payments. The other is to pay each year the return on the whole investment, and also to put aside $1/s_n(i)$ of the investment into a fund. By the end of the life, the fund grows to the full sum of the investment (the accumulated sum of a series). The longer the life, the less needs to be put aside, and the larger the part of the payment that is the return on capital.

The capital recovery factor at a rate of 10% and what it is made of

Life $n$ , years: 3
$\mathrm{CRF}_n$: 0.4021
of it, the return of capital $1/s_n$: 0.3021
share of the return on capital: 25%
$1/n$: 0.3333

Life $n$ , years: 5
$\mathrm{CRF}_n$: 0.2638
of it, the return of capital $1/s_n$: 0.1638
share of the return on capital: 38%
$1/n$: 0.2000

Life $n$ , years: 7
$\mathrm{CRF}_n$: 0.2054
of it, the return of capital $1/s_n$: 0.1054
share of the return on capital: 49%
$1/n$: 0.1429

Life $n$ , years: 10
$\mathrm{CRF}_n$: 0.1627
of it, the return of capital $1/s_n$: 0.0627
share of the return on capital: 61%
$1/n$: 0.1000

Life $n$ , years: 15
$\mathrm{CRF}_n$: 0.1315
of it, the return of capital $1/s_n$: 0.0315
share of the return on capital: 76%
$1/n$: 0.0667

Life $n$ , years: 20
$\mathrm{CRF}_n$: 0.1175
of it, the return of capital $1/s_n$: 0.0175
share of the return on capital: 85%
$1/n$: 0.0500

Life $n$ , years: 30
$\mathrm{CRF}_n$: 0.1061
of it, the return of capital $1/s_n$: 0.0061
share of the return on capital: 94%
$1/n$: 0.0333

Life $n$ , years: endless
$\mathrm{CRF}_n$: 0.1000
of it, the return of capital $1/s_n$: 0.0000
share of the return on capital: 100%
$1/n$: 0.0000

Life $n$ , years	$\mathrm{CRF}_n$	of it, the return of capital $1/s_n$	share of the return on capital	$1/n$
3	0.4021	0.3021	25%	0.3333
5	0.2638	0.1638	38%	0.2000
7	0.2054	0.1054	49%	0.1429
10	0.1627	0.0627	61%	0.1000
15	0.1315	0.0315	76%	0.0667
20	0.1175	0.0175	85%	0.0500
30	0.1061	0.0061	94%	0.0333
endless	0.1000	0.0000	100%	0.0000

The return on capital in all rows is $i = 0.1000$ a year, and $\mathrm{CRF}_n = 0.1000 + 1/s_n$ . The last column is the factor without discounting. The amount by which it is smaller than $\mathrm{CRF}_n$ is the amount by which simple division of the investment by the life understates the charge.

CalculatedThe capital recovery factor at a rate of 10% against the service life. The dashed line is the return on capital

i

, which the factor approaches when the life is long. The lower curve is

1/n

, the factor without discounting. The life of 7 years is marked.

Worked exampleCapital charge for an oil treatment plant

Given (assumed data)

The oil treatment plant of the example project of this library: US$60 million, part of the capital expenditure of year 2
Service life: 7 years of production, years 3 to 9; after year 9 the plant is not needed and is worth nothing
Rate and timing: 10% (real); the investment and the payments are in the middle of the year, and the first payment comes a year after the investment

Solution

The annuity factor is $a_7(10\,\%) = (1 - 1.1^{-7})/0.1 = 4.86842$ . The capital recovery factor is $\mathrm{CRF}_7 = 1/4.86842 = 0.205405$ .
The annual charge is $A = 60\cdot 0.205405 = 12.324$ million dollars a year.
In the first payment, the return on capital is $0.1\cdot 60 = 6.000$ million dollars, and the return of capital is $12.324 - 6.000 = 6.324$ million dollars. In the last payment they are 1.120 and 11.204 million dollars: the return is now charged on a small balance.
Without discounting: $60/7 = 8.571$ million dollars a year, which is 30% less.

Answer: The plant “costs” the project 12.324 million dollars in each of the seven years of production. Simple division by the life understates this sum by almost a third.

What it means. For the investment to pay back with a return of 10%, the plant must bring the project at least 12.324 million dollars a year, in processed and sold oil, above its own operating costs and before tax. Division by the life without interest is straight-line depreciation in the accounting sense. It returns only the capital itself, and it does not take into account that the money was tied up in the plant for seven years. The data in this example are assumed for illustration.

Investments of several years and the reference date

A large asset is built over several years, and the sum of the estimates of all the years is a poor measure of the investment: a dollar spent in the first year of construction is tied up in the asset longer than a dollar spent in the last. So the investments are brought to one date: the earlier ones are compounded forward (grown with interest up to that date), and the later ones are discounted back to it (moving sums in time). The extra that the compounded sums give over what was actually paid is called interest during construction. The capital recovery factor assumes that the first payment comes a year after this date. So it is most convenient to bring the investments to a date one year before the first payment.

You can also do without choosing a date: take the investments brought to any origin, and divide them by the sum of the discount factors of the years in which payments are made. Moving the origin multiplies the numerator and the denominator by the same factor, so the result does not depend on it. The formula also works for payments that are not consecutive.

The annual charge for the investments of several years

A = \frac{\sum_{t} C_t\,(1+i)^{-t}}{\sum_{t \in P} (1+i)^{-t}}

where:

$A$ equal payment in each year of the set $P$ , US$ million a year
$C_t$ the investment of the year that is $t$ years from the origin, US$ million
$P$ the years in which payments are made, usually the years of operation of the asset, —
$i$ discount rate, fraction per year

Units: any currency; $t$ of all the sums is counted from one origin

Holds when

The investments and the payments are assigned to moments by one convention, for example the middle of the year.
If payments are made for $n$ years in a row and the first is a year after the chosen date, the denominator, brought to this date, equals $a_n(i)$ . Then the formula becomes $A = C\cdot \mathrm{CRF}_n(i)$ with the investments brought to this date.

The numerator is the investments in money of the origin. The denominator is the cost, in the same money of the origin, of a payment of “one a year” in each year of operation. Their ratio is the equal payment with the same present value.

Worked exampleThe annual capital charge of the example project

Given (assumed data)

Capital expenditure: 50, 90 and 20 million dollars in years 1–3, in the middle of each year
Years of payments: the years of production, years 3 to 9, in the middle of the year ( $t = 2 \ldots 8$ counted from the middle of year 1)
Rate: 10% (real)

Solution

The date a year before the first payment is the middle of year 2: $50\cdot1.1 + 90 + 20/1.1 = 163.18$ million dollars.
The charge is $A = 163.18\cdot 0.205405 = 33.52$ million dollars a year.
A check by the general formula: the investments brought to the middle of year 1 are $50 + 90/1.1 + 20/1.1^{2} = 148.35$ million dollars. The sum of the discount factors of years 3–9 is 4.4258. So $148.35/4.4258 = 33.52$ million dollars.
Simple division: $160/7 = 22.86$ million dollars a year, which is 32% less.

Answer: The capital of the project costs 33.52 million dollars in each year of production. Each of the seven years must bring this much above the operating costs, before tax, just to return 160 million dollars with a return of 10%.

What it means. By the middle of year 2, the investments “weigh” 163.18 million dollars, not 160. The 50 million dollars of year 1 have been tied up a year longer and have grown to 55, and the 20 million dollars of year 3 are paid after this date and are worth less. The capital charge is a threshold, not a forecast. The project is worthwhile if the present value of its margin (revenue less operating costs) is larger than the present value of such charges and of abandonment, that is, if the net present value (NPV) before tax is positive (Net present value (NPV)). The data in this example are assumed for illustration.

Capital charge per barrel

Divide the annual capital charge by the annual volume, and you get the capital part of the cost per unit of product: per barrel of oil, per thousand cubic metres of gas, per tonne of oil pumped. If the volume is the same every year, as with a pipeline or a plant loaded under a contract, the answer is unambiguous. For a field, production changes from year to year, and the question arises: which production should we divide by?

Capital charge per barrel

c_K = \frac{A}{q_\text{eq}} = \frac{\sum_t C_t\,(1+i)^{-t}}{Q_\text{disc}}

where:

$c_K$ the capital charge per barrel: the constant fee that returns the investment with a return $i$ , $/bbl
$A$ annual capital charge in the years of production, US$ million a year
$q_\text{eq}$ level-equivalent production: the level yearly production over the same years that has the same discounted production as the real profile, million bbl a year
$Q_\text{disc}$ discounted production at the same rate and with the same origin as the investments, million bbl

Units: US$ million divided by million bbl gives $/bbl

Holds when

The charge is paid in the years of production, and $q_\text{eq}$ is calculated for the same years (level-equivalent production).
With level production $q$ , the denominator equals $q$ , and $c_K = A/q$ .
This is part of the costs before tax. The operating costs per barrel are added separately.

The two fractions are equal. The charge $A$ is the discounted investment divided by the sum of the discount factors of the production years, and $q_\text{eq}$ is the discounted production divided by the same sum. So the charge per barrel of level-equivalent production is the levelised capital cost per barrel (technical cost per barrel).

Worked exampleHow much capital falls on each barrel treated by the plant

Given (assumed data)

Charge for the treatment plant: 12.324 million dollars in each year of production (the example above)
Production: a plateau of 2.4 million bbl a year; 12 million bbl over 7 years; the discounted production at 10% is 7.8987 million bbl (middle of year 1)
Level-equivalent production at 10%: 1.7847 million bbl a year

Solution

Per barrel of the plateau: $12.324/2.4 = 5.14$ $/bbl. This is the lowest of the possible values: at the plateau level the plant is loaded for only two years out of seven.
Per average barrel: $12/7 = 1.7143$ million bbl a year, and $12.324/1.7143 = 7.19$ $/bbl.
Per level-equivalent production: $12.324/1.7847 = 6.91$ $/bbl.
A check: the investment brought to the middle of year 1, $60/1.1 = 54.545$ million dollars, divided by the discounted production gives $54.545/7.8987 = 6.91$ $/bbl, the same number.

Answer: The constant tariff that would return the cost of the plant with a return of 10% is 6.91 dollars for each barrel treated. Division by the production of the plateau understates it to 5.14, and division by the average production overstates it to 7.19 dollars.

What it means. Dividing by the average production overstates the answer, because the production profile of the project is front-loaded (most of the oil comes early): the early barrels are worth more, and their contribution to the payback is larger. For all the 160 million dollars of the investment of the project, the same division gives $33.52/1.7847 = 18.78$ $/bbl. This is the capital part of the levelised cost per barrel from the lesson Unit costs per barrel. The data in this example are assumed for illustration.

The same calculation can be used to test an offer to lease. In the lesson on the cash flow model, the plant worth 60 million dollars was replaced by a lease of 12 million dollars a year over the seven years of production (buy or lease). The capital charge at 10% is 12.324 million dollars a year. So the owner, who receives 12 million dollars, earns a little less than 10%, about 9.2%, as that lesson showed. To earn its 10% and to cover the cost of servicing the plant as well, the owner would have to ask for more.

Equivalent annual cost: options with different service lives

When we choose between options for one asset, such as pumps, compressors or pipes (for example flowlines, which carry the product from the wells to the gathering point), we compare their full costs: investment, operation, repairs and residual value (what the asset is worth at the end of its life). If the service lives are the same, it is enough to compare present values. If they differ, a direct comparison is unfair: a cheap option with a life of 5 years and an expensive option with a life of 10 years serve for different times. To equalise the periods, all the costs of each option are turned into an equal yearly charge over its own service life. This is the equivalent annual cost (EAC).

Equivalent annual cost

\mathrm{EAC} = \mathrm{CRF}_n(i)\left[C + \sum_{t=1}^{n}\frac{O_t}{(1+i)^{t}} - \frac{S}{(1+i)^{n}}\right]

where:

$\mathrm{EAC}$ equivalent annual cost of the option, US$ million a year
$C$ the investment at the start of the service life, US$ million
$O_t$ operating costs of year $t$ of the service life, US$ million
$S$ residual value at the end of the life: the price of sale or of moving to another site, less dismantling, US$ million
$n$ service life of the option, years
$i$ discount rate, fraction per year

Units: any currency; all sums in the same money

Holds when

At the end of its life, each option is replaced by the same one as many times as needed, endlessly or up to the common multiple of the lives. Only then are equal yearly costs comparable between options.
The options provide the same service: the same volume, the same reliability. If not, the difference is taken into account in the flow.
With constant operating costs $O$ , the formula simplifies to $\mathrm{EAC} = \mathrm{CRF}_n(i)\,C + O - S/s_n(i)$ .

EAC is the yearly “rent” that you would have to pay for the option if it could be rented at a price that recovers all its costs with a return $i$ . Choose the option with the lower equivalent annual cost, but only if the asset is really needed for a long time.

Worked exampleSteel or fibreglass flowlines

Given (assumed data)

Option A, fibreglass pipes: US$6.5 million; service life 15 years; inspections US$0.1 million a year
Option B, steel pipes with a corrosion inhibitor (a chemical that slows corrosion): US$3.5 million; service life 7 years; inhibitor and corrosion control US$0.6 million a year
Rate and timing: 10% (real); the investment is at the start of the life, and the costs are at the end of each year; there is no residual value, because pipes in the ground are not recovered
How long the pipes are needed: 7 years of production of the example project (years 3 to 9)

Solution

Equivalent annual costs over each option's own service life: A is $6.5\cdot 0.13147 + 0.1 = 0.955$ , and B is $3.5\cdot 0.205405 + 0.6 = 1.319$ million dollars a year. By this comparison, A is 28% cheaper.
But the field needs the pipes for 7 years. The discounted costs for this time are: A is $6.5 + 0.1\cdot 4.86842 = 6.987$ , and B is $3.5 + 0.6\cdot 4.86842 = 6.421$ million dollars. B is cheaper by 0.566 million dollars. Converted to an equal payment over 7 years, it is 1.435 against 1.319 million dollars a year.
Suppose production is extended to the end of year 13 (the “tail” of production). Then the pipes are needed for 11 years. For B a second set is needed after 7 years, and it serves only 4 of its 7 years: $3.5 + 3.5/1.1^{7} + 0.6\cdot 6.4951 = 9.19$ million dollars, against $6.5 + 0.1\cdot 6.4951 = 7.15$ million dollars for A.

Answer: If the pipes are needed for 7 years, steel is cheaper, by 0.566 million dollars in discounted money. By equivalent annual cost over their own life, fibreglass looks 28% cheaper. If the pipes are needed for 8 years or more, fibreglass is the better choice.

What it means. Equivalent annual cost over 15 years silently assumes that the fibreglass pipes will be needed for all 15 years. But the field needs them for 7 years, and the remaining 8 years of life are worth nothing: pipes in the ground cannot be sold. So when the need has a fixed end, we compare the costs over that period, with the residual value taken into account. The main question of the choice is how many years the field will really work. The data in this example are assumed for illustration.

CalculatedThe equal yearly cost of the two flowline options against how many years the field needs them. Steel pipes are cheaper as long as the period does not exceed 7 years. After that a second set is needed, and fibreglass is better. The period of 7 years is marked: the years of production of the example project.

Capital charge, depreciation and return on capital

The return of capital in each payment is depreciation in the economic sense: it is the amount by which the capital not yet returned to the owner falls. Call this balance $B_t$ . The payment of a year is the return $i$ on the balance of the previous year, plus the depreciation of the year:

The capital charge as a return on the balance plus depreciation

A_t = i\,B_{t-1} + D_t, \qquad B_t = B_{t-1} - D_t, \qquad B_0 = C,\ \ B_n = 0

where:

$A_t$ capital charge of year $t$ , US$ million
$B_t$ capital not yet returned after year $t$ (the book value of the investment), US$ million
$D_t$ depreciation of year $t$ : the part of the investment returned in this year, US$ million
$C$ the investment, US$ million
$i$ discount rate, fraction per year

Units: any currency; the rate is a fraction per year

Holds when

Over the life, depreciation adds up to the investment: $\sum D_t = C$ .
The return is charged on the balance of the previous year at the same rate $i$ that is used for discounting.

Whatever the depreciation scheme, such a series of payments has the same present value, the investment $C$ (the derivation is in the advanced section). The scheme only changes how the payments are spread over the years. An equal payment goes with a depreciation that grows by the share $i$ a year, called annuity depreciation. Straight-line depreciation goes with falling payments.

Worked exampleAn equal payment and straight-line depreciation for the plant

Given (assumed data)

Oil treatment plant: US$60 million, 7 years, rate 10%; the equal payment is 12.324 million dollars a year (the example above)
Comparison: straight-line depreciation of $60/7$ a year, plus a return of 10% on the balance at the start of the year

Solution

Equal payment (the table below): in year 1 the return on capital is $0.1\cdot60 = 6.000$ , and the depreciation is $12.324 - 6.000 = 6.324$ million dollars. In year 7 the return is 1.120 on a balance of 11.204, and the depreciation is 11.204 million dollars. Depreciation grows by 10% a year: $6.324\cdot1.1^{6} \approx 11.20$ .
Straight-line depreciation: $60/7 = 8.5714$ million dollars a year. The payment in year 1 is $8.5714 + 6.000 = 14.571$ , and in year 7 it is $8.5714\cdot1.1 = 9.429$ million dollars (the balance at the start of year 7 equals one year's depreciation).
The present value of both series at 10%, at the date of the investment, is 60.000 and 60.000 million dollars: exactly the investment.

Answer: Both series return 60 million dollars with a return of 10%. Straight-line depreciation only moves part of the charge for capital to the first years: 14.571 million dollars in year 1, against 12.324 with an equal payment.

What it means. The depreciation scheme is a question of distribution, not of the sum: whatever the scheme, once we add the return at the rate $i$ on the balance of capital, the payments have the same present value. Tax depreciation, which is set by law, is a third scheme. It changes when tax is paid, and so the present value of the tax, but not what the capital itself “costs” (capital and operating costs). The data in this example are assumed for illustration.

Return of an investment of 60 million dollars over 7 years at 10%, US$ million

Project year: 3
Balance at the start: 60.000
Return on capital: 6.000
Depreciation with an equal payment: 6.324
Payment with straight-line depreciation: 14.571

Project year: 4
Balance at the start: 53.676
Return on capital: 5.368
Depreciation with an equal payment: 6.957
Payment with straight-line depreciation: 13.714

Project year: 5
Balance at the start: 46.719
Return on capital: 4.672
Depreciation with an equal payment: 7.652
Payment with straight-line depreciation: 12.857

Project year: 6
Balance at the start: 39.066
Return on capital: 3.907
Depreciation with an equal payment: 8.418
Payment with straight-line depreciation: 12.000

Project year: 7
Balance at the start: 30.649
Return on capital: 3.065
Depreciation with an equal payment: 9.259
Payment with straight-line depreciation: 11.143

Project year: 8
Balance at the start: 21.389
Return on capital: 2.139
Depreciation with an equal payment: 10.185
Payment with straight-line depreciation: 10.286

Project year: 9
Balance at the start: 11.204
Return on capital: 1.120
Depreciation with an equal payment: 11.204
Payment with straight-line depreciation: 9.429

Project year	Balance at the start	Return on capital	Depreciation with an equal payment	Payment with straight-line depreciation
3	60.000	6.000	6.324	14.571
4	53.676	5.368	6.957	13.714
5	46.719	4.672	7.652	12.857
6	39.066	3.907	8.418	12.000
7	30.649	3.065	9.259	11.143
8	21.389	2.139	10.185	10.286
9	11.204	1.120	11.204	9.429

The equal payment is 12.324 million dollars a year; it equals the return on capital plus depreciation. With straight-line depreciation, the payment is $60/7$ plus 10% of the balance at the start of the year (the balance in the straight-line scheme is its own: 60, 51.4, 42.9 …). Because of rounding, the sum of the parts may differ from the payment by one unit of the last digit. The data in this example are assumed for illustration.

CalculatedThe equal annual charge for the treatment plant (60 million dollars, 7 years, 10%) and its parts by year: the return on capital falls, and depreciation grows. The dots are the payment with straight-line depreciation and a return on the balance: larger in the first years and smaller in the last, with the same present value.

This shows why the accounting return on capital employed (ROCE), which is profit divided by the book value of the investment, is not the same as the return on the investment. Suppose the plant brings exactly its capital charge, 12.324 million dollars a year above the operating costs. Its internal rate of return (IRR) is exactly 10%: every year it earns 10% on the capital not yet returned (the project account). But in the accounts its cost is written off in a straight line, 8.571 million dollars a year, and the profit of each year before tax is $12.324 - 8.571 = 3.753$ million dollars.

Worked exampleAccounting return and the return on the plant

Given (assumed data)

Profit before tax after straight-line depreciation: 3.753 million dollars in each of the 7 years
Book value at the start of the year: 60; 51.429; 42.857; 34.286 … 8.571 million dollars

Solution

Return on the book value at the start of the year: in year 1, $3.753/60 = 6.3$ %. In year 4, $3.753/34.286 = 10.9$ %. In year 7, $3.753/8.571 = 43.8$ %.
The average over the life, on the average book value $60/2 = 30$ million dollars: $3.753/30 = 12.5$ %.
With the annuity depreciation of the table above, the profit of each year equals the return on capital, and the return is 10% in every year: $6.000/60$ in year 1, $1.120/11.204$ in year 7.

Answer: With straight-line depreciation, the reported return of the plant grows from 6.3 to 43.8%, and on average it is 12.5%, although the return on the investment is 10% in every year.

What it means. In the accounts, new assets look poor earners and old ones look very profitable, although they earn the same. If divisions or projects are judged by accounting return, this pushes managers not to renew equipment. For investment decisions we need cash flows and the NPV. Accounting indicators serve reporting (Valuation methods: book value and market value). The data in this example are assumed for illustration.

Terms and symbols

Annual capital charge $A$: The equal yearly sum that returns an investment over the service life with a return $i$ : $A = C\cdot \mathrm{CRF}_n(i)$ .
Capital recovery factor $\mathrm{CRF}_n(i)$: The factor equals $1/a_n(i) = i/[1 - (1+i)^{-n}]$ . In bank calculations it is called the annuity coefficient. It equals $1 + i$ for one year and tends to $i$ for an endless life.
Return on capital and return of capital: The two parts of a payment: the interest $i$ on the investment not yet returned, and the repayment of the investment itself, which is depreciation in the economic sense.
Interest during construction: An addition to the investments of several years when they are brought to the date the asset starts to work. It is the charge for the money tied up in construction.
Equivalent annual cost (EAC): All the costs of an option over its service life turned into an equal yearly charge. It is used to compare options with different service lives when they are renewed.
Residual value $S$: What the asset is worth at the end of its life: the price of sale or of moving it, less dismantling. Unused service life has value only through it.
Annuity depreciation: Depreciation which, together with the return on the balance, gives an equal payment. It grows by the share $i$ a year.
Return on capital employed (ROCE): Accounting profit divided by the book value of the investments. With straight-line depreciation it grows as the asset ages, and it is not equal to the IRR.

Common mistakes and misconceptions

Wrong

“The annual capital charge is the investment divided by the service life.”

In fact

This is straight-line depreciation without a return on capital. At 10% and 7 years, the charge for a plant worth 60 million dollars is 12.324 million dollars a year, not 8.571: money tied up in an asset must earn a return too.

Wrong

“We compound the investments to the year of first production and multiply by the capital recovery factor.”

In fact

The factor assumes that the first payment comes a year after the date to which the investment is brought. If the payment is in the same year, the factor $(1+i)\,a_n(i)$ is needed. Otherwise the charge is too high by a factor of $1 + i$ : for the example project, 36.87 million dollars instead of 33.52.

Wrong

“The capital charge per barrel is the annual charge divided by the production of the plateau.”

In fact

This division is allowed only for level production. For a profile with a plateau and a decline, divide by the level-equivalent production. This gives the levelised capital cost per barrel. Division by the plateau understates it, and division by the simple average overstates it.

Wrong

“Fibreglass pipes last twice as long, and by equivalent annual cost they are cheaper, so we take them.”

In fact

Equivalent annual cost over each option's own life compares options only if each will be needed for the whole life or will be renewed. If the field will work for 7 years, the second half of the life of the pipes is worth nothing. Compare the costs over the actual period.

Wrong

“By the accounts, the old plant earns 40% on capital, and the new one 6%, so the new one should not be built.”

In fact

With straight-line depreciation, the reported return grows with the age of the asset simply because its book value falls. Both plants in the example of the lesson earn exactly 10%. Decisions are made by cash flows and the NPV.

In depth: advanced detail

Derivations: the make-up of the factor and the growth of depreciation

The factor as a return plus a contribution. Let $v = 1/(1+i)$ . Then $\mathrm{CRF}_n = i/(1 - v^{n})$ . Since $1/(1 - v^{n}) = 1 + v^{n}/(1 - v^{n})$ , the factor equals $i + i\,v^{n}/(1 - v^{n})$ . In the second term, multiply the numerator and the denominator by $(1+i)^{n}$ . We get $i/[(1+i)^{n} - 1] = 1/s_n(i)$ .

With an equal payment, depreciation grows by the share $i$ a year. Write the equality $A = i\,B_{t-1} + D_t$ for two neighbouring years and subtract: $D_{t+1} - D_t = i\,(B_{t-1} - B_t) = i\,D_t$ , that is, $D_{t+1} = (1+i)\,D_t$ . The first depreciation $D_1 = A - i\,C$ equals $C/s_n(i)$ , the contribution to the fund from the previous paragraph. After that, $D_t = (1+i)^{t-1}\,C/s_n(i)$ . The sum of these $n$ terms of the progression is $C\,s_n(i)/s_n(i) = C$ .

The identity for any depreciation. From the two equalities of the section, $A_t = (1+i)\,B_{t-1} - B_t$ . After discounting, $A_t v^{t} = B_{t-1} v^{t-1} - B_t v^{t}$ , and in the sum over the years all the intermediate terms cancel. What is left is $B_0 - B_n v^{n} = C$ , because $B_n = 0$ .

The capital charge with income tax

The capital charge in the main sections is before tax. Suppose the margin of the asset (its revenue less its operating costs) is taxed at the rate $T$ , and the investment reduces the tax through depreciation under the rules of tax accounting. Then the equal margin that returns the investment with a return $i$ after tax is larger:

The equal pre-tax margin that returns the investment after tax

M = \frac{C - T\sum_{k=1}^{n} D_k\,(1+i)^{-k}}{(1 - T)\,a_n(i)}

where:

$M$ equal yearly margin before income tax: the revenue of the asset less the operating costs, US$ million a year
$T$ income tax rate, fraction
$D_k$ tax depreciation of year $k$ of the service life, US$ million
$C$ the investment, assigned to the moment one year before the first year of service, US$ million
$a_n(i)$ annuity factor, years

Units: any currency; $T$ is a fraction, and $i$ is a fraction per year

Holds when

The margin is taxed every year, and the depreciation deductions are used in the year they arise: the owner has enough profit.
All sums are in the same money at one rate. If tax depreciation is calculated from the historical cost in nominal money, its real value is lower under inflation (where inflation is not neutral).

Without tax depreciation, the margin would have to be raised by a factor of $1/(1 - T)$ : tax would take a share of the whole margin. The depreciation deductions return part of this tax, but later than the investment was made, and so not in full. At $T = 0$ the formula turns into the ordinary charge $C\cdot \mathrm{CRF}_n(i)$ .

Worked exampleThe treatment plant with income tax

Given (assumed data)

Plant: US$60 million in year 2, 7 years of service, rate 10%
Tax: 20% of profit; tax depreciation in 5 equal parts from the start of production, US$12 million a year in years 3–7 (the assumed regime of the example project, payments to the state in the model)

Solution

The tax depreciation brought to the middle of year 2 is $12\cdot a_5(10\,\%) = 12\cdot3.79079 = 45.489$ million dollars. The tax that it saves is $0.2\cdot45.489 = 9.098$ million dollars.
The equal margin is $M = (60 - 9.098)/(0.8\cdot4.86842) = 50.902/3.8947 = 13.07$ million dollars a year.
Without tax, the charge would be 12.324 million dollars. With tax, 6% more is needed. Per barrel of level-equivalent production it is $13.07/1.7847 = 7.32$ dollars instead of 6.91.

Answer: For the plant to pay back with a return of 10% after an income tax of 20%, its margin before tax must be about 13.07 million dollars a year. This is 6% more than the capital charge without tax.

What it means. The extra amount is small, because tax depreciation in the assumed regime is fast: 5 years out of 7 years of life. In discounted money, it returns about three quarters of the tax saving $T\cdot C = 12$ million dollars (9.098 of 12). With slower depreciation, over the service life or in proportion to production, the extra amount is larger. If the investment could be deducted from profit at once, it would disappear altogether. The data in this example are assumed for illustration.

The continuous capital charge and a well with a falling production rate

If the payment arrives not once a year but evenly through the year, it is discounted continuously (the discounting is applied at every moment, not once a year) at the rate $j = \ln(1+i)$ (exponential decline with continuous discounting). A uniform flow of $A$ a year for $n$ years is worth $A\,(1 - e^{-jn})/j$ , and the continuous capital recovery factor is $j/(1 - e^{-jn})$ . Since $e^{-jn} = (1+i)^{-n}$ , it differs from the yearly factor by the factor $j/i$ : at 10% and 7 years it is 0.1958 against 0.2054. The yearly factor assigned to the middle of the year gives almost the same: $0.2054/1.1^{0.5} \approx 0.1958$ . The money that arrives during a year arrives on average half a year before its end.

For a well whose production rate falls exponentially, it is more convenient to take the charge not per year but per barrel. A constant fee $c_K$ on every barrel returns the investment $K$ with a return $i$ if $c_K\,Q_\text{disc} = K$ , where $Q_\text{disc}$ is the discounted production of the well:

Capital charge per barrel for a well with exponential decline

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

where:

$c_K$ the capital charge per barrel, $/bbl
$K$ the investment in the well at the moment production starts, US$
$q_i$ the initial production rate, expressed as a volume per year, bbl a year
$D$ the nominal decline rate of the production rate, 1/year
$j$ the continuous discount rate $\ln(1+i)$ , 1/year
$T$ the time the well works until its economic limit, the lowest rate at which it still pays to produce (here not the tax rate), years

Units: $K$ in dollars, $q_i$ in bbl a year, $D$ and $j$ in 1/year, $T$ in years; $c_K$ in $/bbl

Holds when

The production rate falls exponentially from the start, and the investment is made at the moment production starts.
Discounting is continuous. Without an economic limit ( $T \to \infty$ ), $c_K = K\,(D + j)/q_i$ .

Without an economic limit, the capital not yet returned by the time $t$ equals $K e^{-Dt}$ . It falls in the same way as the production rate. The amount collected through the fee in a year, $c_K\,q(t) = (D + j)\,K e^{-Dt}$ , consists of the return of capital $D K e^{-Dt}$ and the return $j K e^{-Dt}$ on the balance. So a constant fee per barrel amounts to depreciation in proportion to production (units of production) plus a return on the balance.

Worked exampleHow much capital is in each barrel of a new well

Given (assumed data)

The well of the lesson on annuities: 1000 bbl/d; $D = 0.25$ 1/year; economic limit 50 bbl/d; rate 10%
Investment in the well $K$: US$5 million (drilling and completion, which means installing the equipment that lets the well produce)

Solution

The working time is $T = 11.983$ years, and $j = 0.09531$ 1/year. The discounted production is $Q_\text{disc} = 1040$ thousand bbl, and the cumulative production is $N_p = 1387$ thousand bbl (that lesson).
The charge per barrel is $c_K = 5000/1040 = 4.81$ $/bbl (the investment is in thousands of dollars).
Without an economic limit: $5\cdot10^{6}\cdot(0.25 + 0.09531)/365{,}000 = 4.73$ $/bbl.
Per undiscounted reserves: $5000/1387 = 3.60$ $/bbl, which is 25% less.

Answer: Each barrel of the well must carry about 4.81 dollars of charge for capital. Dividing the investment by the reserves without discounting understates it by a quarter.

What it means. The economic limit hardly changes the answer: the “tail” cut off after 12 years is worth little at 10%. The fee per barrel is constant, but in dollars per year it falls with the production rate. This is like depreciation in proportion to production (unit-of-production depreciation), by which oil and gas wells are often written off in the accounts. The data in this example are assumed for illustration.

Connections to other topics

Annuities and discounted production — the annuity factor, the accumulation factor of an annuity and the level-equivalent production, on which the capital charge is built.
Unit costs per barrel — the levelised capital cost per barrel is the same capital charge divided by the equivalent production.
Interest and compounding — moving investments in time, and interest during construction.
Cash flow model of an oil and gas project — lease instead of purchase, the assumed tax regime, and the “tail” of production of the example project.
Project cash flow — capital expenditure, depreciation and tax in the cash flow.
Choosing the discount rate — what return the company requires: the rate in the capital charge.
Inflation: nominal and real money — a constant real payment and a constant nominal payment.
Internal rate of return (IRR) — the return on investments, and the sinking fund method.
Acceleration projects and incremental analysis — valuing an investment in a working field by the difference of forecasts.
Valuation methods: book value and market value — book value and market value.

Check yourself

1
A gas compressor costs 2.4 million dollars and will serve for 6 years. What equal payment at the end of each year will return the investment at a rate of 12%?
1. A.0.40 million dollars
2. B.0.58 million dollars
3. C.0.29 million dollars
4. D.0.65 million dollars
Show the answer
Correct answer: B. 0.58 million dollars
$a_6(12\,\%) = (1 - 1.12^{-6})/0.12 = 4.1114$ , $\mathrm{CRF}_6 = 0.24323$ , and the payment is $2.4\cdot 0.24323 = 0.58$ million dollars. The 0.40 is the division by the life without a return on capital. The 0.29 is only the return on capital ( $0.12\cdot2.4$ ), as for an endless life. The 0.65 comes from first compounding the investment by a year ( $2.4\cdot1.12$ ) and then multiplying by the factor for end-of-year payments, so the payment is 12% too high.
2
Why, at a rate of 10%, does an asset with a service life of 30 years need almost the same yearly charge as an endless one, while an asset for 7 years needs twice as much?
Show the answer
The charge is the return on capital $i$ plus the contribution $1/s_n(i)$ to a fund for the return of capital. The return does not depend on the life, and the contribution is small for a long life. One dollar invested at 10% grows in 30 years by a factor of $1.1^{30} = 17.4$ , so there is almost nothing to save: $1/s_{30} = 0.0061$ , and $\mathrm{CRF}_{30} = 0.1061$ against 0.1000 for an endless asset. Over 7 years the investment has to be returned almost entirely from the payments: $1/s_7 = 0.1054$ , and $\mathrm{CRF}_7 = 0.2054$ , which is 2.05 times more.
3
A compressor station costs 30 million dollars, serves for 8 years, and pumps 400 million m³ of gas a year for the whole time. What part of the tariff per 1000 m³ is the capital charge at a rate of 12%?
1. A.about $6.0
2. B.about $15.1
3. C.about $9.0
4. D.about $9.4
Show the answer
Correct answer: B. about $15.1
$\mathrm{CRF}_8(12\,\%) = 0.20130$ , and the yearly charge is $30\cdot0.20130 = 6.039$ million dollars. A year has 400,000 thousand cubic metres, so per 1000 m³ it is $6.039\cdot10^{6}/400{,}000 \approx 15.1$ dollars. The 9.4 is 30 million dollars divided by 8 years, with no return on capital. The 9.0 is only the return on capital, $0.12\cdot30$ . The 6.0 is the yearly charge in millions of dollars, not divided by anything.
4
Pump A costs 1.0 million dollars and lasts 10 years. Pump B costs 0.7 million dollars and lasts 4 years. Their running costs are the same, and the rate is 10%. By equivalent annual cost, A is cheaper. But in 4 years the well will be converted to injection, and the pump will not be needed. Which one should be chosen?
Show the answer
Equivalent annual costs: A is $1.0\cdot0.1627 = 0.163$ , and B is $0.7\cdot0.3155 = 0.221$ million dollars a year. But this comparison assumes that the pump is needed as long as it serves. The pump is needed for 4 years: B will cost 0.7 million dollars, and A will cost 1.0 million dollars. Choose B, unless A can be sold in 4 years, or moved to another well, for more than $(1.0 - 0.7)\cdot1.1^{4} \approx 0.44$ million dollars.
5
Two identical gas compressors were bought one year and six years ago, and they bring the same margin. By the accounts, the return on capital of the first is 7%, and of the second 30%. Which is more profitable?
Show the answer
They are equal. The accounting return divides profit by the book value of the investment, and with straight-line depreciation this value falls with the age of the asset. The same income divided by a smaller value gives a larger percentage. The return on the investment (IRR) is the same for both compressors. If we judge by accounting return, new equipment will always look worse than old.
6
The capital charge of a field is 20 million dollars a year over 6 years of production, with production by year of 3.0, 2.5, 2.0, 1.5, 1.0 and 0.5 million bbl (at the end of each year). What constant tariff per barrel would return the same capital at the same return of 10%?
1. A.$8.3/bbl
2. B.$6.7/bbl
3. C.$10.6/bbl
4. D.$11.4/bbl
Show the answer
Correct answer: C. $10.6/bbl
The discounted production is $Q_\text{disc} = 3.0/1.1 + 2.5/1.1^{2} + \ldots + 0.5/1.1^{6} = 8.224$ million bbl. $a_6 = 4.3553$ . The level-equivalent production is $8.224/4.3553 = 1.888$ million bbl a year, and the tariff is $20/1.888 = 10.6$ $/bbl. The 11.4 is the division by the average production $10.5/6 = 1.75$ : it overcharges, because the early barrels are worth more. The 6.7 is the division by the production of the first year. The 8.3 is the discounted charges $20\cdot a_6$ divided by the undiscounted 10.5 million bbl.

Updated: 17 October 2026