Economic Analysis of Energy Systems: Payback, Present Worth, NPV and IRR, Life-Cycle Cost, Cost of Saved Energy and LCOE

This chapter answers Section XE8.2 of XE8 Energy Science, the sub-heading “Economic analysis of energy systems”: “Simple Payback Period, Time Value of Money, Discount rate, Present Worth Factor, Capital Recovery Factor, Life Cycle Costing, Internal Rate of Return, Net Present Value, Annual Worth, Cost of Saved Energy, Levelized Cost of Energy.” It belongs to the Engineering Sciences (XE) paper, as Section XE8.2 of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam. The chapter teaches how to compare money at different times, the standard interest factors, and the decision measures built on them. The numericals are payback, present worth and capital recovery factors, net present value, internal rate of return by interpolation, annual worth, life-cycle cost comparison, cost of saved energy and levelised cost of energy, with every rate and price stated in the question (1 lakh = 100 000 rupees).

1. Simple payback and the time value of money

The simple payback period is the time for cumulative net savings to repay the investment: SPP = I/A, where I is the first cost and A the annual net saving (energy cost saved minus any extra operating cost). A retrofit costing ₹7 lakh that saves ₹2.4 lakh a year net has SPP = 7/2.4 = 2.92 years. It is quick and needs no rate, but it ignores the time value of money and everything that happens after payback, so a project with a long profitable life can lose to a short one with a good payback.

Time value of money: a rupee today is worth more than a rupee later, because it can earn interest. With a discount rate i per period, a present sum P grows to F = P(1 + i)ⁿ after n periods, and a future sum F has present worth P = F/(1 + i)ⁿ. At 8 % over 10 years, ₹1 lakh becomes 1 × 1.08¹⁰ = ₹2.159 lakh, and ₹1 lakh due in 10 years is worth 1/1.08¹⁰ = ₹0.463 lakh today. The discount rate reflects the investor’s opportunity cost of capital and the risk of the project.

Interest quoted as a nominal rate r compounded m times a year has an effective annual rate i = (1 + r/m)ᵐ − 1; 12 % compounded monthly is (1.01)¹² − 1 = 12.68 %. With inflation f, the real rate is i_real = (1 + i_nominal)/(1 + f) − 1, not the difference of the two: 12 % nominal with 5 % inflation gives 1.12/1.05 − 1 = 6.67 %. Analyse in nominal money with a nominal rate, or in constant money with a real rate, and never mix.

⚠️ Real rate is not i − f
Subtracting gives 7 % in the example above, against the correct 6.67 %. The error is small at low inflation but grows with it; the exam can ask for either, so read whether cash flows are in current or constant rupees.

2. Interest factors: present worth and capital recovery

Standard factors for interest rate i per period and n periods
FactorFormulaUse
Single-payment present worth, (1 + i)⁻ⁿ1/(1 + i)ⁿPV of one future sum
Uniform-series present worth factor (PWF)[(1 + i)ⁿ − 1]/[i(1 + i)ⁿ]PV of n equal year-end amounts A: P = A × PWF
Capital recovery factor (CRF)i(1 + i)ⁿ/[(1 + i)ⁿ − 1] = 1/PWFAnnual equivalent of a present sum: A = P × CRF
Sinking-fund factori/[(1 + i)ⁿ − 1]Equal deposits that grow to a future sum F

The uniform-series factor assumes the first payment comes one period after time zero (end-of-period convention). The CRF is exactly the reciprocal of the PWF, and as i → 0 it tends to 1/n, which is the straight-line repayment with no interest. Worked at i = 10 %, n = 10: PWF = (2.5937 − 1)/(0.1 × 2.5937) = 6.1446, so CRF = 0.16275; a loan of ₹5 lakh at 12 % over 10 years needs ₹5 lakh × 0.17698 = ₹88 492 a year.

When a cost rises each year at a rate g (energy price escalation), the present worth of a first-year amount A₁ growing at g for n years is P = A₁ [1 − ((1 + g)/(1 + i))ⁿ]/(i − g), for i ≠ g. With A₁ = ₹1 lakh, g = 5 %, i = 10 %, n = 5: P = [1 − (1.05/1.10)⁵]/0.05 = ₹4.15 lakh, against ₹3.79 lakh if the amount stayed flat.

🎯 Why CRF is also a loan factor
Buying an asset for P is equivalent to paying P × CRF every year for n years at the same rate i. So CRF turns a capital cost into an annual charge that can be added to annual running costs. This is the step behind annual worth, cost of saved energy and levelised cost.

3. Net present value, internal rate of return and annual worth

Net present value is the sum of all cash flows discounted to time zero: NPV = −I + Σ A_t/(1 + i)ᵗ (plus the present worth of any salvage value). The rule: accept if NPV > 0 at the required discount rate. Annual worth (annual equivalent) is NPV × CRF, or equivalently the annual saving minus the annualised investment, A − I × CRF; it is positive exactly when NPV is positive, and it lets projects of different lives be compared per year.

The internal rate of return is the discount rate at which NPV = 0. For a project of investment I and equal annual savings A over n years, solve PWF(i, n) = I/A. If the table gives bracketing rates i₁ and i₂, interpolate linearly: i ≈ i₁ + (i₂ − i₁)(PWF₁ − I/A)/(PWF₁ − PWF₂). The project is acceptable if IRR exceeds the required rate. The profitability index PI = PV of inflows/I is greater than 1 exactly when NPV is positive.

Worked example: I = ₹10 lakh, A = ₹2.5 lakh per year for 6 years, no salvage, required rate 12 %
MeasureWorkingResult
SPP10/2.54.0 years
PWF(12 %, 6)(1.9738 − 1)/(0.12 × 1.9738)4.1114
NPV2.5 × 4.1114 − 10+₹0.2785 lakh (₹27 852)
Annual worth2.5 − 10 × 0.24323+₹6774 per year
IRRPWF needed = 4.0; PWF(12 %) = 4.1114, PWF(14 %) = 3.888712 + 2 × 0.1114/0.2227 ≈ 13.0 %
⚠️ IRR has limits
A cash-flow stream whose sign changes more than once can have more than one IRR. For mutually exclusive projects, ranking by IRR can contradict ranking by NPV: a small project with a high IRR may add less value than a larger one with a lower IRR. Rank mutually exclusive options by NPV (equal lives) or annual worth (unequal lives).
🧠 NPV against the discount rate
For an investment followed by inflows, NPV falls as the discount rate rises, crossing zero at the IRR. The IRR itself does not depend on the discount rate you are asked to use.

4. Life-cycle costing

Life-cycle cost adds everything an asset costs over its analysis period, in present worth: LCC = capital + PV(operation and maintenance) + PV(energy) + PV(replacements) − PV(salvage). Recurring annual amounts are converted with the uniform PWF, escalating ones with the growing-annuity formula, and one-off future items with (1 + i)⁻ⁿ. The purpose is to stop the first cost from deciding alone: for energy-using equipment, the energy cost over life often exceeds the purchase price several times.

Worked: two pumps are compared over 10 years at i = 10 %, PWF = 6.1446, with no salvage value. Pump A costs ₹1.0 lakh and ₹50 000 a year in energy; pump B costs ₹1.5 lakh and ₹38 000 a year. LCC(A) = 100 000 + 50 000 × 6.1446 = ₹407 228; LCC(B) = 150 000 + 38 000 × 6.1446 = ₹383 494. B has the higher first cost and the lower life-cycle cost, by ₹23 735. The undiscounted saving, 12 000 × 10 − 50 000 = ₹70 000, overstates the gain because it ignores discounting.

  • Use the same analysis period and discount rate for every option; if lives differ, use a common period with replacement, or compare annual worths.
  • Include only costs that differ between options; items common to all drop out of the comparison.
  • Keep money consistent: nominal cash flows with a nominal rate, or constant money with a real rate.

5. Cost of saved energy and levelised cost of energy

The cost of saved energy (CSE) is the annualised cost of an efficiency measure divided by the energy it saves in a year: CSE = I × CRF/E_saved, in ₹/kWh. A measure is worthwhile if its CSE is below the price of the energy it avoids. It treats saved energy like a supply and so puts efficiency and generation on one scale. Worked: a motor retrofit costing ₹90 000 saves 12 000 kWh a year for 10 years at i = 10 %. CRF = 0.16275, so the annual charge is ₹14 648, and CSE = 14 648/12 000 = ₹1.22/kWh. If the avoided tariff is ₹7/kWh the measure pays handsomely.

The levelised cost of energy (LCOE) is the constant price per kWh at which the present worth of revenue equals the present worth of all costs over the life: LCOE = Σ C_t/(1 + i)ᵗ divided by Σ E_t/(1 + i)ᵗ. Energy is discounted too, because a kWh later is worth less revenue. For constant annual cost and output this simplifies to LCOE = (I × CRF + annual O&M + annual fuel)/E_annual.

Worked: a plant costing ₹50 million, life 25 years, i = 10 % (CRF = 0.11017), O&M ₹1 million a year, no fuel, yields 1.8 GWh a year. Annualised capital = 50 × 0.11017 = ₹5.508 million; total annual cost = ₹6.508 million; LCOE = 6.508 × 10⁶/(1.8 × 10⁶ kWh) = ₹3.62/kWh. Raising i raises the LCOE of a capital-intensive plant sharply, because its cost is mostly the CRF term; a fuel-heavy plant is less sensitive to i. Lower output (poor capacity factor) raises the LCOE in inverse proportion.

⚠️ LCOE is not a market price
LCOE compares the cost of producing energy under stated assumptions; it ignores when the energy is delivered, storage and grid costs. Two sources with equal LCOE are not equally valuable if one produces at peak hours and the other only at times when supply is plentiful.

Key takeaways

  • SPP = I/A ignores the time value of money and post-payback savings, so it screens projects but does not rank them reliably.
  • P = F/(1 + i)ⁿ; uniform PWF = [(1 + i)ⁿ − 1]/[i(1 + i)ⁿ]; CRF = 1/PWF turns a capital sum into an equal annual charge; real rate = (1 + i)/(1 + f) − 1.
  • NPV > 0 means accept; IRR is the rate that makes NPV zero, found by interpolating PWF(i, n) = I/A; annual worth = NPV × CRF.
  • Rank mutually exclusive projects by NPV or annual worth, not IRR; IRR can be multiple when cash-flow signs change more than once.
  • Life-cycle cost sums capital and the present worth of energy, maintenance and replacements less salvage, over a common period and rate.
  • Cost of saved energy = I × CRF/E_saved (compare with the tariff); LCOE = (I × CRF + O&M + fuel)/E_annual, with energy discounted like cost in the general form.

Practice questions (15)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which of the following is a genuine limitation of the simple payback period as a ranking tool?

    1. It needs the discount rate to be known exactly before it can be calculated, which is rarely possible in practice
    2. It cannot be calculated at all when the annual savings are constant from year to year
    3. It counts the salvage value twice, once in the investment and once in the annual savings
    4. It ignores savings after the payback year and the time value of money
    Show answer

    Answer: D — It ignores savings after the payback year and the time value of money

    SPP = I/A uses undiscounted cash flows and stops counting when the cumulative saving reaches I. It needs no rate, which is its attraction, and it works perfectly well for constant savings.

  2. A nominal discount rate of 12 % per year is used while general inflation is 5 % per year. The real discount rate to be used for cash flows in constant money is

    1. 7.00 %, the difference of the two rates
    2. 6.67 %, from 1.12/1.05 − 1
    3. 17.6 %, from 1.12 × 1.05 − 1
    4. 5.00 %, the inflation rate
    Show answer

    Answer: B — 6.67 %, from 1.12/1.05 − 1

    The Fisher-type relation is 1 + i_nominal = (1 + i_real)(1 + f), so i_real = 1.12/1.05 − 1 = 6.67 %. The difference 7 % ignores the compounding of the two effects, and the product gives a nominal rate, not a real one.

  3. As the interest rate tends to zero, the capital recovery factor for n equal annual payments tends to

    1. 1/n, since the principal is just repaid in equal parts
    2. 1, since the whole sum would then have to be repaid in the first year itself
    3. n, since each of the annual payments would repay the whole principal
    4. infinity, since the annual payment grows without bound as interest vanishes
    Show answer

    Answer: A — 1/n, since the principal is just repaid in equal parts

    CRF = i(1 + i)ⁿ/[(1 + i)ⁿ − 1] → 1/n as i → 0 (expand (1 + i)ⁿ ≈ 1 + n i). With no interest the repayment is simply P/n a year, which is also the limit of 1/PWF as PWF → n.

  4. Project X has an investment of ₹2 lakh and an IRR of 20 %. Project Y has an investment of ₹10 lakh and an IRR of 15 %. They are mutually exclusive, of equal life, and the required discount rate is 10 %. Which statement is correct?

    1. X must be chosen because its IRR is higher, whatever the sizes of the two investments
    2. Y must be rejected outright because its IRR is lower than that of X
    3. Both are acceptable on IRR, and the choice between them should be made on NPV at 10 %
    4. Neither is acceptable because the two IRRs differ by more than the discount rate
    Show answer

    Answer: C — Both are acceptable on IRR, and the choice between them should be made on NPV at 10 %

    Both IRRs exceed 10 %, so each has a positive NPV. For exclusive projects the aim is to maximise value added, and a larger project at a lower IRR can add more NPV than a small one at a higher IRR, so the rank by IRR may mislead and NPV at the required rate decides.

  5. Over 10 years at i = 10 % (uniform PWF = 6.1446, no salvage), machine P costs ₹1.0 lakh with energy cost ₹50 000 a year, and machine Q costs ₹1.5 lakh with energy cost ₹38 000 a year. Which has the lower life-cycle cost, and by about how much?

    1. P, by ₹50 000, because its first cost is lower and the energy costs are ignored
    2. Q, by about ₹70 000, from the undiscounted ten-year saving less the extra capital cost
    3. Q, by about ₹23 700, from the discounted saving less the extra cost
    4. They are equal, since the extra cost equals one year’s saving
    Show answer

    Answer: C — Q, by about ₹23 700, from the discounted saving less the extra cost

    Annual saving = 12 000, PW = 12 000 × 6.1446 = 73 735, less the extra ₹50 000 gives +₹23 735 for Q. Using the undiscounted 120 000 − 50 000 = 70 000 ignores discounting, and ranking on first cost alone is the error that life-cycle costing is meant to remove.

  6. The levelised cost of energy of a plant is best described as

    1. the total capital cost divided by the rated power of the plant, expressed in rupees per kilowatt
    2. the fuel cost per kilowatt-hour, averaged over the whole operating life of the plant
    3. the constant price per kWh that equates the present worth of revenue with that of all costs over the life
    4. the average tariff that the utility actually charges its customers in the first year of operation
    Show answer

    Answer: C — the constant price per kWh that equates the present worth of revenue with that of all costs over the life

    LCOE = PV of lifetime cost/PV of lifetime energy, the break-even constant price. The ₹/kW figure is a specific capital cost, fuel cost is only one term, and an actual tariff is a market or regulatory quantity, not a cost measure.

  7. A lighting retrofit has a cost of saved energy of ₹1.9 per kWh, calculated with a stated discount rate and life. The avoided electricity price over the same period, on the same money basis, is ₹6.5 per kWh. The measure is

    1. economic, since saving a kWh costs less than buying one
    2. uneconomic, since the cost of saved energy is positive
    3. economic only if the simple payback is below one year
    4. uneconomic, since the cost of saved energy is above the cost of generation at the plant
    Show answer

    Answer: A — economic, since saving a kWh costs less than buying one

    CSE below the avoided price means each kWh saved costs less than the kWh it replaces, so the annualised investment is more than covered (NPV > 0 on the same basis). Any positive CSE is expected; the comparison is with the avoided price, not with zero.

  8. Which of the following statements about NPV and IRR for a conventional project (one initial outflow followed by inflows) are correct?

    1. The IRR is the discount rate at which the NPV is zero
    2. The NPV is positive when the discount rate is below the IRR
    3. The NPV evaluated at the IRR is positive
    4. Raising the discount rate reduces the NPV
    Show answer

    Answer: A — The IRR is the discount rate at which the NPV is zero; B — The NPV is positive when the discount rate is below the IRR; D — Raising the discount rate reduces the NPV

    NPV falls monotonically as i rises and crosses zero at the IRR, so it is positive below the IRR. By definition the NPV at the IRR is zero, not positive.

  9. Which of the following statements about interest factors (end-of-year convention) are correct?

    1. The present worth of one sum F due after n years is F/(1 + i)ⁿ
    2. The CRF is i(1 + i)ⁿ/[(1 + i)ⁿ − 1]
    3. The CRF multiplied by the uniform-series PWF equals one
    4. The uniform-series PWF rises when the discount rate rises
    Show answer

    Answer: A — The present worth of one sum F due after n years is F/(1 + i)ⁿ; B — The CRF is i(1 + i)ⁿ/[(1 + i)ⁿ − 1]; C — The CRF multiplied by the uniform-series PWF equals one

    The present-worth expression, the CRF formula and the CRF-PWF reciprocal relation are the standard definitions. A higher discount rate shrinks the present worth of every future payment, so the uniform-series PWF falls, not rises.

  10. Which of the following statements about life-cycle costing are correct?

    1. It includes the present worth of energy, maintenance and replacement costs along with the capital cost
    2. The option with the lowest first cost always has the lowest life-cycle cost
    3. Escalation of the energy price raises the present worth of the energy cost
    4. Options should be compared over the same analysis period and discount rate
    Show answer

    Answer: A — It includes the present worth of energy, maintenance and replacement costs along with the capital cost; C — Escalation of the energy price raises the present worth of the energy cost; D — Options should be compared over the same analysis period and discount rate

    LCC = capital + PV of running costs and replacements − PV of salvage, over a common period and rate; escalation raises the PV of energy. A low first cost often goes with higher energy cost, which is why it does not guarantee the lowest LCC.

  11. An energy-saving measure costs ₹700000 and saves 40000 kWh of electricity a year at ₹7.5 per kWh. It adds ₹60000 a year of extra maintenance cost. The simple payback period is ____ years (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.92

    Gross saving = 40 000 × 7.5 = ₹300 000; net annual saving = 300 000 − 60 000 = ₹240 000. SPP = 700 000/240 000 = 2.917 = 2.92 years. Using the gross saving gives 2.33 years, which ignores the added maintenance.

  12. A project requires an investment of ₹1000000 and gives net savings of ₹250000 at the end of each year for 6 years, with no salvage value. Take the discount rate as 12 % and the uniform-series present worth factor as 4.1114. The net present value is ₹ ____ thousand (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 28

    PV of savings = 250 000 × 4.1114 = ₹1 027 850. NPV = 1 027 850 − 1 000 000 = ₹27 850 ≈ ₹28 thousand (the unrounded factor 4.111407 gives 27 852, the same to the nearest thousand). Since NPV > 0, the project is acceptable at 12 %.

  13. For the same project (investment ₹1000000, net savings ₹250000 at the end of each of 6 years, no salvage) the uniform-series present worth factors are 4.1114 at 12 % and 3.8887 at 14 %. Using linear interpolation between these two rates, the internal rate of return is ____ % (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 13.0

    Required PWF = I/A = 1 000 000/250 000 = 4.0000. i = 12 + 2 × (4.1114 − 4.0000)/(4.1114 − 3.8887) = 12 + 2 × 0.1114/0.2227 = 13.00 %. The exact root is 12.98 %, so the interpolated answer to one decimal is 13.0.

  14. A plant costs ₹50 million and has a life of 25 years. The capital recovery factor at the chosen discount rate is 0.11017. Annual operation and maintenance cost is ₹1 million, there is no fuel cost, and the plant delivers 1.8 × 10⁶ kWh each year. The levelised cost of energy is ₹ ____ per kWh (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 3.62

    Annualised capital = 50 × 10⁶ × 0.11017 = ₹5.5085 × 10⁶. Total annual cost = 5.5085 + 1.0 = ₹6.5085 × 10⁶. LCOE = 6.5085 × 10⁶/1.8 × 10⁶ = ₹3.616 ≈ 3.62 per kWh. Omitting the O&M term gives 3.06.

  15. A heat-recovery unit costs ₹240000 and has a life of 12 years. The capital recovery factor for the chosen rate and life is 0.14676. It saves 30000 kWh of electricity a year and has no operating cost. The cost of saved energy is ₹ ____ per kWh (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 1.17

    Annual charge = 240 000 × 0.14676 = ₹35 222. CSE = 35 222/30 000 = ₹1.174 ≈ 1.17 per kWh. Dividing the investment by the energy saved over the life without discounting gives 240 000/360 000 = 0.67, which ignores the cost of capital.