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CLOSE THIS BOOKVillage Electrification (SKAT, 1992, 128 p.)
Part 8: Tariffs & financial evaluation
VIEW THE DOCUMENT(introduction...)
VIEW THE DOCUMENT1. Determine tariffs
VIEW THE DOCUMENT2. Financial evaluation
VIEW THE DOCUMENT3. Example

Village Electrification (SKAT, 1992, 128 p.)

Part 8: Tariffs & financial evaluation

Tariffs are a crucial point in any electrification scheme. Often they are economically not viable and lead to the project's failure. Pricing is on one hand pure economy on the other pure politics and they often counteract. Is for instance a national grid close by, it will be very difficult to have another pricing than 'they' have, unless you are cheaper. Very often the national supply is highly subsidized and economically hard to beat.

The lever to really get low prices is the load factor: the closer generation reaches 100% of the rated energy production, the more competitive your tariffs are. Besides rising the load factor from let's say 40% to 80% almost nothing else counts. But also this is true; the load factor is a vicious circle. If it decreases, your prices will rise, you will lose customers, lose more load ... many MHP electrification projects vanished through this mechanism.

It is difficult to give comprehensive strategies for a successful pricing but we present some methods and calculations to estimate tariffs and some criterias to establish a tariff structure.

1. Determine tariffs

A traditional approach would start with a stocktaking and evaluation of all assets from which, after applying certain depreciation rules, the annual capacity related or kilowatt-related costs are derived. Then there is an evaluation of the various running costs, like fuel, which leads to the energy-related or kilowatt-hour-related costs. For an electrification based on renewable energies like MHP, this heading might become almost negligible. Some costs, like maintenance, have both fixed and variable components and are allocated accordingly. Finally there are some costs, like meter reading, which are consumer-related and not related with either capacity or energy demands.

All these costs have now to be charged to the consumers as fairly as possible through an appropriate tariff structure. With some research in the consumer demand pattern (load research), the supplier is able to identify each consumer class' contribution to the peak consumption, thus, to the capacity-related costs. To these the energy- and customer-related costs are added and a 'cost-based' tariff is formulated for each consumer class.

On a typical bill therefore several elements may appear: for instance a fixed or minimum charge to cover customer costs; a kilowatt charge to reeover the capacity costs and, of course, a kilowatt hour charge.

Often some simplification is needed. For instance metering might be too expensive, at least for some consumer classes. In this case, a fixed maximum power distribution (through current limitation with a fuse (see part 11: The Salleri Chialsa Venture) might be used, to determine an average energy consumption and define a consumer class. This allows for instance to drop kilowatt-hour-charges altogether and include kW-charges (as they are fixed) into the fixed charges. The consumers' monthly bill would include only one constant figure.

This traditional approach, as simple as it is, has its drawbacks:

- It is backward-looking, adjusting tariffs based on 'old' data. But to steer the consumption would need a forward-looking approach.

- It generates tariffs based on averages rather than marginal costs.

Simplifying bills, as discussed before, 'blinds' the consumer. He can't influence the bill directly with his consumption behavior. Especially when peak energy is scarce, one would like to steer the customers behaviour, so they divert some consumption to the off-peak hours (or even off-peak season). It needs the application of different tariffs during day (or season), inviting consumers to benefit from cheaper energy during off-peak. Unfortunately this involves costly (time) meters. This also shows a general relation: to keep the customer informed about the energy costs and influence his consumption pattem, you need to know the details about the consumption of individuals which inevitably involves additional costs.

2. Financial evaluation

To determine the tariffs, partly the same calculation tools can be applied as for the financial analysis of investments. Therefore a very brief introduction to those methods might be helpful. The following is a rather mathematical collection of evaluation methods presented in 'A Guide to the Financial Evaluation of Investment Projects in Energy Supply' (see references). Additional reading could be necessary. In the example presented at the end of this part only annual costs are of interest,soonly formulae {6), {10) and {16b) are actually used. You may skip the rest.

2.1 Interest and Inflation Rates

Depending on the kind of financing an appropriate interest rate has to be chosen:

- For external financing of the investment costs it should be fixed at an effective rate of interest for debt during an equal loan period.

- For internal financing of the investment costs it should be fixed at an effective rate of interest on capital deposits during an equal period.

- For mixed financing a weighted mean value can be used as interest rate.

In case of high price stability a constant interest rate i can be assumed and the market interest rate p is used for calculations. For many countries, however, high inflation rates must be included to ensure accurate calculations. In this case an actual interest rate i* is used. It can be determined as follows:

with inflation rate a (inflation factor e = l+a) and market interest rate p (market interest factor r = 1 + p) and actual interest rate i* (actual interest factor q = 1+i*) and

q = r/e {1}

we get

example: with p = 32% and a = 22%

{2}

Especially for projects in energy supply it might be necessary to assume different inflation rates. Prices for fuel for instance may vary more than the rest.

2.2 Investment Costs

To calculate the economics of an electrification and establish tariffs, the used and required investment is an important factor. It is necessary to collect and quantify all cost components, indexed according to their time of occurrence. The following table (see Table 1) could be used for that. It is necessary to predict the service life of such an installation as accurately as possible, both to determine the depreciation and the residual value.


Table 1 Investments stocktaking along the project's life span.

The first decreases with the service (depreciation) life time, the latter with residual value = investment(1-servicelife/techn life) {3}

is zero if service and technical life are equal and increases if the plant or parts are able to operate longer than the service life time. Often in a power plant, technical life spans of machinery and buildings are very different and residual values (book value) might be high for premises and zero for the equipment.

2.3 Operating or Running Costs

This term includes all foreseeable costs to operate the project. It does not include investment costs like capital interest and depreciation. Following a list of possible operating costs:

manpower costs Administrative and operating staff may form a considerable part of the total operating costs. They must be estimated carefully and exactly, including number, qualifications and employment periods of required staff and market usual wages, salaries, associated (social security contributions... ) and any other standard (holiday pay, bonuses...) personnel costs. Inflation has to be taken into account!

repair & maintenance costs This term is very difficult to estimate accurately and one has normally to rely on manufacturers' information about their experiences. These estimates, however, tend to be optimistic and it is advisable to adjust them upwards.

energy related costs They are nil for renewable energies like solar, wind, hydro and thermal energy.

auxiliary material costs They include lubricants (grease, oil...) etc. and could be included under maintenance.

administration costs (excluding manpower)This includes office rents and supplies, communications etc.

taxes and duties They may be imposed on water utilisation, energy generation, land property etc. and have to be according to local tax laws and duty regulations. other costs Whatever cannot be included above.

2.4 Income

Income is generated by: a) revenues from energy sales b) savings on commercial energy c) increased/ improved production of goods Fairly simple to estimate is case b). a) and c) might be extremely difficult, but very important to know the plant's profitability. Imperative is an investigation on the power demand = > see example at the end of this chapter. d) Services: Connection fees, electrical installations, selling of electrical appliancies and consultancies could be services offered to the consumers and charged individually. Basic services like the consumers' connection to the distribution line, might partly be paid by increased kWh charges. e) Subsidies: They are often the only way to have an electrification scheme started. Governments or foreign donors may grant substantial amounts to electrification programs. However, subsidies always distort the real generation costs. It has to be carefully considered which costs of the program should be subsidised. In the long run subsidies should be avoided!

2.5 Returns & Profit

The annual returns are the difference of the total incomes and the total operating costs. return = total income - operating costs

The profit is the return minus the depreciation (the periodic reduction in the value of the plant and grid).

The depreciation does not influence the periodic returns but the periodic profit! profit = income- operating costs- investment costs

2.6 Static Procedures for Financial Evaluations

2.6.1 Basic Cost Comparison Calculation

We include in the total costs two major headings: the operating costs OC (for which we will use an annual average <OC> in the beginning) and the invested capital costs which have two components, a depreciation and a (compound) interest part. To start with we will simplify the capital costs with:

I0/T linear depreciation

(I0/2)·i* interest {4}

linear depreciation: amortisation of the bound capital per time period in equal chunks. T is the number of periods.

Interest: interest payments per time period at an actual interest rate i*, which is applied to the average bound capital (initially invested) for the duration of the project. In the beginning the bound capital is lo, at the end 0, so we get the average (I0-0)/2).

CT = (OQ+ I0/T + I0/2 ·i*

CT total costs per time period (including depreciation and interest)

<OC> average operating costs per time period

I0 initial investment costs
I* actual (assumed) interest rate {5}

If the plant and distribution system are not completely written off at the end of the service life (residual value > 0), a positive liquidation yield L has to bc deducted from the investment costs and CT is recalculated as follows: (with <1> = average invested capital per time period)

<I> = (I0 - L)/2 + L = (I0 + L)/2

CT = <OC> + (I0 - L)/T + (I0 - L)/2 · I* +L·I* = <OC>+ (I0 - L)/T +<I>·i*

In the last term we have to consider the interest of the tight-up capital L, which adds a constant interest.


Fig 1 Depreciation of an initial investment.

2.6.2 Static Cost Annuity Comparison

(annuities = equal annual payments) Before we present this method, we will introduce two important factors, the Present Value factor PV(q,t):

PV(q,t) = = (qt-1)/qt(q-1) = (1-qt)/(q-1) {7}

The present value factor is a function of time and interest rate; it takes compound interest into account. Multiplied with the periodical, constant payment, it gives the present value of the total investment. If we need to know the present value of some value in the future (for instance the liquidation yield L in T periods), we simply skip the summation E

and get PV(q, T) = q-t 7b}

example: a monthly payment of $200 terminates after 7 years. What is a) the present value of this investment if the interest rate is 8%? And what is b) the present value of a liquidation yield of $5'000 (value in 7 years from now)? q = 1 + 0.08/12 = 1.00667 (interest per month!) t = 7 · 1 2 = 84 (number of payments) PV(1.00667, 84) = 64.159 (use formula {7}) q-T = 1 08 - 7 = 0.583 (hr L)

a) PV of investment = $200·64.159 = $12'832
b) PV of liquidation yield = $5'000·0.583 = $2'917.

and the Recovery factor R(q,t):

R(q.t) = 1/(P·V(q,t)) = (qt·(q-1))/(qt-1)

The recovery factor is a function of time and interest rate and is the inverse of the present value factor; it takes compound interest into account. Multiplied by the present value of an investment, it gives the periodical, constant payment to recover this investment (depreciation and interest). exemple: an initial investment of $15'000 must be recovered after 7 years. What is the constant, monthly payment, if the interest rate is 8%? q = 1 + 0.08/12 = 1.00667, t = 7 12 = 84,

R = 0.0156 PV(1.0067,84) 64.159
monthly payment = $15'000 · 0.0156 = $234

Now the static cost annuity comparison method. The difference to the method above is the inclusion of compound interest on the investment (exact calculation) instead of using a simple average.

Ca = Co+(I0-L)·R(q,t)+L i* {10}

Ca total annual Costs

example: (see also example at the end of this pad)

<OC> = $6'500 (for 1-25 years of operation)
l0 = $70'000 (initial investment)
L = 10'000 (yield after 25 years)
I* = 8% (actual interest rate)
R(1.08, 25) = 0.0937 (use formula {9})
Ca = $6'500+($70'000)-0.0937+$10'000-0.08 = $12'921
2.6.3 Static Pay-Back Period

It is used to determine the time when the invested capital is paid back. There are two methods: a) cumulative method find k so that

-I0 + return1 + return2+. . .+returnk > = 0 {11a}

b) averaging method capttal invested k = annual return

k = capital_invested/ anual_return

example:
k = 70000/(12100-5600) = 11 years

2.6.4 Calculation of Profitabiiity

If an investment produces a profit, the ratio of this profit and the average capital invested measures the profitability. This ratio is expressed in % and called the Return On Investment (ROI).

ROI = (NPT/<I>)[% per time period] {12}

NPT net profit per time period

ROI = 8000/40000 = 20% per year

Including a net profit can be seen as an increase in the annual costs and needs more income.

This calculation can also be used to select the most attractive between two (or more) options.

2.7 Dynamic Procedures for Financial Evaluations

Hereunder we repeat more or less the calculations made before with the difference that future values (running costs, investments... ) are reduced to present values.

2.7.1 Net Present Value (NPV)


This method gives the exact present value of all cash flows and the liquidation yield. For each period it considers the net cash flow, which is the return R minus the investment I in this period. It includes investment costs (depreciation & interest)


If all investments are done at the beginning, this is simplified to ...

2.7.2 Internal Rate of Return (IRR)

Determine the interest rate (IRR) at which NPV = 0. It must then be IRR > i. This is in any case a complicated calculation and has to be solved with a programmable calculator or computer.


For instance with x = I+IRR and using formula {13b}

To find the solution for {14c) for instance is to find x by searching the roots of this equation (Newton algorithm...).

- I0+R·((1-x-T)/(x-1))+L·x-T

2.7.3. Annuity Method

It converts the net present value of the project into annual payments of equal amounts.

A = NPV·R(q,T) {15A}

A = annual payments; it must be A>0


without knowing NPV, A can be found by stepwise calculating (for b)

2.7.4 Dynamic Cost Annuity Comparison


This is similar to the above method, but instead of using the net cash flows, it works only with the expenses (no income). It is necessary to replace in the static method the constant, average operating costs by its dynamic counterpart:

The liquidation yield is also adjusted and is the same as AL before.

In {1 6a}the annual costs are based on only an initial investment lo, otherwise they would have to be adjusted to the second equation {16b}.

2.7.5 Dynamic Pay-Back Period

It is the static method but the returns of each period have to be taken as present values (dynamic). Therefore the dynamic payback period is longer than the static one because here not only the invested capital has to be recovered but also the interests on it.
find k so that:

- I0+return1·q-1+...+returnk·q-k> = 0 {17}

3. Example

This example shall explain how to determine tariffs and where to use the preceding formulas. It is based on the small grid already described in the chapter 'power factor correction'. There are three consumer groups, the households, the mill and the workshop. We try to follow step by step different strategies to determine tariffs for these groups.

First we establish the load curve during one day (24 hours). Table 3 and Figure 2 (see at the end of this part) show the result. It is assumed that this load curve is valid throughout the year, but mill and workshop run only 80% of the year. In this way we find the total daily and yearly energy consumption, shown in Table 3a.


Table 3a Example: the daily and annual energy consumption

This value (together with others like investment, service life etc.) is included in the key data of this electrification, shown in Table 4.


Table 4 Example: electrical and financial key data for this electrification scheme.

Table 5 shows the results of three different cost calculation methods, all described above: basic cost calculation, static and dynamic cost annuity comparison. The annual costs are different as each method handles the depreciation and interest payment differently. As none of this calculation includes profit, the results are just covering the expenses. The costs (and so the tariffs) could be increased by some % to have a certain buffer for contigencies.

The following calculations work with the concept of keys to distribute the annual costs among the consumers. It ensures that the revenues recover the annual costs. We will use two different keys: one to set the distribution of the annual costs among the consumer groups (-> key 1) and the other to set the distribution for the consumers' bill headings (-> key 2).

Table 6 proposes four alternatives for key 1 (a-d). The legend describes their definition and use. Any other key I could be defined and used!

Table 7 displays four alternatives (variant 14) to determine each connection's electricity bill (key 2). Basically three headings are assumed: a fixed charge, a kW charge and a kWh charge, but these are not always used in order to simplify the billing procedure.

The calculation is rather simple once the keys are set: An 'exclusive' price, which is the price each connection would have to pay, if its group would be alone (no other consumer groups on the grid), is calculated. The total on the bill is found by multiplying this 'exclusive' amount with key 1 and key 2 (and to sum the results if more than one bill heading is used).


Table 5 Example: results of cost calculations based on the methods explained at the beginning of this chapter.


Table 6 Example:four proposals for key 1.

(all houses/mill/workshop = x% / y% / z%).

a) based on running costs: this key is assuming that the running costs are unevenly distributed among the connections. Therefore connections are weighted (here arbitrarily chosen 1 house counts as 1 connection, 1 mill as 3 connections and 1 workshop as 4 connections) = > key 1a = 78.1%19.4%/12.5%.

b) based on the installed power (load side): key1b = 39.2%125.5%/35.3%

c) based on the consumed energy: key1c = 43%135.5%/26.6%

d) based on a weighted average between a, b & c: it is assumed that none of the above keys is
appropriate, but some average would distribute the costs fairer. A weight based on the annual total costs is used (about 50% are operating costs (manpower, repair), 40% are kW costs (depreciation and interest) and 10% are kWh costs) to average keyl a-c = > keyld = 59%/17.9%/23.0%.

We could of course figure out any other keys and understandable and accepted by all. However, as combination to produce tariff structures. Using the shown in this part economics should come first to shown methods all will ensure enough income to make electrification viable. As a thumb rule we recover the expenses. How to select the right one? might base a tariff structure on: first cost analyses Finally we need one which appeals to all. To achieve (done), second revenue requirements (done) and this, perceived fairness is required. Fairness which is third faimess (which could be applied to select the tariff structure). If we decide to use for instance altemative four (Variant 4), we get the following tariff structure which obviously distributes the costs unevenly among the consumer groups. Is it fair? will they pay?:

Table 8 Example: Tariffs if Variant 4 is selected.



house

mill

workshop

fixed charge

[$]

23.82

90.39

116.09

kW charge

[$]

-

-

-

kWh charge

[$/kWh]

-

0.0220

0.0323



Table 7 Example: four alternatives for electricity bills.

Each using one of the above defined key1 (a-d) and an arbitrarily fixed key2.

variant 1: energy consumption determines key1 (-> key1 c) and the bill is including one fixed charge for all consumer groups (no meters needed).

variant 2: installed power determines key1 (-> key1 b) and the bill includes one kW charge for all consumer groups (no meters needed, but a group can influence its share by changing its installed power).

variant 3: the energy consumption determines key 1 (-> key 1 c) and the bill includes three charges, distributing the bill's total the same as the annum total costs (about 50% are operating costs (man power, repair) ->fixed charge, 40% are kW costs (depreciation and interest) -> kW charge and 10% are kWh costs ->kWh charge) . This is applied for all consumer groups. Meters are necessary for all connections and allow the customer to influence his bill by controlling the consumption.

variant 4: the average of key1 a-c is used to determine key1 (-> key1 d). For the household the bill shows a fixed charge (no meters needed). For the mill and the workshop half of the bill is afixed charge and half of the bill is consumption-related (meters are needed).


Table 3 Example: consumption during one day (24 hours)


Fig 2 Example: hypothetical load curve during one day (24 hours)

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