Engineering Economics for Public Water Utilities

Max Maurer

Eawag: Swiss Federal Institute of Aquatic Science and Technology

Imprint

Author Prof. Dr. Max Maurer, Institute of Environmental Engineering, ETH-Zürich, Switzerland [email protected]

Aim

This script constitutes part of my lecture ‘Infrastructure Systems in Urban Water Management’ given at ETH Zürich, Switzerland

Version 4.7, Jan 2019

License Creative Commons Attribution 4.0 International (https://creativecommons.org/licenses/by/4.0/)

Title photo Wastewater Treatment Plant (Mönchaltorf, ZH) Maurer: Engineering Economics

Contents

1. Aim of this script ...... - 5 -

Inflation, Interest & Discount Rate ...... - 7 -

2. The strange concept of money ...... - 8 -

3. Inflation...... - 8 -

4. Interest rates ...... - 10 -

5. Discount rate ...... - 10 -

6. Replacement value ...... - 13 -

Engineering Economic Analysis ...... - 17 -

7. Engineering economic analysis ...... - 18 -

8. Assumptions and simplifications ...... - 20 -

9. Present value analysis ...... - 20 -

10. Annual cash flow analysis (annuity method) ...... - 23 -

11. Sensitivity analysis ...... - 24 -

Depreciation & Hidden Assets ...... - 27 -

12. Depreciation ...... - 28 -

13. Hidden assets ...... - 30 -

Cost Calculations Debt-to-Asset Ratio ...... - 33 -

14. Cost calculations ...... - 34 -

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15. Debt-to-asset ratio ...... - 37 -

The Value of Flexibility ...... - 41 -

16. The future is uncertain ...... - 42 -

17. Plant utilisation and idle capacity ...... - 44 -

18. Specific net present value (SNPV) ...... - 46 -

19. Considering uncertainty ...... - 47 -

20. Estimating the value of modularisation ...... - 50 -

Formulas, Tables & References ...... - 52 -

21. Formulas & Tables ...... - 53 -

22. References und further reading ...... - 58 -

Examples & Exercises ...... - 61 -

23. Exercises ...... - 62 -

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1. Aim of this script

This script was written to support my lecture ‘Infrastructure Systems in Urban Water Management’ at ETH Zürich. The target audience are engineers aiming to gain a basic understanding of the tools and methods used to manage water infrastructures. Asset management is meaningless without an idea of the strange concept of money and engineering economics. This script provides the key formulas and concepts to deal with monetary values so that engineers can make fair cost comparisons of various project variants and develop basic cost projections within the scope of asset management plans.

The focus is clearly on public utilities providing water supply and sewerage services. This is not only due to it being my principal domain of experience, but also because it enables some substantial simplifications to be made, so that I can ignore weighty issues such as taxes and profits. Moreover, public utilities are a very common organisational form in the water sector.

This is not a script of the entire course, nor is it a script on asset management in general. It focuses only on the monetary part and is taught in the first 4 weeks of the course. Considerable attention is also devoted to Swiss-specific issues, as many of my students will find work in Switzerland. I try to bring this aspect out clearly, so that the reader can identify the generic concepts involved.

I hope you find this script useful. Please let me know if you have any suggestions for improvement or find any errors. I am still in the phase of improving and changing the course content, and this script will be adapted accordingly. If not very quickly, then at least continuously.

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Inflation, Interest & Discount Rate

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2. The strange concept of money

For engineers, monetary values are very strange. They change over time in unpredictable ways. Today CHF 100 is not the same as CHF 100 next year or CHF 100 fifty years ago, even if we still talk about CHF 100. We can describe these changes with three basic concepts:

1. Inflation: For some dubious reason, the general price level of goods and services in an economy changes over time. Each unit of money buys a different amount of goods and services over time. If it is less then we speak about inflation, if it is more, then it is called deflation. We also say that the purchasing power of a unit of money changes over time.

2. Interest is a fee payable for the right to use somebody else’s money. 3. The discount rate roughly describes the fact that we value having money today more than having it tomorrow. This fact is independent of inflation. If given the option of paying their tuition fees at the beginning of the semester or the same amount after the end of the semester, most people choose the second option. This preference can be quantified and is expressed in the discount rate.

We cannot go into too much economic detail about these three concepts here. In the following chapters, we will present the most basic formulas that enable us engineers to navigate these tricky monetary waters.

3. Inflation

To keep it short and easy: don’t do any mathematical operations with two monetary values unless they are in- Convert all monetary dexed to the same year. We will call this the reference year. As pointed out earlier, a unit of money changes its values to a reference purchasing power over time. A pump valued at CHF year before you do any 1000 today will have had a different monetary value last calculations. year than it will have next year. We therefore need to be aware which year the CHF 1000 is associated with. There are two ways to adjust a monetary value to the reference year: firstly, by using an average inflation rate, and secondly by using cost indices that try to capture the change of purchasing power over time.

3.1. Using average inflation rates

Calculating the present value P from a past value V with an average inflation rate f and a time difference of n years:

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n PV=⋅+(1 f) (1)

P = Present value [CHF] V = Past value [CHF] f = Average inflation rate [a-1] n = Number of periods between P and V [a] where the present value P is set as the reference year. If the value we would like to convert to the reference year is in the future, then we naturally invert equation (1).

3.2. Using cost indices

This method only works for values in the past. You need access to a relevant cost index list. Many different lists are available. They measure the cost development of a specific number of items or services over time and for a specific area. You will consequently need to find a cost index that is as specific as possible for the application you need. An example for the construction cost index for the canton of Zürich is given on page - 56 - (21.8 Construction Cost Index). The baseline in this index is the year 1954 and the index I1954 is defined as 100%.

The following formula converts a past value V of the year v into a present value P of the year p: I PV= ⋅ p (2) Iv P = Present value [CHF] V = Past value [CHF]

Example 1 Convert the following values into 2013 CHF: a. CHF 1000 spent in 1972 b. CHF 1000 that needs to be spent in 2047 1. Average inflation rate of 2.4% per year:

n (2013− 1972) 1a. PV=⋅+(1 f) = 1000 ⋅+( 1 0.024) = 2644.22 CHF(2013)

−−n (2013 2047) 1b. VP=⋅+(1 f) = 1000 ⋅+( 1 0.024) = 446.48 CHF(2013)

2. Use the cost index:

2a. From the construction cost index on Page - 56 -: I2013 = 544.2, I1972 = 241.3

I p 544.2 PV=⋅=1000 ⋅ = 2255.28 CHF(2013) Iv 241.3

2b. There are no cost indices for future values. Use approach 1.

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Ip, Iv = Cost indices for the years p and v, respectively The average inflation rate over a specific period can be calculated from these cost index tables by using equation (1) and performing a parameter estimation for the rate of inflation f.

4. Interest rates

4.1. Nominal interest rate

The nominal interest rate is the fee for borrowing money. It is composed of two parts: inflation and real interest. It is important to keep in mind that in the real world only nominal interest rates exist. This is the rate that banks charge their customers for a loan, for example.

4.2. Real interest rate

The real interest rate is the real cost of borrowing money without inflation. This is what we need to use for our calculations, according to the simplification defined in the last chapter. The real interest rate (i) can be estimated from the nominal interest rate (i’) and an estimated inflation (f) by: i′ − f i = (3) 1+ f

The real interest rate contains the profit plus a risk com- pensation expected by a potential investor. Generally for Use real interest rates public investments the risk of not repaying the debt is assumed to be small. In this case, a nominal risk-free in- with ‘real CHF’, i.e. terest rate, e.g. for government bonds or similar, can be monetary values in- used to estimate the real interest rate. Alternatively, a dexed to one specific social discount rate can be used for the calculations. year.

5. Discount rate

The literature contains two concepts for the (social) discount rate (from Abay, 2006):

• The social time preference rate, which indicates the value put on consuming something today instead of later.

• The social opportunity cost rate measures the threshold rate of return on private investments, or what would you expect to get back if you lend out some of your own money risk- free.

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The two concepts are measured differently, but indicate a similar idea: for most people money los- es value over time. We prefer immediate gratification over waiting and this is expressed by the discount rate.

Different countries use different discount rates to analyse their public investments, typically between 3% and 10% (Table 1). For Switzerland, a discount rate of 2% is suggested, based on the analysis of the social time preference rate. In Germany, a discount rate of 3% is generally stipulated.

Country Discount rate Time horizon (yr)

South Africa 8% 20-40 Germany 3% Variable Australia 6-7% 20-30 Canada 5-10% 20-50 Denmark 6-7% 30 United States 3-7% Variable Italy 5% France 8% 30 Hungary 6% 30 Japan 4% 40 Mexico 12% 30 Norway 4% Variable New Zealand 10% 25 Netherlands 4% 30 Portugal 3% 20-30 Czech Republic 7% 20-30 United Kingdom 3.5% 30 Sweden 4% 15-60 European commission 5% World Bank 10-12%

Table 1: Discount rates as listed by Commissariat Général du Plan 2005 (from OECD, 2007)

The discount rate is crucially important in projects where different costs are incurred over time. High discount rates are an incentive to postpone investments into the future. The higher the rate, the more worthwhile it is to delay costs and investments. The present value of a future investment can be calculated by:

− P = F ⋅ (1+ i) n (4)

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1.0 0.9 0.8 0.7 0.6 n - 0.5

(1+i) i= 0.4 2% 0.3 0.2 3% 0.1 5% 0.0 10% 0 10 20 30 40 50

Periods, n Figure 1: The impact of annual discount rates on the present value of a future investment. At a discount rate of 2% a-1, 5% a-1 and 10% a-1 the present value of an investment in 10 years is 82%, 61% and 39%, respectively.

Figure 1 shows this strong influence of the discount rate on the present value of a future investment. The further the investment is postponed into the future, the lower the present value. Accord- ingly, the higher the discount rate, the stronger this effect is. This is highly relevant, because in engineering economics decisions are based on present values, as you can see in Chapter 9 ‘Present value analysis’ (on page - 20 -).

5.1. Difference between discount rate and real interest rate

From a mathematical point of view, the discount rate is used in exactly the same way as the real interest rate. Either the discount rate or the interest rate has to be defined at the beginning of the calculations. If in doubt and no specifications are given in the assignment, then use 2%a-1 for Switzer- land (Abay, 2006). Public utilities often have access to loans with risk-free interest rates. An example is the SBI domestic government bond in Switzerland that municipalities use for their infrastructure investments. In such cases, the differences between the real interest rate and the discount rate are marginal, as can be seen in Figure 2.

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Figure 2: The difference between the SBI domestic (Swiss) government bond and consumer price inflation is given in the dashed line with triangles. The average of this difference over these 10 years is exactly 0.02 a-1, which corresponds nicely with the discount rate reported by Abay (2006)

6. Replacement value

The concept of replacement value is fairly straightforward. It is the amount that you would have to pay to replace an asset at the present time. For items that can be bought on the market, this value is easily determined: you go and ask the supplier. For custom-built items such as wastewater treatment plants or a drinking water pipe, this is a little trickier. Three widely used ways to calculate replacement values are:

1. Use a detailed cost estimate for the replacement of the specific asset in question, e.g. from a tender bid. This method has the advantage of considering the cost of the latest available technology and being very site-specific. However, it is quite laborious, and different engineers will often have different ideas about the latest suitable technology to be used. For example, would you propose to use an ultrafiltration membrane or a quick sand filter for a modern lake-water treatment plant?

2. Use the original construction cost and add inflation to get the current construction price (for the calculation, see Chapter 3 on Page - 8 -). This method produces a very reproducible result if the original construction costs are accessible. However, it might not yield reasonable outcomes for many older assets because the applicable conditions evolve and change over time. Technical progress changes the costs of machinery, legal requirements (e.g. for safe-

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ty) impact the complexity, and the asset may no longer be located in the countryside but in the middle of a city due to population growth, etc.

3. Use a proxy, such as the construction cost of a similar plant or a typical price for a similar pipe. For example: the Canton of Bern in Switzerland provided a standard formula for the municipalities to calculate replacement values of sewer lines (AWA, 2011). Figure 3 gives an example of such a generic approach.1 This method can be quite reproducible as long as a comparable calculation approach is used. The disadvantage is that it might also lead to systematic deviations in a very specific case. For example, the city of Bern identified a substantial deviation between the quoted AWA approach (AWA, 2011) and their real observed construction costs for sewers.

None of these three approaches is very precise and they Always reference where all have considerable uncertainties. Method 2 for recent- and how you derived ly constructed assets may be an exception. This makes it the replacement values essential to reference the replacement values used and make it very transparent as to where the numbers were you are using! derived from or how this was done.

Figure 3: Average replacement value for sewers below streets for different diameters and construction depths (from Fritsche et al., 2014 and AWA, 2011).

1 Further typical replacement values for Swiss infrastructures can be taken from Fritzsche et al. (2014)

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Engineering Economic Analysis

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7. Engineering economic analysis

7.1. Decision support

The task of an engineer is to find and design solutions to problems. Good engineers typically come up with several alternative solutions that could satisfy the needs of the client. The challenge is to identify the best solution amongst all these potential alternatives. Engineering economic analysis is about supporting this kind of decision-making. It focuses on the costs, reve- The main goal of an nues, and monetary benefits that occur during the plan- economic engineering ning horizon of a specific project. It can distinguish quantitatively between the costs of different alternatives. analysis is to obtain a Engineering economic analysis is only a part of a com- ranking of the costs for prehensive decision-making process. It complements the alternatives and gain other decision-making aids such as a cost-benefit analysis or a multi-criteria decision analysis. For a good and an idea about the rela- comprehensive introduction, see the text book ‘Rational tive difference between Decision Making’ (Eisenführ et al., 2010). them. It is important to realise the limits of engineering economic analysis:

• It only includes monetary terms and does not consider social welfare, environmental pro- tection or any other benefits. Decisions based on engineering economic analysis alone assume that the benefits of all options are equal and that costs are the most important factor.

• The main goal of the analysis is to obtain a ranking of the alternatives and gain an idea about the relative distance between them. This approach is clearly distinguished from a detailed financial analysis, where you get a clear idea about the absolute costs and can derive tariff changes, for example. Engineering economic analysis can achieve this precision, but does not have to do so.

• The standard methodologies presented here are static and do not consider the monetary values of the flexibility or adaptability of an alternative. See also the chapter ‘The Value of Flexibility’ (p - 41 -).

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Different ways to calculate costs Confusingly, there are several different ways to calculate costs and all of them might be correct. There are three main approaches to distinguish: 1. Engineering economic assessment: In this case, the rules are clear and laid out in this chapter 7 “Engineering economic analysis”. There we calculate with indexed money values and real interest rates. The goal is to produce reproducible and comparable costs that characterise a specific technology – mostly ignoring different financing alternatives. It is the engineers’ way to look at costs and by far a more generic approach. 2. Managerial or business point of view: This approach also includes different ways to finance the project. Here one usually calculates with nominal values and nominal interest rates. The financing can come from different sources, such as banks, tariffs, insurance or subsidies. 3. Municipal bookkeeping: Municipalities usually have strict rules how they have to do their bookkeeping and how to set tariffs. Typical examples are the neglect of inflation for the calculation of the book value or the definition that anything below a certain value is operating costs. An example might be the pro-active rehabilitation of an old water supply pipe. Looking at the overall costs (engineering economics) it might be more cost effective to replace the pipe even if it is not damaged yet. However, from a business point of view it might be cheaper to wait until the pipe bursts, because the costs of the damage are paid by the property owners and their insurances and not by the water supplier. Another example for approaches 1 and 3 can be found in exercise 5 (p. - 69 -) and for approaches 1 and 2 in exercise 9 (p. - 76 -). It is important to realise that the difference between the three approaches is mainly based on the choice of different system boundaries.

7.2. Public water utilities

Public water utilities are a special breed of business. They operate as natural monopolies and perform an essential public service. This means that no market exists to set an independent price for their services and that tariff- or price-setting is heavily regulated. The revenues are therefore set proportionally to the costs. As a consequence, most sophisticated tools used in engineering economic analysis, such as ‘rate of return analysis’ or ‘breakeven analysis’, are not applicable.

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8. Assumptions and simplifications

For practical purposes, we use several simplifications and assumptions: • End-of-year convention2: Generally, it is assumed that all costs and revenues occur at the end of a time period, typically a year. • Sunk costs: Events and investments that took place in the past, and do not affect the present or future, are called sunk costs. They have no relevance for deciding between alternatives and are excluded from the analysis. Only current and future differences between the alternatives are relevant and, therefore, considered. • Borrowed money viewpoint: The analysis focuses on investing money and not on how the money is obtained. We therefore assume that the money is borrowed at an interest rate. How the money is obtained or financed is not part of the economic analysis. • Real cost approach: We ignore inflation or deflation for the analysis. All calculations are car- ried out with real money that is indexed to a specific year.

9. Present value analysis

9.1. Introduction & definition

The present value analysis is the most robust and also most widely used approach. The main goal is to calculate the total sum of money that needs to be set aside today to cover all the expenses during a defined time span, known as the planning horizon. The lower this sum is, the cheaper the alternative and the better the ranking.

The net present value (NPV) is defined as Net present value = Present value of benefits + Present value of costs NPV = PV of benefits + PV of costs

Please be aware that costs can have a negative or positive sign, depending on how they are defined. Because we deal predominantly with costs, in this manuscript costs are assigned a positive sign and benefits a negative sign.

Step 1: Define the planning horizon At the end of the planning horizon, all alternatives should be at the end of their useful life. This can be achieved by choosing the least common multiple of the useful lifespans of all considered items as the planning horizon.

2 Or end-of-period convention, if the useful time period is not a year.

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If the least common multiple leads to an unreasonably long planning horizon, the analysis period can be shortened by assuming that the depreciated value3 of the investment can be recovered at the end of the planning horizon. It is important to realise that this is not a realistic approach for most investments in the water sector, as the re-sale value is usually zero, i.e. a used sewer cannot be sold and therefore has no monetary value, but only a remaining useful life. Step 2: Calculate the present value of all costs and benefits

The present value (Pk) of a future cost/benefit (Fk) at time nk can be calculated on the basis of the interest rate (i) with:

− nk (4) Pk = Fk ⋅(1+ i)

For a series of annual recurring costs Ak from the beginning of the planning horizon to the year nk, the following simplification can be used:

(1+ i) nk −1 (5) Pk = Ak ⋅ i ⋅ (1+ i) nk

The net present value of an alternative with i cost elements is then defined as:

i l − nk (6) NPV = ∑ Pk =∑ Fk ⋅ (1+ i) k =1 k =1

If costs have a positive sign, then the alternative with the lowest NPV is the cheapest.

3 Typically, a linear depreciation model based on the lifespan can be used.

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Example 2 Compare two pumps with a 5-year lifespan. Which one is cheaper to operate? Assume a discount rate of 3%a-1. Pump 1: Initial costs CHF 1000 with annual costs of CHF 100/a Pump 2: Initial costs are CHF 800 with annual costs of CHF 120/a (1+ 0.03)5 −1 P = 100 ⋅ = 100 ⋅ 4.58 = 458 1 ⋅ + 5 0.03 (1 0.03) (5) (1+ 0.03)5 −1 P = 120 ⋅ = 120 ⋅ 4.58 = 550 2 0.03⋅ (1+ 0.03)5

i (6) NPV1 = ∑ Pk =1000 + 458 = 1458 CHF k =1 i NPV2 = ∑ Pk =800 + 550 = 1350 CHF k =1 Pump 2 is cheaper to operate despite the higher annual costs.

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10. Annual cash flow analysis (annuity method)

The annuity method converts all costs into a homogenous cash flow over the entire planning horizon. This is equivalent to the average amount of money you need to spend each year during the planning horizon to cover all expenses.4 A typical unit is [CHF per year] compared with [CHF] for the net present value. It is important to realise that the NPV can always be converted into an annuity, and vice versa. The benefit of an annuity is that it is easier to communicate and slightly faster to calculate under specific simple circumstances. A typical case is the comparison of two wastewater treatment plants with different expected annual operating costs.

Based on a present investment Pk, the annuity Ak for an interest rate i and the planning period n can be calculated as (analogous to eq. (5)): i ⋅ (1+ i) n A = P ⋅ (7) k k (1+ i) n −1

The annual cash flow then can be calculated with:

ACF = Sum of all annuities + sum of all uniform annual expenses

If costs have a positive sign, then the alternative with the lowest ACF is the cheapest. Equation (7) can also be used to convert the NPV into an ACF and vice versa.

Example 3 Compare two pumps with a five-year lifespan. Which one is cheaper to operate? Assume a discount rate of 3%a-1. Pump 1: Initial costs CHF 1000 with annual costs of CHF 100/a Pump 2: Initial costs are CHF 800 with annual costs of CHF 120/a i ⋅ (1+ i) n 0.03⋅ (1+ 0.03)5 A = P ⋅ = 1000 ⋅ = 1000 ⋅ 0.218 = 218 CHF/a 1 k (1+ i) n −1 (1+ 0.03)5 −1 (7) 0.03⋅ (1+ 0.03)5 A = 800 ⋅ = 800 ⋅ 0.218 = 175 CHF/a 2 (1+ 0.03)5 −1

ACF1 = 218 +100 = 318 CHF/a

ACF2 = 175 +120 = 295 CHF/a And not surprisingly, Pump 2 is again cheaper to operate despite the higher annual costs.

4 Or alternatively, the annual earnings you make over the planning horizon. In this text I focus on costs, but of course the same would be valid for earnings or benefits.

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Example 4 Cross comparison of Example 2 and Example 3: i ⋅ (1+ i) n 0.03⋅ (1+ 0.03)5 ACF = NPV ⋅ = 1458⋅ = 318 CHF/a 1 1 (1+ i) n −1 (1+ 0.03)5 −1 (7) 0.03⋅ (1+ 0.03)5 ACF = NPV ⋅ = 295 CHF/a 2 2 (1+ 0.03)5 −1

(1+ 0.03)5 −1 NPV = ACF ⋅ = 1458 CHF 1 1 ⋅ + 5 0.03 (1 0.03) (1+ 0.03)5 −1 NPV = ACF ⋅ = 1350 CHF (5) 2 1 ⋅ + 5 0.03 (1 0.03)

This shows that both methods produce exactly the same results.

11. Sensitivity analysis

All the numbers used in an engineering economic analysis are associated with an uncertainty. In the case of building projects, the uncertainty of the cost estimate is given by the engineers. However, parameters such as interest rates or population trends are more difficult to assess for future events. It is crucial to investigate the influence of these uncertainties with a sensitivity analysis that should achieve the following goals:

• Are the differences between the alternatives significant? For example, if all cost estimates are +/-25% then a 5% difference between two alternatives is not significant at all.

• Is the ranking robust? Which combination of parameter values influences the ranking of the alternatives? The result of the sensitivity analysis helps to characterise the quality and robustness of the engineering economic analysis and to define the next steps. Thus it may be worth investing in a more precise cost estimate for the top three alternatives in order to achieve a more robust ranking. Or information about the strong impact of the interest rate can influence how the investments are financed.

Performing good and reliable sensitivity analyses is an art in itself. A good overview can be found in Saltelli’s book “Sensitivity Analysis” (Saltelli et al., 2000). In practice, simple screening designs are most often applied. They aim to isolate the most important factors that may affect the outcome of

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an analysis. This is usually done by setting reasonable lower and upper boundaries for all inputs and parameters, then recalculating the analysis by varying a single factor (one-at-a-time experi- ment) and comparing the results.

A sensitivity analysis must be performed for any engineering project that is worth the salaries of those involved.

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Depreciation & Hidden Assets

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12. Depreciation

Depreciation aims to identify the value of an investment or asset over time. It basically involves two concepts:

a) The decrease of the value of an asset with respect to its current market value. Thus the re- sale value of a car decreases over time.

b) The allocation of the cost of the asset over its lifespan, or the period during which the asset is used.

Concept a) rarely applies to water infrastructures. Water mains, sewer lines or treatment plants do not usually have a re-sale value that would represent the real use value.

Many different ways are available to allocate the asset costs over time. There is no fundamental wrong or right way of doing it. However, legal requirements may insist on a certain depreciation method for utilities. We will only present three important types here: linear depreciation, the declining balance method and annuity depreciation.

12.1. Linear depreciation

Linear or straight-line depreciation has the following mathematical representation: age PC=⋅−1 (8) lifespan P = Present value [CHF] C = Cost of the asset [CHF] age = Current age of the asset [a] lifespan = Estimated useful life of the asset [a]

The annual depreciation D [CHF a-1] can be calculated from this with: C D = (9) lifespan

Linear depreciation shows a constant decrease of the value over the lifetime (see Figure 4). It requires an adequate lifespan to be assessed for every asset and the use value to be tracked individ- ually.

12.2. Degressive depreciation (or declining balance method of depreciation)5

The use value for year n+1 is calculated with:

5 German translation: Degressive Abschreibung. Another term used is ‘declining balance method of depreciation’.

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(10) Pnn+1 =−⋅(1 jP)

(11) PC0 = P = Present value at year n+1 or year n [CHF] C = Cost of the asset value [CHF] j = Depreciation rate [a-1] Or alternatively, equations (10) and (11) can be written as:

n (12) Pn =−⋅(1 jC)

More is charged near the beginning of the lifetime and less at the end (see Figure 4). This method is very simple as it does not require any identification of lifespan or age. No separate accounting for the different assets is needed either, as they can all just be lumped together in one big sum. So this method is very attractive for bookkeeping, as no inventory is needed.

Figure 4: Relative value of the asset cost due to linear or degressive depreciation.

12.3. Annuity depreciation

This is a special case of linear depreciation where the total capital cost of the asset is equally distributed over the lifespan. The total capital cost includes the investment and the interest due over the entire period. For the calculation, see equation (7).

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12.4. Book value

Again, in this script we make the extreme simplification that the book values do not include any cor- rection for inflation. This means that we use equations (8) or (10), for instance, for calculating book values and apply them to nominal values and not to a real monetary value that is indexed to a specific year.

13. Hidden assets

A hidden asset is an asset that is not represented adequately on a balance sheet. This can be either:

• an asset that was ‘given’ to the operator. For example, a municipality connects their network to the trunk sewer of a wastewater treatment plant, adding a combined sewer overflow and a stormwater tank at the connection. The community decides that this was part of the connection cost, but they transfer its operation, maintenance and ownership to the WWTP. So the treatment plant now owns a hidden asset that has no value (= no cost) in their books.

• a difference between the ‘real’ or ‘operational’ value and the book value. Several reasons can contribute to this difference. The most prominent ones are subsidies and a depreciation scheme that does not correspond to the lifespan of the asset.

A typical Swiss case relates to the sewer networks built in the 70s and 80s of the last century. Their construction was heavily subsidised and then depreciated with a 10% degressive (declining balance) approach. See Exercise 5 in this script (p. - 69 - ) for the possible consequences.

Figure 5 shows an example of the value of hidden assets over time for a sewer. It assumes that an equal distribution of the value over the lifespan of the asset (linear depreciation over 80 years) offers a better representation of the ‘real opex’ value of the sewer. If the book value of the sewer is depreciated with a 10% degressive approach, then the hidden assets vary substantially over time. Thus at age 20 the two values are:

• linear: (80-20)/80 = 75% • 10% degressive: 0.920 = 12.2% The difference of 62.8% of the original value of the sewer is then defined as the hidden asset. This disappears at the end of the lifespan or when the asset is replaced.

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Figure 5: An example of hidden assets, due to depreciation, in a sewer with an assumed lifespan of 80 years. In this example it is assumed that linear depreciation is best suited to represent the ‘operational value’ of the sewer. Therefore, the dashed line is the value of the hidden asset over time if degressive depreciation is instead used.

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Cost Calculations Debt-to-Asset Ratio

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14. Cost calculations

14.1. The fundamentals

It is characteristic of many water services that they represent natural monopolies6 with no competitive (waste-)water market. The price for these services, and ultimately their tariffs, are determined by the costs, so the overall revenue is proportional to the expenses.7 Being able to identify the full costs is thus essential for determining the prices and tariffs. A common keyword here is full cost re- covery.

Determining full costs is straightforward: Full Costs = Capital Expenditure + Operating Expense FC = CAPEX + OPEX

Capital expenditures (CAPEX) comprise depreciation and interest paid, whereas operating expenses (OPEX) include all the costs needed to run the utility, such as costs of labour, electricity, chemi- cals or sludge disposal8.

Although this approach might seem to be clear and straightforward, it is in practice complicated by several issues. The main point of contention is the fair valuation of depreciation if substantial hidden assets (see Chapter 13 on p. - 30 - ) are present. The word ‘fair’ is deliberately chosen as the discussion is not a technical one but concerns intergenerational fairness. Sewer networks in particu- lar were rapidly built with substantial subsidies during the 70s and 80s in response to mounting environmental concerns. In Switzerland, they were additionally depreciated with a 10% degressive approach. As a consequence, many sewer networks have hardly any book value and therefore hardly any CAPEX.

For sewer networks with expected lifespans of 80 years, this means metaphorically speaking that one generation builds and pays for the infrastructure, while the second generation can benefit from it and the third generation needs to pay for its rehabilitation. This is basically the core of the discussion, and has led - at least in Switzerland - to two different approaches for calculating the costs relevant to tariffs. I have named them the ‘Zürich’ and ‘Bern’ models, after the first cantons that formu- lated explicit laws to regulate these cost calculations. The first is a more specifically bookkeeping point of view that focuses on real cash flow. The latter considers intergenerational fairness by taking an operator point of view and uses a linear depreciation based on replacement value to calculate the CAPEX.

6 Natural monopolies show one or more of the following properties: (i) great economies of scale over a large range of services provided (marginal costs decrease with scale); (ii) high ratio of fixed to variable costs; (iii) one utility can produce a desired service at a lower (social) cost than two or more companies. Network-bound water and wastewater services are clas- sic natural monopolies. 7 For an in-depth discussion of marginal vs. average price setting, see Vickrey, 1948, 1987 and Hotelling, 1938 8 OPEX is also termed running costs. Please be aware that OPEX and CAPEX should not be confused with variable and fixed costs.

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It is important to emphasise that these two models are only relevant when substantial hidden assets are present. Both models yield the same results in the long run.

14.2. The bookkeeper view (‘Zürich’ model)

The ‘Zürich’ model takes a bookkeeper view and focuses on the (expected) cash flow. It demands a linear depreciation for the CAPEX. The remaining book value of the asset is thus depreciated linearly until the expected end of life. New assets will go in the books and will also be linearly depreciated.

This approach lets the users benefit from the hidden assets until they are gone. It simply accepts the fact that the financing scheme has changed over time. Substantial increases in costs are to be expected once the hidden assets have been ‘consumed’ and need to be rehabilitated (see Exercise 5 on p. - 69 - for the possible consequences). The effects are not as dramatic as Figure 5 might im- ply, as the end of life for most assets is subject to wide variability. In addition, careful planning can also involve the targeted build-up of reserves, which can have an additional smoothing effect on costs and therefore tariffs (see also the next chapter on debt-to-asset ratio). Material on the regulatory details can be found in AWEL (2007).

14.3. The operator view (‘Bern’ model)

The ‘Bern’ model has a clear operator viewpoint and aims to provide a stable CAPEX over time. The basic philosophy is that large variations in cost for the same service are not acceptable. This is achieved by calculating a CAPEX based on a linear depreciation of the replacement value, hence making sure that hidden assets do not distort the CAPEX that a user has to pay.

As a consequence, there is a discrepancy between the CAPEX derived from these two approaches. To balance it out, the regulation creates a special account9 where the difference is deposited. This account acts as a reserve from which rehabilitation and investment costs can be paid. Material on the regulatory details can be found in AWA (1999).

The great benefit of this approach is that with a constant replacement value the CAPEX remains constant over time and only needs to be corrected for inflation. It basically turns the hidden assets into cash that is then available for the rehabilitation and renovation of the system.

9 In Canton Bern this is called ‘Konto Spezialfinanzierung’.

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Example 5: Uncle Rich gifts you a car Let us assume the following situation. You are planning a start-up company that offers on-site support for water monitoring campaigns. Your uncle Rich decides to support your endeavour by gift- ing you his used car for your site visits. Let us assume that the replacement value of the car is CHF 40,000 and it is the only asset you need for your start-up. It is five years old and you expect a lifespan of ten years. For the sake of simplicity, we ignore inflation and assume that we can get loans at zero interest. Case 1: The bookkeeper view As the car was a gift, there is nothing on the books for the first four years. At the end of year five, the car is at its end of its life and needs to be replaced by a new one. The value of CHF 40,000 for the car goes onto the books and at the same time a loan of CHF 40,000 needs to be paid off with the CAPEX of CHF 4,000 per year (40,000/10). The following two figures show the book value (left) and the development of the CAPEX and debts over time.

Case 2: The operator view At the time of the gift, the car has an operational value of CHF 20,000. This is what goes onto the asset accounts. Every year CHF 4,000 CAPEX is due and forms the reserves for buying a new car (in German: ‘Rückstellungen’). At the end of year five, the new car is bought with CHF 20,000 reserves and CHF 20,000 debt.

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15. Debt-to-asset ratio

The debt-to-asset ratio10, or short debt ratio, is simply defined as: Debt Ratio = Total Liabilities / Total Assets or Debt Ratio = Total Debts / Total Assets

A debt ratio of 1 means that the assets are completely financed by loans. Conversely, a debt ratio of 0 means that the assets are completely financed by subsidies and tariffs.

The debt ratio is a key decision-making factor for setting tariffs. Besides the full costs, the utility can decide to increase its tariffs in order to provide cash for investments. This reduces the need for loans and correspondingly also the future CAPEX due to lower interest payments. The ‘Bern’ model of cost calculation has this in mind by converting hidden assets into cash.

From an economic point of view, however, replacing loans with money from customers is not free either. It creates opportunity costs for the customer that can be approximated with a discount rate (for calculation, see Chapter 5, p. - 10 - ). Public utilities generally have access to loans with risk-free interest rates that are close to the discount rate (see also Figure 2 on p. - 13 - ). In these situations, the decision to set a debt ratio follows political rather than economic arguments. In other cases, where there is a difference between the discount and interest rates, it might make sense to carefully optimise the debt-to-asset ratio.

10 In German: ‘Fremdfinanzierungsgrad’. Please be aware that in Swiss practice the term ‘Eigenfinanzierungsgrad’ is also often used. Eigenfinanzierungsgrad = 1-Fremdfinanzierungsgrad

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Example 6: Debt-to-asset ratio The following figure shows the debt ratio for the two cases in Example 5:

It is clear that in Case 1 (bookkeeper model) the debt ratio is 0 during the first four years and 1 at the end of year 5 (remember the end-of-year convention), as the new car is entirely financed by a loan. In Case 2, the CAPEX charged during the first five years, based on the activation of the ‘operational’ value of the car in the book, created a reserve fund. This reduces the debt ratio to 50% in year 5 and decreases over time. Be aware that the debt ratio can only have values between 0 and 1. Question: Why does the debt-to-asset ratio stay at 1.0 after year 5 in the ‘bookkeeper’ model?

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The Value of Flexibility

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16. The future is uncertain

16.1. Uncertainty matters

Water infrastructures are built to last for decades. The average lifespan of the entire system for water supply and wastewater treatment (including treatment plants and private installations) in Swit- zerland is around 56 years. A typical planning horizon for a new treatment plant is 30 years. Many things change over a 30 year period, and so such forecasts are always wrong. For example, an error of 0.005 a-1 in estimating the average demand growth rate (e.g. 2.5%/a instead of 2.0%/a) leads to a 16% error in built capacity. This error means that the plant will reach its capacity limit in 24 years instead of 30 years. Depreciation will therefore increase by 25%.

In Switzerland the importance of uncertainty is clearly recognised. Figure 6 shows the result from a survey of over 200 Swiss engineering companies that work in the water sector. More than 50% of the respondents stated the importance of uncertainty in their decision-making.

Figure 6: Answers to the question “In your company, how important are uncertainties when finding a solution in decision-making problems?”. UWIP = companies working in urban water infrastructure planning (figure taken from FitzGerald, 2015).

The survey also gave some hints about the kind of uncertainty that matters most for Swiss practi- tioners. Figure 7 shows, not surprisingly, that for companies working with urban water infrastructure, planning the demographic development and the water demand are essential. Also upcoming new technologies and energy costs are high on the list. These are all aleatory or scenario uncertainties. They are system immanent and cannot be reduced with any engineering measures (at least not in most democratic countries).

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Figure 7: Answers to the questions: Which future uncertainties do you consider when finding a solution in general (25) and for urban water infrastructure planning respectively (26)? The values are given as fractions of the number of companies (figure taken from FitzGerald, 2015). See Figure 6 for the definition of UWIP.

16.1. Lack of methodologies to deal with uncertainty

In engineering economics, a new treatment plant is planned and evaluated in several steps. The terms and plant requirements, such as the planning horizon, capacity, performance, footprint, and others, are evaluated first. Second, this profile is passed on to the design office or used in a call for tenders. Third, the alternatives generated in this way are evaluated by various criteria, in which an economic comparison is usually prominent.

De Neufville R. and Scholtes (2011) characterises this standard engineering design method as follows: “The standard process designs projects based on a limited set of assumptions and then considers uncertainties. The focus is on creating robust designs that will perform satisfactorily under various uncertainties and possible stresses. This reflects a bunker mentality: Will we be able to survive adverse futures?” This bunker mentality manifests as ‘safety factors’ or worst-case scenarios that are applied to compensate for uncertainty. In contrast, modern engineering design should aim to include flexibility in the design. In order to find the optimum level of flexibility, we need to start developing measures

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to quantify the ‘right’ amount of flexibility. More flexibility might be more expensive in the initial investment phase, but may be more cost-effective overall. The planning approach described above commonly leads to overcapacities, except in the rare cir- cumstance where the load remains constant over the planning horizon of 30 years. In any other case, involving growth or decay, there will be overcapacities at some point in the lifespan of the plant. Depending on the growth rate, these can be quite substantial and raise the question as to the conditions under which it would be beneficial to apply modularisation.

Many water services are natural monopolies characterised by the absence of a competitive (waste-) water market. The price for their services is usually determined by their costs, so that the overall revenue is proportional to the expenses. The lack of a cost-independent price for water services substantially invalidates the information value of the commonly used net present value (NPV) and renders other methods, such as the internal rate of return (IRR), pointless. Consequently, only the present value (PV) of the costs is used in the economic evaluation of water projects, excluding the revenues.

This means that the commonly used NPV (as introduced in Chapter 9, p. - 20 - ) is not really adequate to evaluate water infrastructures in a growth scenario. In the following discussion, I will intro- duce a new evaluation approach, called the specific net present value (SNPV) method. This method is not (yet) commonly used for engineering economic comparisons. However, I hope that I can convince you that in situations where demand grows over the planning horizon, the SNPV approach is more suitable for comparing different treatment approaches with different adaptation capacities, whereas the conventional NPV method is restricted to very similar technical configurations.

This chapter is not a comprehensive manual of flexible engineering design. For that purpose, you might want to have a look at De Neufville R. and Scholtes (2011) for generic approaches, and the doctoral thesis of Dominguez (2008) more specifically for the wastewater sector. This chapter will show you how to estimate the costs you can invest into staging a treatment plant and how to assess the impact of uncertainty on cost. I also hope to convince you that for treatment plants the inclusion of safety factors is probably the most expensive way of preparing for an uncertain future.

17. Plant utilisation and idle capacity

One of the consequences of centralised treatment plants is that they have to deal with changing demand in their catchment area. This is usually done by setting up some kind of forecast about the expected maximal demand the plant has to cover. For example, Figure 8 shows the demand curve for a catchment with an exponential growth of 0.03 a-1. The plant is then designed to handle 100% demand. So when the construction of the plant is finished, it starts with 40% utilisation.

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Figure 8: The development of an exponential demand (0.03 a-1) over the lifespan of a treatment plant.

The form of the curve does not matter. Any plant with this design approach (design for 100% demand) will have idle capacity over time. The effective plant utilisation (PU) over the entire planning horizon can be calculated with:

TP Pdt ∫ t (13) PU = 0 TPP⋅ Dim

PDim = Installed capacity in population equivalents [PE]

Pt = Demand (or load) in population equivalents at time t [PE]

TP = Planning horizon [a] PU = Plant utilisation [-]

In the case of exponential growth, the demand Pt is:

λ⋅t (14) Pt = P0 ⋅ e

P0 = Population equivalents at time t=0 [PE] λ = Growth rate of the demand [a-1] t = Time [a]

This defines the installed capacity PDim with:

λ⋅ P=⋅> PeTP for λ 0 Dim 0 (15) PPDim = 0 for λ ≤ 0

And the plant utilisation PU from equation (13) is then:

TP λt P⋅ e dt − λ ∫ 0 1 TP Pe0 ⋅−λ ( 1) (16) PU =0 = for λ >0 λTP TPP⋅ Dim TPeP ⋅⋅0

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−λ 1− e TP PU = for λ > 0 λTP λ e TP −1 (17) PU = for λ ≤ 0 λTP

Figure 9: Plant utilisation (PU) vs. exponential growth rate (λ). High growth rates lead to low plant utilisation and therefore a large amount of idle capacity over the lifespan of a treatment plant.

For any growth rate other than 0, the plant utilisation is below 100%, as can be seen in Figure 9. Higher growth rates translate into large idle capacity over the planning horizon of a treatment plant. Having substantial idle capacity means that the cost per utilised capacity is higher than indicated by the present value of the installed performance. As a consequence, two plant designs with the same NPV may still have different average costs if their utilisations differ.

18. Specific net present value (SNPV)

Idle capacity increases the average cost of a serviced unit. This cost can be quantified with the specific net present value or SNPV: NPV

SNPV = T 1 P (18) ⋅⋅ ∫ Pt dt TP 0 NPV = Net present value according to eq. (6) [CHF]

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For the other nomenclature, see eq. (14)

Substituting eq. (13) into eq. (18) gives the main definition of SNPV: NPV SNPV = (19) PU⋅ PDim

This formula means that the plant utilisation, and therefore the growth rate, has a substantial impact on the specific costs that a potential user has to pay. To minimise the overall cost to the consumer, it is better to minimise the SNPV than the NPV. Figure 10 shows clearly that the SNPV can change substantially with changing plant utilisation, while the NPV would stay constant. This shows that in a demand growth situation the SNPV may well favour a modular system with higher NPV costs. See Exercise 7 (on p. - 74 -) to calculate the cost of a staged construction of a treatment plant.

Figure 10: How the specific net present value changes with exponential growth rates. This figure corresponds to Figure 9.

19. Considering uncertainty

19.1. Adding uncertainty

Various strategies are available to deal explicitly or implicitly with the uncertainties of future development. They include: deliberately building over-capacities to construct a robust plant that can deal with a higher demand than predicted, a staged design that allows the design to be reconsid- ered based on an updated knowledge of the effective parameters, or decentralised small-scale systems that can keep pace with the actual demand (build as you go).

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Here I will present a simulation approach to quantify and illustrate the economic consequences of these uncertainties for wastewater treatment plants. As a measure for the specific costs, we apply the specific net present value (SNPV) as introduced above. We discuss the economic value of flexibility and draw conclusions for planning appropriate and cost-efficient water and wastewater infrastructures. Figure 11 (left) shows the same growth rate as in Figure 8 over a planning horizon of 30 years, but together with possible alternative outcomes. These alternatives were calculated by assuming an annual exponential growth rate of 0.03 a-1 that follows a normal distribution with a standard deviation of 0.03 a-1. The histogram on the right of the figure indicates the uncertainty in reaching a certain demand at the end of the 30 years. Details can be found in Hug et al. (2010).

Figure 11: Simulated random scenarios of future demand over a planning horizon of 30 years. The y-axis represents the relative demand compared to the demand at time zero (note the different scales). Darker regions of the graph contain more growth series and therefore indicate a higher possibility that the plant experiences this specific demand. The curves were simulated by randomly sampling each annual exponential growth rate from a normal distribution (left: λ =0.03a-1, σ=0.03a-1; right: λ=-0.01a-1, σ=0.03a-1). The histograms indicate the distribution of the final population at t=30a. The bold lines indicate the predictions of population growth that were used for the plant design (50% and 90% percentiles of the distribution in each case). From Hug et al. (2010).

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Figure 12: Effect of uncertainty shown as the cumulative probability distribution of the specific costs as SNPV. The vertical line indicates the SNPV based on the predicted demand development without uncertainty. The curves represent the distribution of SNPV due to the increasing uncertainty of future demand. The annual growth rates of the scenarios shown are sampled from normal distributions with λ=0.03a-1 and increasing standard deviations σ from 0.00a-1 to 0.05a-1. The bold line stands for the illustrative example with σ=0.03a-1. The values are normalised to SNPV100%,ref of a reference plant with PDim=10,000 PE and 100% plant utilisation. From Hug et al. (2010).

Again, as in Figure 8, we assume that the average growth prediction determines the capacity of the treatment plant built. Now, following one random possible development curve, the plant may run out of capacity after 20 years, when it will need to be extended. This will incur additional costs. On the other hand, there is also a chance that the plant is never actually utilised to its full capacity. Both scenarios will lead to potentially higher costs than the average case would suggest. Again, the relevant quantity we use here for comparison is SNPV (see eq. (19)). Figure 12 shows the results of all possible development scenarios. First, and as Figure 10 also indicates, one can see that even without any uncertainty the specific costs will be 20% higher with 3% annual growth compared to a plant without growth.11 Of course adding uncertainty also makes this outcome uncertain; i.e. there is a small probability that the SNPV will actually be cheaper, but in the majority of cases the costs will be substantially higher. Thus at a growth rate of 0.03 a-1 and a standard deviation of 100% there is an 80% chance that the costs will be more than 30% higher than for a plant of the same size with a certainty of no growth. The higher the uncertainty, the greater the probability of high costs (the curves shift and lean to the right in Figure 12).

19.2. The fallacy of ‘safety’ factors

Safety factors have their place in engineering, especially when we are dealing with safety issues. This chapter introduces the factors used in the design of non-safety critical infrastructures. Thus one way to compensate for the uncertainty in Figure 11 is to increase the capacity of the plant. We

11 There is a small difference between Figure 10 and Figure 12 because the operating costs are included in the latter.

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could increase the design capacity by 24% in order to obtain a 90% probability that the plant will last the entire 30 years (Figure 13, left).

Figure 13: The effect of over-sizing shown as a cumulative probability distribution of the specific costs as SNPV. The solid lines indicate the SNPV based on the standard prediction of demand development (50th percentile of simulated distribution). The dashed curve shows the resulting SNPV distribution if the plant design is based on a higher growth prediction (90th percentile of simulated distribution). The annual growth rates of the scenario shown are sampled from normal distributions with λ=0.03a-1 and a standard deviation σ=0.03a-1. The values are normalised to SNPV100%,ref of a reference plant with PDim=10,000 PE and 100% plant utilisation.

Figure 13 (right) shows the result of this decision. The dashed curve shows very clearly that there is a good probability that the specific costs will increase substantially without any visible gains. This is the consequence of having large idle capacities for most of the operating time in most development scenarios. Adding safety factors to the demand forecast just increases the specific costs.

20. Estimating the value of modularisation

The SNPV introduced in Chapter 18 can be used to estimate the additional CAPEX for the staged expansion of a treatment plant. Exercise 7 on p. - 5 - gives an example of how to do that. At low projected rates of demand growth (< 0.02 a-1), a modularised project should involve less than 20% additional CAPEX compared to the single-stage alternative. For fast-growing situations, this additional cost can be substantially higher and can even reach 100%. It is worth mentioning that these results are independent of the size of the project and the estimates are relatively conservative. Hence the title of this chapter: ‘The value of flexibility’. This approach provides a way of quantifying the value of flexibility, e.g. in the form of staging or modularising the construction of a treatment plant.

One of the interesting results is that demand growth has a similar effect on average costs as economies of scale. High growth rates lead to low plant utilisation (PU). The higher the growth, the smaller PU becomes, with direct consequences for the number of customers sharing the CAPEX of a plant. So far, this factor has not been explicitly included in typical economic engineering deci-

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sions made for treatment plants in the water sector. This script argues that because the SNPV approach considers plant utilisation, it represents the most appropriate criterion to be used in engineering economic decisions for water services.

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Formulas, Tables & References

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21. Formulas & Tables

A collection of useful formulas are given below. Please be aware that the units are only given as examples.

21.1. General

r = (1+ i) (20) i = interest rate [a-1]

21.2. Interest

Calculates the Future value based on a Present value and an interest rate:

F = P ⋅ r n (21) r = see eq. (20) [-] F = Future value [CHF] P = Present value [CHF] n = Number of periods [a]

21.3. Annuity

Based on a present value, the annuity can be calculated as: (r −1)⋅ r n A = P ⋅ a = P ⋅ (7) r n −1

A = PMT(i;n; P;0;0)

P = PV (i;n; A;0;0)

The future value of an annuity is given with: r n −1 F = A⋅ (22) r −1

F = FV (i;n; A;0;0) r = see eq. (20) [-] -1 A = Annuity [CHF a ] P = Present value [CHF] F = Future value [CHF]

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n = Number of periods [a] a = Annuity factor, see Table 2 (page - 55 -) PMT, FV, PV = Corresponding Excel formula

21.4. Other Excel formulas

These Excel formulas may also be of interest:

IPMT = Returns the interest payment for a given period

CUMIPMT = Returns the cumulative interest paid on a loan for a fraction of the payment period PPMT = Returns the payment on the principal for a given period

CUMPRINC = Returns the cumulative principal paid on a loan for a fraction of the payment period

21.5. Real interest rate

The real interest rate (i) can be calculated from the nominal interest rate (i’) and the inflation (f) with: i′ − f i = (3) 1+ f i’ = Market (nominal) interest rate [a-1] i = Real interest rate, discount rate [a-1] f = Inflation rate [a-1]

21.6. Typical lifespan

These are typical lifespans of technologies and equipment relevant to water infrastructures: Item Lifespan [a] Wastewater transport Public sewer lines 80 Private sewer lines 50

Wastewater treatment Entire wastewater treatment plants 30

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21.7. Annuity Table

Lists the annuity factor a

Interest Number of Periods, n Rate, i 8 10 15 25 30 50 80 100 0.0% 12.500% 10.000% 6.667% 4.000% 3.333% 2.000% 1.250% 1.000% 1.0% 13.069% 10.558% 7.212% 4.541% 3.875% 2.551% 1.822% 1.587% 1.2% 13.184% 10.672% 7.324% 4.654% 3.989% 2.671% 1.951% 1.723% 1.4% 13.300% 10.786% 7.438% 4.768% 4.105% 2.794% 2.086% 1.864% 1.6% 13.417% 10.901% 7.552% 4.885% 4.223% 2.921% 2.225% 2.011% 1.8% 13.534% 11.016% 7.667% 5.003% 4.343% 3.050% 2.368% 2.163% 2.0% 13.651% 11.133% 7.783% 5.122% 4.465% 3.182% 2.516% 2.320% 2.2% 13.769% 11.249% 7.899% 5.243% 4.589% 3.318% 2.668% 2.482% 2.4% 13.887% 11.367% 8.017% 5.366% 4.714% 3.456% 2.823% 2.647% 2.6% 14.006% 11.485% 8.136% 5.490% 4.842% 3.597% 2.983% 2.816% 2.8% 14.126% 11.604% 8.256% 5.616% 4.971% 3.740% 3.145% 2.989% 3.0% 14.246% 11.723% 8.377% 5.743% 5.102% 3.887% 3.311% 3.165% 3.2% 14.366% 11.843% 8.498% 5.872% 5.235% 4.035% 3.480% 3.343% 3.4% 14.487% 11.964% 8.621% 6.002% 5.369% 4.187% 3.652% 3.524% 3.6% 14.608% 12.085% 8.744% 6.133% 5.505% 4.341% 3.826% 3.708% 3.8% 14.730% 12.207% 8.869% 6.267% 5.643% 4.497% 4.003% 3.893% 4.0% 14.853% 12.329% 8.994% 6.401% 5.783% 4.655% 4.181% 4.081% 4.2% 14.976% 12.452% 9.120% 6.537% 5.924% 4.816% 4.362% 4.270% 4.4% 15.099% 12.576% 9.247% 6.675% 6.067% 4.978% 4.545% 4.460% 4.6% 15.223% 12.700% 9.376% 6.813% 6.212% 5.143% 4.730% 4.652% 4.8% 15.347% 12.825% 9.504% 6.954% 6.358% 5.309% 4.916% 4.845% 5.0% 15.472% 12.950% 9.634% 7.095% 6.505% 5.478% 5.103% 5.038% 5.5% 15.786% 13.267% 9.963% 7.455% 6.881% 5.906% 5.577% 5.526% 6.0% 16.104% 13.587% 10.296% 7.823% 7.265% 6.344% 6.057% 6.018% 6.5% 16.424% 13.910% 10.635% 8.198% 7.658% 6.791% 6.542% 6.512% 7.0% 16.747% 14.238% 10.979% 8.581% 8.059% 7.246% 7.031% 7.008% 7.5% 17.073% 14.569% 11.329% 8.971% 8.467% 7.707% 7.523% 7.505% 8.0% 17.401% 14.903% 11.683% 9.368% 8.883% 8.174% 8.017% 8.004% 8.5% 17.733% 15.241% 12.042% 9.771% 9.305% 8.646% 8.512% 8.502% 9.0% 18.067% 15.582% 12.406% 10.181% 9.734% 9.123% 9.009% 9.002% 9.5% 18.405% 15.927% 12.774% 10.596% 10.168% 9.603% 9.507% 9.501% 10.0% 18.744% 16.275% 13.147% 11.017% 10.608% 10.086% 10.005% 10.001% Table 2: Annuity table

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21.1. Construction Cost Index Zürich (Wohnbaupreisindex)

See the next page.

Source:

https://www.stadt-zuerich.ch/prd/de/index/statistik/themen/bauen- wohnen/wohnbaupreise/zuercher-index-der-wohnbaupreise.html#daten

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Zürcher Index der Wohnbaupreise (ZIW) • Basis August 1957 = 100 Indice zurichois des prix de la construction de logements – base août 1957 = 100 ► Gesamtkosten (Totalindex) und Sondergliederungen/Frais totaux (indice général) et classifications supplémenta

Stichtag Gesamt- Rohbau Innen- Übrige Stichtag Gesamt- Rohbau Innen- Übrige kosten ausbau Kosten kosten ausbau Kosten Date de l'enquête Frais Gros Travaux Autres Date de l'enquête Frais Gros Travaux Autres totaux oeuvre intérieur frais totaux oeuvre intérieur frais

1957 1. August 100.0 100.0 100.0 100.0 1984 1. April 314.8 341.4 284.5 350.9 1958 1. Februar 99.8 99.3 100.1 100.0 1. Oktober 314.7 341.0 285.4 349.7 1. August 101.3 100.7 101.8 101.3 1985 1. April 321.6 347.0 292.0 358.6 1959 1. Februar 101.0 99.8 101.9 101.1 1. Oktober 321.6 347.4 292.2 358.0 1. August 102.5 102.7 102.5 102.2 1986 1. April 331.1 357.9 300.7 368.5 1960 1. Februar 103.2 103.5 103.1 102.8 1. Oktober 332.7 359.6 302.5 369.4 1. August 106.6 108.2 105.5 106.3 1987 1. April 337.6 364.5 309.2 371.8 1961 1. April 111.3 113.7 109.6 110.7 1. Oktober 339.7 366.0 310.3 376.5 1. Oktober 114.5 116.0 113.3 114.4 1988 1. April 352.3 384.1 317.1 391.9 1962 1. April 121.1 124.7 118.4 121.1 1. Oktober 355.3 388.7 318.9 394.8 1. Oktober 122.5 125.7 120.2 122.5 1989 1. April 371.0 402.9 332.6 420.1 1963 1. April 130.3 137.5 124.6 131.4 1. Oktober 376.8 406.1 337.2 435.0 1. Oktober 133.4 140.9 127.5 134.2 1990 1. April 403.1 434.5 355.1 477.8 1964 1. April 139.7 149.8 131.1 143.0 1. Oktober 406.9 436.5 361.0 481.4 1. Oktober 141.9 152.8 132.7 144.6 1991 1. April 427.8 456.5 384.1 501.9 1965 1. April 145.9 158.2 136.0 147.6 1. Oktober 424.0 449.3 384.0 496.8 1. Oktober 146.2 157.2 136.6 148.9 1992 1. April 425.0 446.3 391.0 493.8 1966 1. April 150.1 162.2 140.4 151.8 1. Oktober 410.5 425.5 383.0 477.8 1. Oktober 149.6 160.9 140.3 152.3 1993 1. April 405.6 419.8 383.2 462.9 1967 1. April 151.5 163.6 141.0 155.4 1. Oktober 402.0 419.6 378.3 453.9 1. Oktober 150.5 163.3 139.2 155.0 1994 1. April 400.6 416.0 375.5 460.6 1968 1. April 152.8 165.5 141.2 159.1 1. Oktober 402.1 418.2 376.5 461.7 1. Oktober 152.3 165.1 141.3 155.8 1995 1. April 410.5 414.0 387.8 493.3 1969 1. April 155.6 169.3 143.8 159.7 1. Oktober 406.3 408.2 386.3 486.4 1. Oktober 161.8 178.6 147.5 165.0 1996 1. April 404.4 405.0 384.9 486.2 1970 1. April 175.9 191.6 158.8 192.8 1. Oktober 402.5 403.3 382.7 484.3 1. Oktober 180.4 197.4 161.9 198.0 1997 1. April 397.9 400.5 376.7 478.3 1971 1. April 197.8 220.3 173.4 218.4 1. Oktober 396.8 400.8 374.9 475.2 1. Oktober 200.7 222.9 176.4 222.2 1998 1. April 396.2 401.1 375.7 469.9 1972 1. April 218.8 246.2 190.3 239.6 1. Oktober 395.8 400.8 375.4 468.7 1. Oktober 221.0 246.9 193.6 242.4 1999 1. April 401.2 405.4 379.3 479.8 1973 1. April 240.8 267.9 212.3 263.0 2000 1. April 416.5 426.9 384.9 504.0 1. Oktober 240.8 263.5 215.5 266.0 2001 1. April 436.3 454.1 399.6 520.3 1974 1. April 261.8 284.6 235.3 291.6 2002 1. April 435.9 448.1 406.8 516.6 1. Oktober 257.6 277.0 232.6 293.0 2003 1. April 422.2 428.6 401.1 497.5 1975 1. April 251.4 268.1 226.2 295.8 2004 1. April 426.1 440.3 398.6 498.2 1. Oktober 242.8 261.1 216.8 283.7 2005 1. April 436.4 451.7 408.2 508.3 1976 1. April 235.0 253.5 210.2 271.1 2006 1. April 443.5 463.7 410.4 515.9 1. Oktober 237.3 255.6 213.1 272.5 2007 1. April 463.6 488.0 422.2 544.4 1977 1. April 241.9 262.4 217.2 271.8 2008 1. April 482.1 508.2 432.4 575.1 1. Oktober 246.2 268.5 220.6 275.3 2009 1. April 484.0 505.6 436.4 580.6 1978 1. April 249.5 272.7 225.8 274.4 2010 1. April 489.6 517.1 436.8 585.8 1. Oktober 250.3 273.9 227.8 272.9 2011 1. April 497.6 524.5 444.7 596.1 1979 1. April 257.9 288.1 231.8 277.3 2012 1. April 501.2 527.7 446.9 602.7 1. Oktober 263.2 294.9 235.5 283.5 2013 1. April 498.2 524.8 443.0 600.7 1980 1. April 281.7 316.7 247.1 310.3 2014 1. April 500.6 524.1 445.7 607.4 1. Oktober 286.7 319.6 253.1 317.1 2015 1. April 494.5 516.0 441.7 600.3 1981 1. April 307.1 342.7 266.6 346.7 2016 1. April 485.7 501.2 434.4 597.0 1. Oktober 313.6 342.9 271.6 365.6 2017 1. April … … … … 1982 1. April 328.0 366.2 281.6 375.3 1. Oktober 323.7 362.8 280.6 363.8 1983 1. April 314.7 342.3 281.3 354.6 1. Oktober 313.4 341.2 281.7 350.1 Erläuterungen siehe Seite 2/Voir explications en page 2 > > > > > >

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22. References and further reading

22.1. References

Abay G. (2006) Diskontsatz in Kosten-Nutzen-Analysen im Verkehr (Discount rate in cost benefit analysis in transportation). Forschungsauftrag VSS 2003/201 des Schweizerischen Verbandes der Strassen- und Verkehrsfachleute (VSS). AWA (1999) Abwasserentsorgungsreglement mit Gebührenreglement. Amt für Gewässerschutz und Abfallwirtschaft des Kantons Bern (GSA), Muster 1999, http://www.bve.be.ch/bve/de/index/wasser/wasser/abwasser/organisation_abwasser/organisa tion_dokumente.html (especially: Art. 28 – 32, p. 13ff).

And: Kantonale Gewässerschutzverordnung vom 24. März 1999 des Kantons Bern. http://www.bve.be.ch/bve/de/index/umwelt/umwelt/altlasten/gesetzlichegrundlagen.html (especially: Art 31 – 35).

AWA (2011) Vorgehen zur Bestimmung der Kosten von Abwasserkanälen. Merkblatt des Amt für Wasser und Abfall des Kantons Bern. Download: (last accessed: 20 April 2014)

AWEL (2007) Wegleitung zum finanziellen Führungssystem der Wasserver und Abwasserentsor- gung. Baudirektion Kanton Zürich, Amt für Abfall, Wasser, Energie und Luft, http://www.awel.zh.ch/dam/baudirektion/awel/wasserwirtschaft/finanzierung/fwa/FWA_Wegle itung.pdf.spooler.download.1286096757142.pdf/FWA_Wegleitung.pdf

Access and other information on the website: http://www.awel.zh.ch/internet/baudirektion/awel/de/betriebe_anlagen_baustellen/abwasserr einigungsanlagen/finanzierung.html

De Neufville R. and Scholtes S. (2011) Flexibility in engineering design. MIT Press, Cambridge USA, ISBN: 978-0-262-01623-0 ;ISBN: 978-1-61344-411-5 (ebook).

Dominguez D. (2008) Handling future uncertainty. strategic planning for the infrastructure sector. Doctoral thesis of ETH Zürich, Switzerland. http://dx.doi.org/10.3929/ethz-a-005779391.

Eisenführ F., Weber M. and Langer T. (2010) Rational decision making. 3rd edition. Springer, Berlin, Germany, 447 p., ISBN 978-3-642-02850-2.

FitzGerald D. (2015) Internship report. Part III: on-line survey. Eawag, Dübendorf, Switzerland. Con- tact: [email protected] Fritzsche C., Scholten L. and Maurer M. (2014) Selected Prices and Costs for Swiss Urban Water In- frastructure. Unpublished report for internal use at ETH and Eawag only. Contact: [email protected]. Manning L. J., Graham D. W. and Hall J. W. (2016) Wastewater systems assessment. In: Hall, J.W., Tran, M., Hickford, A.J. and Nicholls, R.J. (eds) The Future of National Infrastructure: A System- of-Systems Approach. Cambridge University Press, Cambridge. ISBN 978-1-107-06602-1. http://dx.doi.org/10.1017/CBO9781107588745

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Hug, T., Dominguez, D. and Maurer, M. (2010) The cost of uncertainty and the value of flexibility in water and wastewater infrastructure planning. In: Proceedings 1st WEF-IWA Cities of the Fu- ture Conference, 7-10 March 2010, Boston, MA, USA.

OECD (2007) Use of Discount Rates in the Estimation of the Costs of Inaction with Respect to Se- lected Environmental Concerns. Report ENV/EPOC/WPNEP(2006)13/FINAL. Download on http://search.oecd.org/officialdocuments/displaydocumentpdf/?cote=ENV/EPOC/WPNEP(200 6)6/FINAL&docLanguage=en (last access: 22.04.2014)

Saltelli A., Karen Chan K. and Scott E.M. (2000) Sensitivity Analysis. John Wiley & Sons, New York, USA, 475 p., ISBN 0-471-99892-3.

Vickrey, W., 1948. Some objections to marginal-cost pricing. Journal of Political Economy 56 (3), 218–238. Vickrey, W., 1987. Marginal- and average-cost pricing. In: Arnott, R., Arrow, K., Atkinson, A., Dre`ze, J. (Eds.), Public Economics: Selected Papers by William Vickrey. Cambridge University Press, Cambridge 1994.

22.2. Further reading

Newnan D.G., Eschenbach T.G. and Lavelle J.P. (2011) Engineering economic analysis. 11th ed, Oxford University Press, USA, p. 672, ISBN-13: 978-0199778126.

Sander T. (2003), Ökonomie der Abwasserbeseitigung – Wirtschaftlicher Betrieb kommunaler Ab- wasseranlagen, Springer-Verlag, Berlin, p. 320, ISBN 3-540-00675-3.

OECD (2011), Meeting the Challenge of Financing Water and Sanitation: Tools and Approaches, OECD Studies on Water, OECD Publishing. DOI: 10.1787/9789264120525-en

Marlow, D., Beale, D. and Gould, S. (2014) Practitioner’s Guide for Economic Decision Making in Asset Management, WERF / IWA, IWAP ISBN: 978-1-78040-528-5/ 1-78040-528-6

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Examples & Exercises

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23. Exercises

23.1. Exercise 1: Interests

Framing: The following examples simply play around with possible interest rates and types in order to highlight a number of key points occurring in the context of an engineering economic analysis.

1. Cost comparison: Repaying a loan

Assume that you have just received CHF 5,000 that you need to pay back within the next five years. The annual interest rate is 8%. Calculate the different payments for the following five repayment schemes and the total payments after 5 years. You can use a spreadsheet or any suitable analytical method to do the calculations.

Plan A: At the end of each year, pay CHF 1000 principal plus interest due

Plan B: Pay the interest due at the end of each year and the principal at the end of 5 years

Plan C: Pay in five equal end-of-year instalments

Plan D: Pay in equal monthly payments due on the last day of each month Plan E: Pay the principal and interest in one single payment at the end of the 5 years

Assuming that the inflation rate is constant 3%, calculate the real costs (in current CHF) of these five payment plans.

2. Inflation

After how many years does the nominal value of a real value double (or after how many years does a burger cost twice as much) with the following inflation rates? a) 3% b) 5% c) 10%

3. Leasing example From a consumer loan advertisement (leasing a car): Conditions: Cash price: CHF 40,000, calculatory salvage value after 4 years: CHF 10,000 Payment: 48 easy leasing rates of CHF 763 You have the option to buy insurance that will pay your rates if you are unable to pay. Additional costs: 48 rates of CHF 59.70

Questions a) What is the effective annual interest rate of the leasing contract? b) What is the effective annual interest rate if you include the insurance premiums?

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23.2. Exercise 2: Cost comparison of alternatives

A small community needs to renew its wastewater treatment plant now. The expected population growth requires more capacity as well as an additional post-filtration unit in the effluent when the load exceeds 106 m3 wastewater per year. The expansion is expected to be needed in 10 years and the filter in 15 years. Assume a nominal interest rate of 5% and an inflation rate of 3%.

Alternative A: Construction of the WWTP with full capacity and a filtration system for CHF 6 million.

Alternative B: Upgrading the current performance of the WWTP for 4 million and realising the expansion in 10 years for 1.6 million, and the filtration system in 15 years for CHF 1 million.

All values used stem from quotes (bids) and are precise within a margin of 10%. They are given in current values.

Questions 1. Compare the two alternatives using the Net Present Value method.

2. Which alternative should be the preferred one? What is the reasoning behind this decision?

3. What will be the outcome of this situation if the expansion is needed within five years and the filter in eight years?

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23.3. Exercise 3: Digester – to build or not to build is the question

Framing: The numbers in this exercise are taken from a realistic example 12 and show that EEA does not always result in a clear outcome. For this specific case the operator decided to build the digester. However, for a very similar case the decision maker decided not to invest in a digester. Can you provide the argumentation for both cases? Additionally, this is a nice example of where the simpli- fied annual cash flow method (or annuity method) can be applied.

Two wastewater treatment plants (WWTP) are merged into a new central unit of 150,000 population equivalents (PE). The decision was facilitated by the fact that both plants were old and would have needed to undergo a complete rehabilitation anyway. It was decided to demolish the existing digester and to utilise the space for the new extended aeration tanks. There is plenty of spare land to build a new digester. The core question is:

Does it really make sense to build a new digester and utilise the gas for energy production? Or should the sludge be brought directly to the incinerator after dewatering?

Tasks

1. Make an engineering economic comparison of the two options.

2. List all the non-economic advantages and disadvantages for both options.

3. Make a suggestion as to which option you would prefer. Give good reasons!

4. How does the interest rate influence the costs? Make a simple sensitivity analysis.

5. What would the costs look like if the current digester was not demolished but rehabilitated? In that case, the investment costs for the buildings and the digester would only be 30% of the former value (CHF 650,000); all other costs would remain the same.

6. Prepare a little presentation in which you show your results and explain your preferred option. Assume that your audience is the Board of Directors that needs to decide on the final plans.

Fundamentals / Data

Option without digester By omitting a primary clarifier, the sludge can be aerobically stabilised so that it can be stacked for several days. An aerobic sludge age (sludge retention time) of 12 to 15 days is sufficient to allow the sludge to be stored for one week without any adverse effects. The dewatering properties are somewhat worse than with digested sludge. A dry matter content of around 24% can typically be achieved. The costs for sludge disposal depend on the water content. For this exercise, the difference in energy consumption for the aeration and for the sludge dewatering process can be ne- glected. All cost values below are given in current values.

12 Bitterli C. und Koch G. (2004) Schlammbehandlung und Ressourcennutzung – Praxis im Kanton Basel-Landschaft. In: Un- terlagen zum VSA-Fortbildungskurs 2004: Abwasserreinigung der Zukunft. VSA, Zürich.

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Sludge disposal costs (24% dry weight, DW): CHF 540/tDW

Transport of sludge (15m3/trip): CHF 150/trip

-3 Specific weight of sludge (density) 1100 kg m

Sludge production per year 2500 tDW/a

Additional costs for stacking tank and biology CHF 170,000 Lifespan of tanks (stacking and aeration tank) 30 a

(Real) interest rate 3%

Option with digester The digester gas or biogas from the mesophilic digester is used in a combined heat and power unit to produce electricity and thermal energy. The dewatering properties of the digested sludge are fairly good and 31% dry weight can typically be achieved. Digesting the sludge decreases the amount (total weight) and volume. Additionally, there is an option to use the digester by adding high carbon waste products (co-digestion). This increases energy production and produces an additional source of income. Some of the electricity produced is consumed again in the nitrification of the digester supernatant, so that only the net electricity production can be used for other purposes. For this exercise, we neglect the difference in costs and energy consumption of the sludge dewatering for the two options. All cost values below are given in current values.

Sludge disposal costs (31% dry weight, DW): CHF 470/tDW

Transport of sludge (15 m3/trip): CHF 150/trip

-3 Specific weight of sludge (density) 1200 kg m

Sludge production per year 2000 tDW/a

Investment costs for buildings and digester CHF 1,950,000

Investment costs for electromechanical gear

(thickener, digester, gas treatment) CHF 2,800,000

All control and electronic equipment

(EMSRL, electronics, instrumentation, control, control engineering) CHF 940,000

Engineering fees and unforeseen costs CHF 1,480,000

Lifespan of buildings 30 a Lifespan of electromechanical gear 15 a

Lifespan of control and electronic equipment 8 a

(Real) interest rate 3% Estimated net earnings from co-digestion CHF 50,000/a

Net electricity production 780,000 kWh/a

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Electricity price CHF 0.1/kWh

Labour costs for operating the digester CHF 50,000/a

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23.4. Exercise 4: Cost estimate for wastewater service

General case: Switzerland

The following data were extracted from a recent survey:

Replacement value, public sewers CHF 55.2·109 Replacement value, wastewater treatment plants (>500 PE) CHF 10.1·109

Annual operating costs, public sewers CHF 287·106/a

Annual operating costs, wastewater treatment plants (>500 PE) CHF 440·106/a Population (P) of Switzerland connected to sewers 7.28·106

Population equivalents (PE) in Switzerland 10.4·106

In 50% of the communities, the tariff for wastewater is CHF 120/P/a or lower. Assume that the communities can obtain loans at an interest rate of 3% (market rate).

Task

What are the average annual costs in Switzerland? How will they change in the future? Give good reasons for your calculations.

Specific case: Small wastewater organisation

The following example concerns a small wastewater organisation in Switzerland. It operates a WWTP for nine communities with 15,000 inhabitants. There are no wastewater-relevant industries in the catchment area. The sewer system and WWTP were formally financed with tax money. The WWTP will be completely rebuilt next year.

The following data were extracted from the yearbook and from the sewer planning documents (GEP):

Wastewater treatment plant, replacement cost CHF 15·106

Wastewater treatment plant, annual operating costs CHF 880,000/a

Public sewers, replacement value CHF 141·106

Public sewers, annual operating costs CHF 107,000/a

Public sewers, rehabilitation needs in the next 10 years (total) CHF 17.5·106 Population (P) 15,000

Until now the average annual wastewater tariff was approx. CHF 70 per person. Neither the WWTP nor the communities have any reserves, and all new investments for rehabilitation and construction need to be funded externally at an interest rate of 3%.

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Task

Assume that the loan for the new WWTP has to be taken out today and that the costs for the sewer rehabilitation are spread out evenly over the next ten years. How do the costs develop over the next ten years for the following four cases?

1. Ignoring the actual evaluated rehabilitation needs, determine the average annualised costs that you would expect for this catchment area (lifespan: 80 years for the sewers and 30 years for the WWTP).

2. Assume a linear depreciation model, with lifespans of 80 years for the sewers and 30 years for the WWTP.

3. Assume 10% annual depreciation on the current book value.

4. Assume that a private company will be taking over and promises to shave 40% off the operating costs while providing the same service level. They have also solved the problem of financing by having their own investors for the rehabilitation and renewals. These investors expect 5% interest on their long-term investments.

Make a graph showing the four trends. How would you present your results to the relevant community leaders?

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23.5. Exercise 5: Effect of depreciation on annual cost

Framing: This example shows how bookkeeping and subsidies can lead to substantial hidden reserves. Together with the silent impact of inflation this leads to distorted views on costs. Once new investments need to be financed, these might lead to unpleasant political discussions.

A town has 10,000 population equivalents (PE), a number that has remained constant during the last 50 years. The following investments were made in the past: Year What Amount

1960 Investment in sewers CHF 16.5 million

Federal subsidies 60%

1970 Investment in WWTP (13,000 PE) CHF 4.0 million

Federal subsidies 40%

2003 Investment in renewing WWTP (13,000 PE) CHF 11.0 million

Federal subsidies 0%

For the conversion of nominal to real values, use the extrapolated Zürich construction cost index (inflation of construction prices) on the next page. No interest rates are paid. The lifespans can be assumed to be 80 years for the sewers and 30 years for the WWTP. To simplify the calculations, it can be assumed that the installations are built within the given year.

The operating costs were CHF 230,000 for the sewer and CHF 522,000 for the WWTP in 2006 and can be assumed to have remained constant since then. The wastewater tariff is based on the operating costs and a 10% degressive depreciation of the book value.

Questions:

1. Calculate the operating costs and the capital costs for the years 1960 to 2010. Plot the real and nominal costs in a graph. 2. Calculate the real costs of the system based on a linear depreciation of the entire wastewater infrastructure (i.e. including the subsidised fraction).

3. After building the new WWTP in 2003, the local council complained that the wastewater tariff had increased by more than a factor of five: “This is unprecedented in the entire histo- ry of the wastewater infrastructure. Why is such an expensive wastewater treatment plant needed?” As the responsible engineer you need to have a good explanation. Is the WWTP plant really so much more expensive? Are the tariffs really unprecedented?

4. The finalised GEP suggests substantial rehabilitation needs. It is expected that a minimum of CHF 12.5 million (2006CHF) will need to be invested in the sewers in the period from 2007 to 2011. No federal subsidies can be expected. Can you predict the future cost trend?

5. Make a graph showing both developments over 50 years. Prepare a 10 minute presentation in which you explain the reasons for this development to the local council.

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Zürcher Index der Wohnbaukosten Statistisches Amt Kanton Zürich Extrapolation 1985-2005 Jahr Basis 54 Rate Jahr Basis 54 Rate 1954 100.0 2006 486.7 1.6% 1955 102.4 2.4% 2007 494.6 1.6% 1956 105.4 2.9% 2008 502.6 1.6% 1957 109.3 3.7% 2009 510.7 1.6% 1958 110.3 0.9% 2010 518.9 1.6% 1959 111.7 1.3% 2011 527.3 1.6% 1960 115.1 3.0% 2012 535.8 1.6% 1961 123.9 7.6% 2013 544.4 1.6% 1962 133.7 7.9% 2014 553.2 1.6% 1963 144.7 8.2% 2015 562.1 1.6% 1964 154.5 6.8% 2016 571.2 1.6% 1965 160.2 3.7% 2017 580.4 1.6% 1966 164.5 2.7% 2018 589.8 1.6% 1967 165.7 0.7% 2019 599.3 1.6% 1968 167.4 1.0% 2020 609.0 1.6% 1969 174.2 4.1% 2021 618.8 1.6% 1970 195.5 12.2% 2022 628.8 1.6% 1971 218.6 11.8% 2023 638.9 1.6% 1972 241.3 10.4% 2024 649.2 1.6% 1973 264.2 9.5% 2025 659.7 1.6% 1974 285.0 7.9% 2026 670.3 1.6% 1975 271.2 -4.8% 2027 681.1 1.6% 1976 259.2 -4.4% 2028 692.1 1.6% 1977 267.8 3.3% 2029 703.3 1.6% 1978 274.3 2.4% 2030 714.6 1.6% 1979 286.0 4.3% 2031 726.2 1.6% 1980 311.9 9.1% 2032 737.9 1.6% 1981 340.6 9.2% 2033 749.8 1.6% 1982 357.6 5.0% 2034 761.9 1.6% 1983 344.7 -3.6% 2035 774.2 1.6% 1984 345.4 0.2% 2036 786.7 1.6% 1985 353.0 2.2% 2037 799.4 1.6% 1986 364.2 3.2% 2038 812.3 1.6% 1987 371.6 2.0% 2039 825.4 1.6% 1988 388.3 4.5% 2040 838.7 1.6% 1989 410.3 5.7% 1990 444.5 8.3% 1991 467.5 5.2% 1992 458.5 -1.9% 1993 443.1 -3.4% 1994 440.5 -0.6% 1995 445.9 1.2% 1996 442.7 -0.7% 1997 436.1 -1.5% 1998 434.6 -0.3% 1999 440.3 1.3% 2000 457.2 3.8% 2001 478.9 4.7% 2002 478.4 -0.1% 2003 463.4 -3.1% 2004 467.7 0.9% 2005 479.0 2.4% mittlere Teuerungsrate seit 1985: 1.61%

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23.6. Exercise 6: Merger of two wastewater treatment plants

Framing: Treatment plants show strong economies of scale (as seen for example in the figures below). This can be used to make rough estimations as to whether a merger could make financial sense. However, be aware that there are large uncertainties involved and that for a specific case the costs might deviate substantially.

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Assumptions for all calculations

• Estimates of the WWTP investment and operating costs are Swiss averages, as given in the figures on the previous page

• (Real) interest rate = 0.03 a-1

• Lifespan of WWTP: 30 years • Lifespan of sewer line: 80 years

-1 -1 • Operating cost of sewer line: 3 CHF a m

• The construction size in population equivalents (PEDim) is 40% greater than the load in pop-

ulation equivalents (PECOD) = PEDim : PECOD = 1.4

1. Specific case: City with two WWTPs

A city with two WWTPs plans to merge these plants. WWTP(1) has a load of PECOD = 130,000 PE;

WWTP(2) has PECOD = 380,000 PE. The interconnection sewer line is 5300 m long. The estimated construction costs of the sewer line are CHF 183∙106. Can the costs of this connecting sewer be jus- tified by economic considerations alone? Use the assumptions given above for your calculations.

A. Do your analysis based on the Swiss average costs provided in the figures. Use the replacement value given above.

B. A careful follow-up project estimates the cost for WWTP(1) to be CHF 215·106. How does the reasoning for task A change in light of this?

2. General case

From an economic point of view, it is attractive to merge WWTPs. For this exercise, we assume that only financial criteria matter and that technical feasibility and political acceptance are given. Then the only remaining factor is the cost of the interconnecting line.

Question: Calculate the distance for the connection of one WWTP to the next one, for which the costs of the interconnection lines are equal to the savings generated by the merger (i.e. the breakeven sewer distance). Use the following approximation for the investment costs of an interconnection line [13]:

0.5 -1 Costs = 11⋅ (PEDim ) [CHF m ]

Make calculations for:

A. Equal sized WWTPs B. WWTP(2) is 10 times larger than WWTP(1)

C. WWTP(2) is 100 times larger than WWTP(1)

[13] Please be aware that in reality the costs can diverge substantially from this assumption. . See Task 1 (specific case) above where the specific costs for the interconnection line were CHF 34,500/m, a factor of eight higher than the value given in this formula.

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D. Compare your results with the following figure:

Figure 14: Copied from from Manning et al (2016).

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23.7. Exercise 7: SNPV

Framing: Here we give an example of how the additional costs for staging the expansion of a WWTP can be estimated based on different growth scenarios. Please observe that an option might have a larger NPV (= higher absolute costs) but a smaller SNPV (smaller specific costs per provided service unit) when compared with another option.

A WWTP needs to be rehabilitated. The electromechanical elements are failing and the safety and security of the plant need to be updated. The plant is too small and the project needs to be execut- ed as quickly as possible.

The catch for you as the planning engineer is that in five years’ time the military will surrender a large patch of grassland. It is the explicit intention of the town to develop it into a new neighbour- hood, as this will be the only opportunity for the town to develop. Due to its close proximity to the space-deprived city and the airport, this new area will attract people and businesses as soon as it becomes accessible.

Here’s what the expected wastewater load (in population equivalent PE) for the WWTP is expected to look like:

Year Load (PE) 0 20,000 5 20,000 10 25,000 15 30,000 20 35,000 30 35,000

Three different options are currently being discussed:

1. Build a WWTP with 35,000 PE capacity 2. Build for 27,500 PE and then expand to 35,000 PE

3. Build for 25,000 PE and then expand in two stages to 30,000 and then to 35,000 PE. Assume that no time is needed for construction and planning, and that the specific operating costs

(CHF/PE/a) are constant and the same for all three options. Tip: In the SPNV formula, PDim can be set to 35,000 PE for all three options.

Question: How much can you invest in the staging of options 2 and 3?

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23.8. Exercise 8: Feasibility studies

Framing: This typical task for consulting engineers highlights the impact of the discount rate in an EEA. The higher the discount rate, the stronger the options in the future are favoured. See also Fig- ure 1 (p - 12 -).

A region with three wastewater treatment plants A, B, and C is thinking about the future options to cooperate. You are commissioned to make an engineering economic analysis of the three different options for how to proceed into the future:

• Option 1: Status quo; i.e. all three WWTPs are kept separate.

• Option 2: Merge WWTP A and B by connecting B to A, and keep WWTP C separate. • Option 3: Merge all three together at location A. Questions: 1. Compare the three options in an engineering economic analysis. Which one is the cheapest? By how much? 2. Explain (quantitatively) how the numbers and the ranking change if a higher or no discount rate is considered. 3. Compare Option 1 and Option 3 and discuss the advantages and disadvantages of the two options. Give at least three pros and cons.

Background information: What Unit WWTP A WWTP B WWTP C Cost to rebuild CHF 33'000'000 14‘000'000 17'000'000 Remaining useful life a 0 8 12 Book value CHF 0 2'000'000 3'750'000 Annual O&M costs CHF/a 1'280'000 870'000 377'000

What Unit WWTP A+B WWTP A+B+C Cost to build CHF 40‘000'000 57'000'000 Annual O&M costs CHF/a 2'175'000 2'170'000 Cost of interconnection line CHF 2'750'000 4’000’000

All values are given in real 2011 CHF. You are requested to use a discount rate of 4 %/a. Currently there are no savings or debts. Inflation is assumed to be constant at 2.3 %/a.

Hint: You might want to consider using the NPV method. Use a reasonable planning horizon of more than 30 years and assume that you can sell the infrastructure at the end of the planning horizon.

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23.9. Exercise 9: Feasibility studies

Framing: This exercise gives an example, specifically in the context of an engineering economic analysis, of how a monetary comparison can be influenced if financing is considered. See also the blue box in section 7.1 (p. - 18 -). A pure EEA would use the investment costs with standard discount rates, indicating which of the options would objectively rank better. However, from a business perspective, including cheaper financing option (e.g. subsidies) might make more sense. Please be aware that somebody is still paying for subsidies and therefore this is mainly an issue of setting the system boundaries of the analysis differently.

A WWTP with a capacity of 100‘000 PE has to implement a further treatment step to eliminate mi- cro-pollutants. The operator and the authorities want to consider two different treatment options: ozonation and pulverized activated carbon (PAC). You have been put in charge of carrying out an economic comparison of the two options. Both systems have an average expected lifespan of 25 years. Assume a discount rate of 4%.

To estimate the investment costs and the annual expenses for operation and maintenance, the following cost estimations are available:

Investment costs Operation and maintenance

6 6 -1 [10 ·CHF] [10 ·CHF a ]

Ozonation 14.0 2.0

PAC 9.0 2.2

Questions:

1. How do the two options compare in an engineering economic analysis? 2. How does the picture change if you include financing (subsidies)? Assume that 75% of the investment costs are covered by federal subsidies and that the real interest rate you get is 4%. Neglect inflation. 3. What happens to the result in task 2 if the interest rate changes?

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23.10. Exercise 10: Cost calculations (tariffs)

Framing: This example highlights the key differences between the operator (‘Bern’) and bookkeeper (‘Zürich’) models (see also Chapter 14 on page - 34 -) and their impact on cost and ultimately on customer tariffs. The main difference between these models is the way in which they treat hidden assets. Significant cost differences can result from the two models if substantial hidden assets are present, e.g. due to previous degressive depreciation or large subsidies. The exercise is taken from a previous exam question.

You have been given the task of doing the rehabilitation planning for a small municipality. The given planning horizon is 20 years. The entire sewer system has 2’000 sewer pipes and was built in 1985. So far, nothing has been rehabilitated. Rehabilitating a sewer pipe costs on average 150’000 CHF and operating costs are 120’000 CHF a-1; these cost values are referenced to the current year of 2017. The regulation states that in the future all depreciation needs to be linear, based on lifespan.

Assume that based on a previous investigation, the time T from construction until a sewer pipe reaches the worst condition class is Gamma distributed:

~ ( = 0.8, = 100)

𝑇𝑇 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝛼𝛼 𝛽𝛽 The following figure depicts the same distribution.

1.00

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.20 Cumulative distribution function distribution Cumulative 0.10

0.00 0 10 20 30 40 50 60 70 80 90 100 T (age at which a sewer pipe reaches the worst condition class) [years]

Figure 15: Cumulative distribution function (CDF) of the Gamma distribution with the parameters =0.8 and =100. α β

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Hint: Excel provides the function GAMMA.DIST to compute the cumulative distribution function. For example, GAMMA.DIST(10;0.8;100;TRUE) calculates the cumulative distribution at the age of 10 years for the parameters given above.

Questions:

1. What is the replacement value of the sewer infrastructure? 2. Assuming that the real costs for building sewers stayed constant, what were the original construction costs? 3. What is the current book value, assuming 10% degressive depreciation per year? 4. How do the nominal rehabilitation costs develop over the planning horizon? Assume that the rehabilitation need of the past is evenly distributed over the entire planning horizon. Assume a reasonable inflation rate and that the newly rehabilitated sewers will not break during the planning horizon. 5. Calculate the total costs that have to be covered by tariffs for the given planning horizon. For the investment costs use the rehabilitation costs calculated in task 4. All new investments are financed by loans with a nominal interest rate of 0.04 a-1. Perform your calculations for each of the following two models: a) Start with the ‘Bern Model’ (Operator Model), which is slightly simpler. This model uses the replacement value to calculate depreciation based on the lifespan. This depreciation is used to pay back loans or is deposited into the account “Spezi- alfinanzierung”. Note that for this model it is more convenient to calculate with real values. b) Now use the ‘Zürich Model’ (Bookkeeper Model), which is a little more demanding. You need to calculate the depreciation for (i) the existing infrastructure based on its current book value (calculated in task 3) and remaining lifespan and (ii) the new investments based on the full lifespan. For this model it is more convenient to calculate with nominal values. 6. Compare the costs (that have to be covered by tariffs) from the two models.

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