Financial methods in environmental decisions
Financial methods in environmental decisions

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Financial methods in environmental decisions

Discounted cash flow methods

Rate of return and payback period calculations, described in the previous subsections, provide a good starting point for appraising an investment, but they ignore what is often termed the time value of money.

If you were offered the choice of accepting £100 now or in one year’s time, you would almost certainly take the money now to spend and enjoy the proceeds or perhaps invest and earn some interest. In contrast, if the choice were between £100 now and £200 in a year’s time you might well decide that it is better to wait and get the higher sum – assuming that you are confident the offer will still stand in a year and that you don’t have an urgent need for the money now. But if the choice were between £100 now and £110 or maybe £120 in a year’s time, the situation is less clear-cut.

Organisations have to take decisions like this one all the time and one way of making the decision is to look at the interest rate that could be earned on the money. Suppose that the interest rate on a savings account offered by a reliable and trustworthy bank is 3%.

Investing the £100 for one year would give

equation left hand side 100 multiplication open one plus three divided by 100 close equals right hand side prefix pound of 103

If the £103 is reinvested for a further year the result would be

equation left hand side 103 multiplication open one plus three divided by 100 close equals right hand side prefix pound of 106.09

You may recognise this processing as compounding. Calculating compound interest has the following general formula

total sum equation left hand side equals right hand side cap p multiplication open one plus r divided by 100 close super n

where

P = the initial amount invested (the principal)

r = the percentage interest rate

n = the number of years.

Or in financial language you can say that if the interest rate is 3% the future value of £100 in one year is £103 and in two years is approximately £106, and so on.

The converse of this is known as the present value (PV) and, rearranging the above formula, the present value of £P in n years’ time is given by

equation left hand side cap p times cap v equals right hand side cap p multiplication open one plus r divided by 100 close super negative n

Activity 6 Present value

Timing: Allow 15 minutes to complete this activity

What is the present value of £215 in three years’ time if the interest rate is 5%?

Answer

Using the above equation for the present value

multiline equation line 1 equation left hand side cap p times cap v equals right hand side 215 multiplication open one plus five divided by 100 close super negative three line 2 equation left hand side equals right hand side 215 multiplication 1.05 super negative three line 3 equation left hand side equals right hand side 215 multiplication 0.8638 line 4 equation left hand side equals right hand side prefix pound of 185.73

Table 3 shows the present value of £1 receivable at the end of each year for periods ranging from 0 to 30 years and interest rates ranging from 0 to 10%. If you have access to a spreadsheet package, you might find it interesting to recreate this table yourself.

Table 3 Present values of £1

Time(years)Interest rate (%)
12345678910
10.9900.9800.9710.9620.9520.9430.9350.9260.9170.909
20.9800.9610.9430.9250.9070.8900.8730.8570.8420.826
30.9710.9420.9150.8890.8640.8400.8160.7940.7720.751
40.9610.9240.8880.8550.8230.7920.7630.7350.7080.683
50.9510.9060.8630.8220.7840.7470.7130.6810.6500.621
60.9420.8880.8370.7900.7460.7050.6660.6300.5960.564
70.9330.8710.8130.7600.7110.6650.6230.5830.5470.513
80.9230.8530.7890.7310.6770.6270.5820.5400.5020.467
90.9140.8370.7660.7030.6450.5920.5440.5000.4600.424
100.9050.8200.7440.6760.6140.5580.5080.4630.4220.386
110.8960.8040.7220.6500.5850.5270.4750.4290.3880.350
120.8870.7880.7010.6250.5570.4970.4440.3970.3560.319
130.8790.7730.6810.6010.5300.4690.4150.3680.3260.290
140.8700.7580.6610.5770.5050.4420.3880.3400.2990.263
150.8610.7430.6420.5550.4810.4170.3620.3150.2750.239
160.8530.7280.6230.5340.4580.3940.3390.2920.2520.218
170.8440.7140.6050.5130.4360.3710.3170.2700.2310.198
180.8360.7000.5870.4940.4160.3500.2960.2500.2120.180
190.8280.6860.5700.4750.3960.3310.2770.2320.1940.164
200.8200.6730.5540.4560.3770.3120.2580.2150.1780.149
210.8110.6600.5380.4390.3590.2940.2420.1990.1640.135
220.8030.6470.5220.4220.3420.2780.2260.1840.1500.123
230.7950.6340.5070.4060.3260.2620.2110.1700.1380.112
240.7880.6220.4920.3900.3100.2470.1970.1580.1260.102
250.7800.6100.4780.3750.2950.2330.1840.1460.1160.092
260.7720.5980.4640.3610.2810.2200.1720.1350.1060.084
270.7640.5860.4500.3470.2680.2070.1610.1250.0980.076
280.7570.5740.4370.3330.2550.1960.1500.1160.0900.069
290.7490.5630.4240.3210.2430.1850.1410.1070.0820.063
300.7420.5520.4120.3080.2310.1740.1310.0990.0750.057

The principles of discounted cash flow (DCF) allow possible investments to be reviewed by using the table of present values (Table 3), or discount factors as they are also known, to determine the time in a project’s life when payments are made and when income is earned. The term discount rates is used rather than interest rates. You can think of the discount rate as being the rate of return or profit that a project will make. In general you can assume that a project will not go ahead if investing in it, which carries some risk, is estimated to give a lower rate of return than investing the money in a secure bank account.

The use of discounted cash flow is best illustrated by means of a worked example, as demonstrated in Activity 7.

Activity 7 Discounted cash flow

Timing: Allow 25 minutes to complete this activity

A landfill site operator is considering installing an engine to generate power by burning landfill gas. It is estimated that the capital cost of the engine, generator, control equipment and connections to the grid will be £1000 000. The annual staffing, operating and maintenance costs will be £300 000. The annual income from the sale of power will be £315 000 in the first year and £630 000 in each following year. The capital costs have to be paid at the start of the project and the total project life is expected to be 5 years. If company policy states that a rate of return of 5% must be achieved, is the project worth pursuing?

Answer

The first stage in the process involves drawing up a cash flow statement. In the example, this is relatively straightforward. However you should appreciate that because I have said that the capital investment takes place at the start of the project, its present and future values are the same. This is indicated by placing this expenditure in year 0 – more complex projects will involve staged payments over more than one year. I have also assumed that the income starts to arrive in year 1.

The cash flow statement is shown in Table 4.

Table 4 Cash flow statement (all money values in £000s)

Year012345Total
Capital cost–1000–1000
Operating cost–300–300–300–300–300–1500
Income3156306306306302835
Net cash flowTotal –1000Total 15Total 330Total 330Total 330Total 330Total 335

The next stage is to adjust the net cash flow in each year by multiplying it by the appropriate discount factor. This is shown in Table 5 for a discount rate of 5% (with the values taken from Table 3).

Table 5 Discounted cash flow (all money values in £000s)

Year012345Total
Capital cost–1000–1000
Operating cost–300–300–300–300–300–1500
Income3156306306306302835
Net cash flow–100015330330330330335
Discount factor1.000.9520.9070.8640.8230.784
Discounted cash flow–100014.28299.31285.12271.59258.72129.02

The final cell in the right-hand column (£129 020) represents the net present value (NPV) of the project. This is positive, so the project has met the target rate of return of 5%.

It is often useful to calculate the rate of return of a project, which is represented by the discount rate that achieves an NPV of zero. Again, this is something you might want to try if you have access to a spreadsheet. If you do so, you will find that the NPVs of 8% and 9% are £25 928 and –£5406 respectively, so the rate of return is somewhere between 8% and 9%. If you use a ‘goal seek’ function on a spreadsheet, you will find that the rate of return is 8.82%.

In principle all DCF calculations can be performed in this way, but it can get long-winded for projects with long lifetimes, such as a reservoir or road. If it is clear that a cash flow will occur over a long period, it is possible to sum the discount factors and multiply this sum by the cash flow.

Example 1

What is the present value of £300 000 received annually for a period of 5 years if the discount rate is 4%? Use Table 3 to look up the discount factors.

multiline equation line 1 cap p times cap v equals 300 zero width space 000 postfix multiplication left parenthesis sum with, 5 , summands 0.962 plus 0.925 plus 0.889 plus 0.855 plus 0.822 right parenthesis times line 2 equals 300 000 multiplication 4.453 line 3 equation left hand side equals right hand side prefix pound of one 335 900

Table 6 shows the present value of £1 received annually for different time periods.

Table 6 Present value of £1 received annually

Time (years)Interest rate (%)
12345678910
10.9900.9800.9710.9620.9520.9430.9350.9260.9170.909
21.9701.9421.9131.8861.8591.8331.8081.7831.7591.736
32.9412.8842.8292.7752.7232.6732.6242.5772.5312.487
43.9023.8083.7173.6303.5463.4653.3873.3123.2403.170
54.8534.7134.5804.4524.3294.2124.1003.9933.8903.791
65.7955.6015.4175.2425.0764.9174.7674.6234.4864.355
76.7286.4726.2306.0025.7865.5825.3895.2065.0334.868
87.6527.3257.0206.7336.4636.2105.9715.7475.5355.335
98.5668.1627.7867.4357.1086.8026.5156.2475.9955.759
109.4718.9838.5308.1117.7227.3607.0246.7106.4186.145
1110.3689.7879.2538.7608.3067.8877.4997.1396.8056.495
1211.25510.5759.9549.3858.8638.3847.9437.5367.1616.814
1312.13411.34810.6359.9869.3948.8538.3587.9047.4877.103
1413.00412.10611.29610.5639.8999.2958.7458.2447.7867.367
1513.86512.84911.93811.11810.3809.7129.1088.5598.0617.606
1614.71813.57812.56111.65210.83810.1069.4478.8518.3137.824
1715.56214.29213.16612.16611.27410.4779.7639.1228.5448.022
1816.39814.99213.75412.65911.69010.82810.0599.3728.7568.201
1917.22615.67814.32413.13412.08511.15810.3369.6048.9508.365
2018.04616.35114.88713.59012.46211.47010.5949.8189.1298.514
2118.85717.01115.41514.02912.82111.76410.83610.0179.2928.649
2219.66017.65815.93714.45113.16312.04211.06110.2019.4428.772
2320.45618.29216.44414.85713.48912.30311.27210.3719.5808.883
2421.24318.91416.93615.24713.79912.55011.46910.5299.7078.985
2522.02319.52317.41315.62214.09412.78311.65410.6759.8239.077
3025.80822.39619.60017.29215.37213.76512.40911.25810.2749.427
3529.40924.99921.48718.66516.37414.49812.94811.65510.5679.644
4032.83527.35523.11519.79317.15915.04613.33211.92510.7579.779

Example 2

A water supply company is proposing to construct a reservoir on a river at point X at a cost of £5.86 million. No further increment in water storage would be needed for 29 years. It has been suggested that three smaller reservoirs costing £3.25 million, £3.60 million and £4.33 million would be financially and environmentally preferable. However this scheme would require the construction of the second reservoir after 13 years and the third 9 years later. Investigate the financial aspects of the two alternatives for discount rates of 6% and 8%. Assume that the cash to build the first reservoir is required in year 0, the cash for the second is required in year 13, and in year 22 for the third reservoir.

Answer

The capital costs of the first scheme are all met at the present time, so the present value is £5.86 million for both rates of return.

In the case of the second scheme at a rate of 6% the present value is given by

multiline equation line 1 cap p times cap v of first reservoir equals one multiplication three full stop 25 prefix multiplication of 10 super six line 2 equation left hand side equals right hand side pound three full stop 25 million
multiline equation line 1 cap p times cap v of second reservoir equals zero full stop 469 multiplication three full stop six prefix multiplication of 10 super six line 2 equation left hand side equals right hand side pound one full stop 69 million
multiline equation line 1 cap p times cap v of third reservoir equals zero full stop 278 multiplication four full stop 33 prefix multiplication of 10 super six line 2 equation left hand side equals right hand side pound one full stop two zero million

giving the total as

equation left hand side cap p times cap v equals right hand side three full stop 25 plus one full stop 69 plus one full stop two zero equals pound six full stop 14 million

Therefore, the one-reservoir scheme is the more financially attractive.

At a rate of 8% the PV of the second scheme is given by

multiline equation line 1 cap p times cap v of first reservoir equals one multiplication three full stop 25 prefix multiplication of 10 super six line 2 equation left hand side equals right hand side pound three full stop 25 million
multiline equation line 1 cap p times cap v of second reservoir equals zero full stop 368 multiplication three full stop six prefix multiplication of 10 super six line 2 equation left hand side equals right hand side pound one full stop 33 million
multiline equation line 1 cap p times cap v of third reservoir equals zero full stop 184 multiplication four full stop 33 prefix multiplication of 10 super six line 2 equation left hand side equals right hand side prefix pound of zero full stop eight zero million

giving the total as

equation left hand side cap p times cap v equals right hand side three full stop 25 plus one full stop 33 prefix plus of zero full stop eight equation left hand side equals right hand side pound five full stop 36 million full stop

At 8% the three-reservoir alternative is preferable, even though the total capital cost is almost twice as great as the single-reservoir scheme. Looked at in another way, if the difference between the cost of the larger and the first-stage reservoir (£5.86m – £3.25m = £2.61m) were to be invested at 8% (say), by the time the second reservoir had to be built the amount available (£6.58m) would be more than sufficient to meet the cost of £3.6m, and the accrued surplus again invested would likewise be more than sufficient to meet the cost of the last reservoir (in fact £5.96m would be available).

T867_1

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