In its simplest form an investment decision can be defined as one which involves a firm making a cash outlay with the aim of receiving future cash inflows

INVESTMENT APPRAISAL

In its simplest form an investment decision can be defined as one which involves a firm making a cash outlay with the aim of receiving future cash inflows. Variations on this definition are possible, for example, investing in energy efficiency measures might require a cash outlay today in order to reduce future spending on energy bills, but all investment decisions can generally be adapted to the initial definition.

Decisions about buying equipment, retrofitting a building, instituting staff training or building a new factory are all examples of investment decisions that may be made by companies. Evidence to inform the decision about whether to make such investments requires a robust method of appraisal which can be applied consistently to a spectrum of investment decisions.

It is important to remember that no investment appraisal method can give a definitive decision. It is sometimes easy to believe that analysis techniques can make decisions for us. This is never true -their job is to provide clarity to help the decision maker towards the best choice but it must be remembered that all decisions about future cash flows are made in the face of uncertainty. Any investment appraisal will always comprise an element of forecasting, estimating or guesswork about an investment’s future performance.

THE PAYBACK METHOD

Probably the simplest appraisal method, payback calculates how quickly the incremental benefits of an investment project ‘pay back’ the initial capital invested. The payback method can be used as a guide to investment decision making in two ways. When faced with a straight ‘accept or reject’ decision, it can provide a rule whereby projects are accepted only if they pay back the initial investment within a certain time frame. In addition, it can allow a comparison to be made between several competing, mutually exclusive investments. In such cases projects can be ranked according to speed of payback, with the project with the fastest payback being most favoured.

Example 1: Project A

Project A is a £4,000 investment which is predicted to generate cash income of

£1,000 in Years 1 and 2, £2,000 in Year 3, £3,000 in Year 4 and £1,000 in Year 5. The company criterion for acceptance is a maximum four year payback. As shown in the table below, this project should be accepted because it pays back the initial outlay within three years;

Example 2: Projects B and C

Cash flows for Projects B and C are mutually exclusive and are shown in the table below.

Project B pays back within 3 years (actually 2 and three-fifths years). Project C has the faster payback and is therefore the preferred investment.

TIME VALUE OF MONEY

Many engineering projects evolve over long periods of time. Costs incurred at the outset may generate benefits for years or even decades to come. The evaluation of whether new projects are worthwhile must therefore compare benefits and costs which occur at different times. The essential problem with this comparison is that money has a time value. A euro, pound or dollar today is not the same as a euro, pound or dollar a year from now. The money will have the same face value but will not have the same buying power as it has today. Because of this difference, we cannot estimate benefits or costs by simply comparing cash amounts that are realized at different times. In order to make a valid comparison we need to translate all cash flows into comparable quantities. Different quantities of money at different points in time are said to be equivalent when they are directly comparable with each other at a given point in time for a given interest rate.

The solution to the equivalence problem is mathematically relatively straightforward. It requires only a few formulae that depend on just two variables; the timespan (or duration) of the project, t; and the interest rate, i. However, from a practical point of view, mathematical solutions should be interpreted with care. Analytical solutions generated by the formulae are entirely dependent on the two variables, which are only rarely known with certainty. For project lifetimes which last more than a couple of years, the sensitivity of the results to changes in i or t mean that investments in high- capital projects requires a combination of analytical and strategic thinking.

INVESTING FOR A SINGLE PERIOD

Suppose you invest £100 in a savings account that pays 10 percent interest per year. In one year you will have £110.

This is equal to your original principal of £100 plus £10 in interest that you earn. £110 is the future value of £100 invested for one year at 10 percent, and it simply means that £100 today is worth £110 in one year, given a 10% interest rate.

In general, if you invest for one period at an interest rate of i, your investment will grow to (1 + i) per £ invested. In our example, i is 10 percent, so your investment grows to 1 + 0.1 = £1.1 per £ invested.

You invested £100 in this case, so you ended up with £100 x 1.1 = £110.

INVESTING FOR MORE THAN ONE PERIOD

Going back to our original £100 investment, what will you have after two years, assuming the interest rate doesn’t change?

If you leave the entire £110 in the bank, you will earn £110 x 0.1 = £11 in interest during the second year, so you will have a total of £110 + £ 11 = £121.

£121 is therefore the future value of £100 in two years at 10 percent. Another way of looking at it is that one year from now you will be investing £110 at 10 percent for a year. This is a single-period problem, so you’ll end up with £1.10 for every pound invested, or £110 x 1.1 = £121 total.

This £121 has four parts.

• The first part is the £100 original principal.
• The second part is the £10 in interest you earned in the first year.
• The third part is another £10 you earn in the second year, for a total of

£120.

• The last £1 is interest you earn in the second year on the interest paid in the first year: £10 x 0.1 = £1.

The process of leaving money and any accumulated interest in an investment for more than one period, thereby reinvesting the interest, is called compounding. Compounding the interest means earning interest on interest. With simple interest, the interest is not reinvested, so interest is earned each period only on the original principal. 

Now go back to how our original £100 and how we calculated the £121 future value.

We multiplied £110 by 1.1 to get £121. The £110, however, was £100 also multiplied by 1.1. In other words:

£121 = £110 x 1.1

= (£100 x 1.1) x 1.1

= £100 x (1.1 x 1.1)

= £100 x 1.1²

= £100 x 1.21

How much would £100 grow to after three years? Once again, in two years, we’ll be investing £121 for a one year period at 10 percent.

We’ll end up with £1.10 for every pound invested, or £121 x 1.1 = £133.10 total. 

This £133.10 is thus: 

£133.10 = £121 x 1.1

= (£110 x 1.1) x 1.1

= (£100 x 1.1) x 1.1 x 1.1

= £100 x (1.1 x 1.1 x 1.1)

= £100 x 1.1³

= £100 x 1.331

You’re probably noticing a pattern to these calculations, so we can now go ahead and state the general result. As our examples suggest, the future value of £1 invested for t periods at a rate of i per period is this: 

The expression (1 + i)^t is sometimes called the future value interest factor (or just future value factor ) for £1 invested at i percent for t periods and can be abbreviated as FVIF( i, t ).

In our example, what would £100 be worth after five years? We can first compute the relevant future value factor as follows:

(1 + i )^t = (1 + 0.1)⁵ = 1.1⁵ = 1.6105

£100 will thus grow to:

£100 x 1.6105 = £161.05

The growth of £100 each year is illustrated in the table below. As shown, the interest earned in each year is equal to the initial amount multiplied by the interest rate of 10 percent.

Total simple interest = £50

Total compound interest = £11.05 Total interest = £61.05

Conversely, if an amount of money is borrowed, the interest is subtracted. In both cases the amount of money that was borrowed or invested has increased with time.

TIME EQUIVALENCE

The term equivalence is based on the time value of money and interest rates. Equivalence implies that different amounts of money at different points in time are equal in terms of their economic value. For example, if a simple rate of 10% per annum is assumed, £1,000 today is equivalent to £1100 in a year’s time. It means that £1000 today has exactly the same economic value as £1100 one year from today.

If a sum of £1000 was lost today and discovered again five years later, its face value would have remained the same but some of its earning power would have been lost; in contrast, £1000 deposited at 10% compounded annually over the same period would have a value of;

£1000 x (1 + 0.1)⁵ = £1610.51 

Therefore £1000 today is equivalent to £1610.51 in five years’ time, assuming an interest rate of 10% compounded annually. An investor would find £1000 paid now and £1610.51 in five years’ time equally attractive.

Equivalence is central to the economic analysis of engineering projects as it permits cash flows, or different amounts of money accruing at different points in time, to be compared at one common discrete time location. 

DISCOUNTED CASH FLOW ANALYSIS

The translation of future money to its present value is called discounted cash flow (DCF) analysis and the interest rate used is called the discount rate (R). The discount rate in DCF analysis differs from the interest rate (i) in that it takes into account not just the time value of money, but also the risk or uncertainty of future cash flows; the greater the uncertainty of future cash flows, the higher the discount rate. We can use the discount rate R to calculate the present value of income as follows;

For N equal annual income amounts (A), the present value V_p is given by;

NET PRESENT VALUE

Discounting is particularly important when revenue is expected over many years, as in the case of a renewable energy scheme. The following example considers a simple case for a wind turbine (ignoring operating and maintenance costs), and only takes into account the capital cost of manufacture and installation and the annual revenue.

Calculate the present value of the revenue generated over a 30 year lifetime from a 3 MW wind turbine. The capital cost of the turbine is £2 million and the discount rate is 6%. The cost of electricity charged by the turbine operator is 4p/kWh. The capacity factor of the turbine is 0.25.

Capacity factor (CF) is a measure of the actual output achieved by any generator over a year, divided by the maximum theoretical output if it operated constantly at its maximum capacity. For example, a gas turbine which is used mainly for system reserve and is only switched on a few times a year might have a CF of less than 0.01 (1%); while a large coal-fired steam plant which runs at near to its maximum output for long periods of time could have a CF of more than 0.8 (80%). For renewable generators like wind turbines or solar PV, their energy output is dictated by the availability of the wind or sun. For onshore wind turbines, a typical CF is 0.25 – 0.35 (25% - 35%).

In this example, at a CF of 0.25 the energy (E) produced per year will be;

E = 24 (hours per day) x 365 (days per year) x 0.25 (CF) x 3 (max output in MW) E = 6.57 x 10³ MWh (6.57 x 10⁶ kWh)

At the price of 4p/kWh, revenue per year (A) is; A = £0.04 x 6.57x10⁶ = £262,800

Substituting in our equation for present value with N = 30 gives the present value V_p of the total revenue as;

V_p = 262,800 [1-(1.06)^-30]/0.06 V_p = £3,617,398

Note that this figure is less than half of what we would have calculated had we neglected the effect of the interest rate