ENG2000 Chapter 17 Evaluating and Comparing Projects: The IRR ENG2000: R.I. Hornsey CM_2: 1
Introduction This chapter introduces a second method for comparing between projects While the result of the process is the same as those already considered, the circumstances determine which method is appropriate The technique we shall discuss now is known as the internal rate of return, IRR ENG2000: R.I. Hornsey CM_2: 2
Internal rate of return (IRR) A major reason for investing in a project is that it will generate future earnings And one way to express this return is as a rate of return per dollar invested which is effectively an interest rate This interest rate is the IRR the word internal indicates that the interest rate is dependent on internal factors, such as the cash flows The IRR is the interest rate at which a project just breaks even ENG2000: R.I. Hornsey CM_2: 3
Example 5.1 If $100 is invested in a project that returns $110 one year later, we can calculate the IRR by determining the interest rate that would give the same benefits where i* is the IRR here, i* = 10% P = F( P / F,i *, N) $100 = $110( P / F,i *,1) $100 = $110 1+i * The IRR is thus the interest rate that makes the present worth of the benefits equal to the first costs ENG2000: R.I. Hornsey CM_2: 4
Formal definition of IRR The internal rate of return on an investment is that interest rate, i*, such that, when all cash flows related to the project are discounted at i*, the present worth of the cash inflows equals the present worth of the outflows causing the project to just break even where R t = cash inflow (receipts) in period t D t = cash outflow (disbursements) in period t T = number of time periods i* = IRR T ( ) R t D t = 0 1+ i * t=0 ( ) t ENG2000: R.I. Hornsey CM_2: 5
We can rearrange the previous equation: T R t 1+ i * = D t 1+ i * t=0 ( ) t so that the IRR an be calculated by setting disbursements = receipts and solving for i* We must ensure that the disbursements and receipts are equivalently expressed as a present worth, annual series or future worth i.e. PW or AW or FW(disbursements) = PW or AW or FW(receipts) IRR is supposed to be positive; if not, the project loses money! t=0 ( ) t Typically IRR is calculated by spreadsheet T ENG2000: R.I. Hornsey CM_2: 6
Interest rate tables In the absence of a computer, interest rate tables are used to determine the values of the various compound worth factors for various interest rates linear interpolation can then be used to find intermediate values Some web pages also present these tables: http://www.pearsoned.ca/park/downloads_3.html http://www.uic.edu/classes/ie/ie201/discretecompoundintere sttables.html also posted on the course web page ENG2000: R.I. Hornsey CM_2: 7
Example for 4% ENG2000: R.I. Hornsey CM_2: 8
Example 5.2 You are interested in buying a tuxedo; the cost is $500 but you expect it to save $160 per year in rental charges over its 5-year life. What is the IRR? what does this calculation neglect? ENG2000: R.I. Hornsey CM_2: 9
Example 5.2 solution ENG2000: R.I. Hornsey CM_2: 10
Example 5.3 The High Society Bean Company is considering a new canner cost: $120 000 life: 10 years scrap value: $5 000 extra sales: $15 000 in first year, increasing by $5 000 each subsequent year extra costs incurred: $10 000 per year MARR: 12% Should they invest in the new canner? ENG2000: R.I. Hornsey CM_2: 11
Using the IRR As before, we need to consider the relationships, if any, between projects to be compared For independent projects, the process is similar to the MARR calculations we did before there, we wanted a MARR = 0 for marginal acceptability now, we want IRR = MARR for marginal acceptability (IRR > MARR is better) again, the projects must have equal lives for this to be valid ENG2000: R.I. Hornsey CM_2: 12
Example 5.3 solution ENG2000: R.I. Hornsey CM_2: 13
Mutually exclusive projects Life becomes a little more tricky when it comes to deciding between mutually exclusive projects see the next two examples If your MARR = 70%, is it better to invest $1 today and get back $2 in a year or to invest $1000 today to get $1900 in a year? Calculate the IRRs: $1+ $2 P /F,i *,1 ( ) = 0 ( P /F,i *,1) = $1 $2 = 0.5 $1000 + $1900( P /F,i *,1) = 0 ( P /F,i *,1) = $1000 $1900 = 0.52631 i * =100% i * = 90% ( P /F,i,N ) = 1 1+ i ENG2000: R.I. Hornsey CM_2: 14
So we choose the first option because it has the higher IRR Not necessarily because it is not the second option alone that must be evaluated, but the incremental investment and return between the two options i.e. is it worth investing the additional $999 required by option 2? ( $1000 $1) + ( $1900 $2) ( P /F,i *,1) = 0 ( P /F,i *,1) = $999 $1898 i * = 89.98% ENG2000: R.I. Hornsey CM_2: 15
In the above case, the use of the incremental approach made little difference except to earn $1898 more by choosing the second option But consider the following example Monster Meats are considering buying a new meat slicer which option is best option 1: Meat slicer for $50 000 save $11 000 per year option 2: Meat slicer plus automatic loader for $68 000 save $14 000 per year option 3: Do nothing, cost $0 save $0 per year life of equipment 8 years, MARR 12% ENG2000: R.I. Hornsey CM_2: 16
System without loader ( ) =0 $50,000 + $11,000P/A,i *,8 ( P/A,i *,8)= $50,000 $11,000 = 4.545 From tables or by some other means ( P / A,14%,8) = 4.6388 ( P / A,15%,8) = 4.4873 giving i * 14.5% Since the IRR > MARR then this is better than doing nothing ENG2000: R.I. Hornsey CM_2: 17
Now consider the second option: ( ) =0 $68,000 + $14,000 P / A,i *,8 ( P/A,i *,8)= $68,000 $14,000 = 4.857 From tables or by some other means ( P / A,12%,8) = 4.9676 ( P / A,13%,8) = 4.7987 giving Also a good choice NOT i * 12.5% ENG2000: R.I. Hornsey CM_2: 18
We must consider the incremental investment ( $68,000 $50,000) + ( $14,000 $11, 000) ( P / A,i *,8) = 0 ( P / A,i *,8) = $18,000 $3, 000 = 6.0000 From tables or by some other means i * 7.0% Since this incremental IRR < MARR, the automatic loader is a poor additional investment essentially, the excellent return from the investment on the basic slicer subsidised the apparent value of the automatic attachment ENG2000: R.I. Hornsey CM_2: 19
loader IRR = 7% slicer IRR = 14.5% slicer IRR = 12.5% ENG2000: R.I. Hornsey CM_2: 20
General approach From these examples it is clear that we need to address the incremental investment as well as the two main options If there are more than two choices, life becomes more complex and we need a systematic approach tot he problem The first step is to rank the projects in order of first cost, lowest first which may be the do nothing option this becomes (ungrammatically) the current best alternative (the IRR for this may be less than the MARR) ENG2000: R.I. Hornsey CM_2: 21
Set: A = 1 B = 2 A = current best B = challenger n = total number of mutually exclusive projects B = n + 1? no IRR of B - A < MARR Reset B = B + 1 yes best option is A MARR Reset A = B B = B + 1 The current best is challenged by the next option (ordered 1 to n in terms of first cost) if option A is not the last in the list, the incremental IRR is calculated between A and B if this IRR < MARR then we try a new B (A remains the best so far) if the incremental IRR MARR then B is the new current best and we try a new B ENG2000: R.I. Hornsey CM_2: 22
Example 5.6 An aircraft company must purchase a new lathe. There are four possible lathes, each with a life of 10 years and no scrap value. With an MARR of 15%, which should be chosen? lathe 1 2 3 4 first cost $100 000 $150 000 $200 000 $225 000 annual savings $25 000 $34 000 $46 000 $55 000 ENG2000: R.I. Hornsey CM_2: 23
Example 5.6 solution ENG2000: R.I. Hornsey CM_2: 24
Complexities Suppose that a project pays $1 000 today, costs $5 000 a year from now, and pays another $6 000 in two years. What is the IRR? $6 000 $1 000 $5 000 By equating the PWs and solving for the IRR: $1000 $5000( P /F,i *,1) + $6000( P /F,i *,2)=0 ENG2000: R.I. Hornsey CM_2: 25
( ) = 1 ( 1+ i * ) N since P /F,i *,N we eventually find i * 3i * + 2 = 0 or ( i * 1) ( i* 2) = 0 This has two solutions, i* = 1 (100%) and i* = 2 (200%) which is correct?! ENG2000: R.I. Hornsey CM_2: 26
General case We have a project with T periods there is a net cash flow in any period, t, given by A t = R t D t where R is a receipt and D is a disbursement Converting to PWs gives A 0 + A 1 ( 1+ i) 1 + A 2 ( 1+ i) 2 +K+ A T ( 1+ i) T = 0 or A 0 + A 1 x + A 2 x 2 +K+ A T x T = 0 where x = ( 1+ i) 1 the solution contains as many real positive solutions as there are changes in sign of A and hence there are as many IRRs as sign changes in A ENG2000: R.I. Hornsey CM_2: 27
Project balances One can think of the project as being composed of a set of project balances, each representing the situation at the end of successive periods A 0, A 1,, A T are cash flows (as before) now there are T + 1 project balances (B 0, B 1,, B T ), one at the end of each period, t = 0, 1, 2,, T each B represents the accumulated future value of all cash flows up to the end of that period, with an interest rate i B 0 = A 0 B 1 = A 0 1+ i ( ) + A 1 B 2 = A 0 1+ i to B T = A 0 1+ i ( ) 2 + A 1 ( 1+ i ) + A 2 ( ) T + A 1 ( 1+ i ) T 1 +K+ A T ENG2000: R.I. Hornsey CM_2: 28
For the previous example we find End of year 0 1 2 $1 000 i = 100% $1 000(1 + 1) $5 000 = $3 000 $3 000(1 + 1) + $6 000 = 0 $1 000 i = 200% $1 000(1 + 2) $5 000 = $2 000 $2 000(1 + 2) + $6 000 = 0 The fact that the totals are zero at the end confirms that both 100% and 200% are valid IRRs ENG2000: R.I. Hornsey CM_2: 29
The issue is that the IRR calculation assumes that the initial $1 000 is invested at one of the IRRs (100% or 200%) however, the $1 000 is provided by the project, not invested in it it is therefore sitting around for a year before being reinvested in the project during that year, it can be invested elsewhere but will probably not make this IRR (more likely MARR) Hence, to resolve the IRR problem, we need to consider what happens to cash not invested in the project usually, we assume that this cash makes a return equal to the MARR ENG2000: R.I. Hornsey CM_2: 30
External rate of return (ERR) The external rate of return, i e *, is defined as the rate of return on a project where any cash flows that are not invested in the project are assumed to earn interest at a predetermined rate typically the MARR in this situation, a project only has one ERR In simple cases, one can determine the ERR precisely However, in more complex cases, it can be difficult to decide where to apply the explicit interest rate and an approximate method is used ENG2000: R.I. Hornsey CM_2: 31
Precise ERR Let s revisit the previous example suppose that a project pays $1 000 today, costs $5 000 a year from now, and pays another $6 000 in two years the MARR is 25% Now the first $1 000 is invested outside the project at the MARR the cumulative cash flow for year 1 is thus: $1 000(F/P, 25%,1) $5 000 = $1 250 $5 000 = $3 750 The cash flow diagram is now $6 000 $3 750 ENG2000: R.I. Hornsey CM_2: 32
We can now determine a precise ERR: $3750 + $6000( * P /F,i e,1 ) = 0 ( * P /F,i e,1 ) = $3750 $6000 = 0.625 1 * 1+ i = 0.625 e i e * = 0.6 ERR = 60% ENG2000: R.I. Hornsey CM_2: 33
Approximate technique In a complex case, we don t necessarily know when to apply the explicit rate of interest because the cash flow can depend on what that rate of interest is Instead,we can use an approximate method take all net receipts forward at the MARR until the time of the last cash flow take all net disbursements forward at an unknown interest rate, i ea *, until the time of the last cash flow equate the future value of the receipts and disbursements from two steps above, and solve for i ea * this value for i ea * is an approximate ERR for the project [net means the total (inflow ± outflow) in a period] ENG2000: R.I. Hornsey CM_2: 34
Example re-revisited Now, the net receipts are the initial $1 000 and the final $6 000 after year 2 hence the amount carried forward at the MARR is $1 000(F/P,25%,2) + $6 000 The net disbursements are the $5 000 at the end of year 1 these are carried forward at the unknown interest rate $5 000((F/P,i ea *,1) Equating these expressions and solving gives i ea * = 51.25% = approximate ERR compared with the exact ERR of 60% ENG2000: R.I. Hornsey CM_2: 35
It turns out that the approximate ERR always lies between the precise ERR and the MARR hence, if the precise ERR > MARR, then the approx. ERR will be too alternatively, if the precise ERR < MARR, so too will be the approx. ERR Hence, the approximate calculation of the ERR will always give the correct decision the approx. ERR underestimates the rate of return for a viable project (since the precise ERR is larger) Hence the approx. ERR is simpler to calculate gives the correct decision provides a lower limit to the actual rate of return ENG2000: R.I. Hornsey CM_2: 36
When to use the ERR? It is generally better to calculate an IRR whenever possible, even though IRR and ERRs result in the same decision it gives a more precise answer that is valuable for project planning for example, it does not depend on the MARR chosen or the assumption of investment at the MARR Simple investments always give one IRR, e.g. techniques to determine whether there is more than one IRR are covered in the textbook appendix 5a ENG2000: R.I. Hornsey CM_2: 37
IRR/ERR versus PW/AW It can be shown (text 5.4.1) that the rate and worth techniques lead to the same decisions so why choose one techniques over the other? The rate methods give numbers in terms of percentages rather than absolute numbers this can be useful where projects are of very different sizes and the absolute numbers of one type of project will always be larger alternatively, it may be desirable to know the absolute profit that results from the decision A general comparison is shown on the next slide ENG2000: R.I. Hornsey CM_2: 38
Method Advantages Disadvantages IRR PW AW Payback period facilitates comparison of projects of different scales commonly used gives actual number for profit annual cash flow figures are easy to interpret very easy to calculate commonly used rapid recovery of capital difficult to calculate difficult to compare projects of differing sizes difficult to compare projects of differing sizes discriminates against long-term projects ignores time value of money ignores expected service life ENG2000: R.I. Hornsey CM_2: 39
Conclusion We have seen that the internal rate of return (IRR) is an alternative to present or annual worth calculations for determining the choice of projects The IRR can be difficult to determine, but is useful in some circumstances The decisions made by IRR are the same as those from AW/PW Rate of return calculations allow straightforward comparison of projects with very different absolute figures ENG2000: R.I. Hornsey CM_2: 40
Important things we haven t discussed Financial accounting how to understand a company s balance sheets Taxes inevitable! Inflation the effects of inflation on long-term projects Uncertainty how sensitive is the decision to estimated future values? ENG2000: R.I. Hornsey CM_2: 41