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ROR: Multiple Alternatives
Previously we looked at determining ROR for a project. Given a series of cash flows, the ROR was the value of i that exactly balanced them. We further expanded this, by recognizing that positive cash flows from an investment could be reinvested at a new rate, the external ROR, and by combining this with the internal ROR, we determined the composite ROR. Now, we take this 1 further step and look at how we compare alternatives. We have done this already with PW, FW and AW analysis, here we do it for ROR analysis. We have $90,000 to invest, and a MARR of 16% / year Alt A requires $50,000 and has irr of i* = 35% / year Alt B requires 85,000 and has irr of i* = 29% / year Which do we chose? Depends on what happens with left over money. As we saw last time with CRR, any positive cash flows were invested at MARR. Assume that is true in this case. Composite rate of return for each alternative Overall ROR a = 50,000(35%) + 40,000(16%) / 90,000 = 26.6% Overall ROR b = 85,000(29%) + 5,000(16%) / 90,000 = 28.3% In this case, B would be the better investment. ROR analysis may not provide the same ranking of alternatives as PW/FW/AW analysis. As it merely tells us the rate of return on a project, but does not consider the magnitude of return. A 20% rate of return on a $100 investment is fine. But a 10% rate of return on a $1000 investment yields more money in one year. One of the first rules we mentioned for engineering economics was that it was only the differences between project that really matter. Incremental Cash Flow Analysis makes use of this rule, by comparing the differences in the cash flows between investments and how those differences affect the money not invested in the project, but invested in an outside project. To conduct an incremental cash flow analysis we begin with a tabulation. In order to compare differences between projects, we need to make sure the cash flows all line up. To do so we may have to use the LCM. If the projects have the same life, then table goes to n = end of service life. If different life span, we must use LCM and table
goes to n = LCM. The ordering of the columns is important. The investments are always ranked from lowest to most expensive initial cost. Such at the cheapest to implement will be option A and the most expensive to implement is option B. Alt B yr 0 cost > Alt A yr 0 cost Example: Machine 1, cost -40K, annual -10, salvage 12 3 year life ALT A Machine 2, cost -65K, annual -8, salvage 25 6 year life ALT B Cash Flow year Alternative A Alternative B Incremental Cash Flow 0 -40,000 -65,000 -25,000 * extra investment 1 -10,000 -8,000 2,000 * extra savings 2 -10,000 -8,000 2,000 * extra savings 3 -10,000+12, -40,000 = -38, -8,000 30,000 * extra savings 4 -10,000 -8,000 2,000 * extra savings 5 -10,000 -8,000 2,000 * extra savings 6 -10,000+12, = 2000
15,000 * extra savings incremental cash flows represent extra investment or cost required if the alternative with the larger first cost is selected. value in year 0 represents the extra initial cost of the most expensive alternative. This savings, if we go with alternative A can be reinvested in another project. We compare the equivalent worth of the cash flows in yr 1-n against the equivalent worth of the extra investment at the MARR t=0 to determine which alternative should be chosen. If the ROR available through incremental cash flow equals or exceeds the MARR, alternative B should be selected. Otherwise, chose alternative A Incremental ROR method Essentially, a value i^ B − A ∗ is calculated which represents the internal ROR of the incremental series. We may get several sign changes. Thus we need to be vigilant and check for the # of real roots. Steps:
We can also see graphically that ∆i* is about 23% So, what does this mean? It means that the extra savings we get from the more expensive alternative B, earns us 23%. The MARR on this project is 15%. The savings are greater than this, meaning the extra cost is justified, and we go with alternative B. What do we do if we have more than 1 possible root? Use composite approach with the incremental cash flow series. Solve it exactly the same as we did previously. However, we don't know exactly what the reinvestment rate will be. 2 options: set reinvestment rate at ∆i*, or specify a value of c, perhaps the MARR Correctly using ROR method gives same value as PW / FW and AW analysis. However for AW, we have options.
- use incremental cash flow over LCM just as we have done so far
- annualize all single cash flow values
- find AW for each alternative's actual cash flow and solve for
- 0 = AWb - AWa 0% 5% 10% 15% 20% 25% 30% 35% 40% 45% 50%
0 5 10 15 20 25 30 i* PW MARR ∆i*
To this point, what have we learned?
- Determine the equivalent worth of cash flows at any point(s) in time
- Determine Present, Future and Annual Worth of an investment.
- Single investments, we calculate PW at the MARR, if > 0, we accept it.
- For alternatives, calculate PW at the MARR and select the largest value.
- compare investments with different lifespans using LCM
- Annual Worth Analysis avoids LCM
- We have determined PW and AW for investments with infinite lifespans
- Determine the ROR of an investment given a cash flow
- Internal Rate of Return on the investment, value for i giving a PW = 0
- select projects with ROR > MARR
- External Rate of Return on positive cash flows not needed.
- revenues
- initial funds not fully invested.
- Composite Rate of Return, weighted average of IRR and ERR.
- Incremental ROR analysis to select between alternatives
- compare the “savings” by going with 1 alternative over another.
- If the difference in ROR between the 2 options is less than MARR, invest in the cheaper option and reinvest the remaining funds in another investment
- if the difference is greater than MARR, you pick the more expensive option.
- Rules
- Requires LCM for alternative comparison.
- Multiple roots must be determined
- External funds (e.g., revenues) should get reinvested at the External ROR Mutually Exclusive Alternatives
- Determine ROR for all alternatives
- if only 1 exceeds MARR, select it.
- If >1 exceed MARR --> Incremental Cash Flow analysis
- Order all alternatives from smallest to largest initial investment
- calculate i* for the first alternative.
- if i* < MARR, eliminate it and go to next investment
- first alternative with i* > MARR is the DEFENDER. Next alternative on list is CHALLENGER
- Determine incremental cash flow between CHALLENGER and DEFENDER.
- Calculate i ∗ . If i ∗
MARR, DEFENDER is eliminated and CHALLENGER becomes DEFENDER. If i ∗ < MARR, eliminate CHALLENGER.
- Repeat until 1 alternative remains standing
PW b-a = -200 + 45(P/A,del i, 5) del i = 4.06%. The extra initial cost with going for the more expensive option yields a greater incremental return than the safe option, so we will pick it. A is eliminated NOTICE del i* b-a <> ib – ia 4.06% <> 7.08 – 7. del i* refers to the yearly difference in revenues between alternatives. Thus, we eliminate A. Now C is CHALLENGER, B is DEFENDER PW C-B = -300 + 50(P/A, del i, 5) del i = -5.8% A negative del i* means that the additional 300 we spend initially (3000-2700) buying C does not compensate for the 50 extra we get per year. In fact, compared to B, C is a losing proposition. This is not surprising if we look at IRR for B, C, we see 7.08% and 5.89%, 7% is much greater than 6%. Repeating this for PW D-B = -500 +75(P/A,del i,5) del i = -8.88%. Again, the extra $500 initially spent on D yields a negative ROR. We see some obvious and not obvious answers here. That C and D were eliminated is somewhat obvious, as they definitely were underperforming compared to A and B. However, B beats A is a little surprising, and is simply due to the fact that the extra 45 earned yearly by buying B offset the initial 200 cost. WHEN COMPARED TO THE SAFE OPTION.
