Applications of Money-Time Relationships And Methods For Evaluating Investments

Applications of Money-Time Relationships

And Methods

For

Evaluating Investments

Overview• Minimum Acceptable Rate of Return (MARR)• Two kinds of choice problems:

– Choose ALL the profitable projects?– Choose ONE most profitable project from many alternatives (mutually

exclusive alternatives)?

• Methods for comparing mutually exclusive alternatives– You actually already know these methods. They are just renamed

slightly.• PW Present Worth uses PV present value

(also called NPV or Net Present Value)• AW Annual Worth uses AV annuity value• FW Future Worth uses FV future value

– In each method, the MARR should be used as i%

• Rates of Return– Calculating IRR (Internal Rate of Return)

• The IRR is the i% where PV=0.

– Calculating ERR (External Rate of Return)• A bit more complicated. Covered after IRR.

– Direct use of IRR and ERR do not allow comparison of mutually exclusive alternatives

– The method for comparing mutually exclusive alternatives

Tunnel Construction Example($ figures in million hk$)

Tunnel projectsSha Tin Tsuen WanLantau NWLantau SW

Initial Cost C0 $8,000 $3,200 $3,500 $2,000Fare $5 $5 $8 $10Daily Traffic (millions) 0.500 0.350 0.150 0.100Annual O&M Ca $200 $400 $100 $150Project Life L 30 20 25 30Salvage Value at end of Life $1,500 $750 $1,000 $500

Tunnel Construction ExampleTunnel projectsSha Tin Tsuen WanLantau NWLantau SW

Initial Cost C0 $8,000 $3,200 $3,500 $2,000Fare $5 $5 $8 $10Daily Traffic (millions) 0.500 0.350 0.150 0.100Annual O&M Ca $200 $400 $100 $150Project Life L 30 20 25 30Salvage Value at end of Life $1,500 $750 $1,000 $500

We are going to consider two problemsA. Choose all profitable projects.B. Choose most profitable project.

We will assume MARR = 6%

Choose all the profitable projects

Several equivalent ways to identify all the profitable projects. Given a MARR% requirement,

Find all projects where IRR > MARR%

OR

Find all projects where PW>0 at i%=MARR%

OR

Find all projects where AW>0 at i%=MARR%

• (assumes repeatability)

Choose most profitable project

Constraints besides money (management time, difficult to obtain resources or equipment, land) may mean that not all projects can be funded.

Most profitable means the highest $ amount. Find project with highest PW (or AW) given i

%=MARR%.Important: Do not compare IRR%. IRR% can be

misleading. Seek the highest $ amount, not the highest % return.

Tunnel life cycle

1. The company builds the tunnel. We will treat this as a year 0 cost.

2. The company operates the tunnel for L years, and receives an annual revenue from vehicle tolls. There is also an annual cost for salaries, repair, etc.

3. After a useful life L years, the company sells the tunnel for a “salvage value” $S

Let’s look at the raw data again

We have the annual cost but not the annual revenue.We are going to need to know the annual revenue.Annual revenue = Fare x Yearly Traffic. Is yearly traffic = Daily Traffic x 365?

(no! the planners tells us it is less, they guess x325)


Initial Cost C0 $8,000 $3,200 $3,500 $2,000Fare $5 $5 $8 $10Daily Traffic (millions) 0.500 0.350 0.150 0.100Annual O&M Cost Ca $200 $400 $100 $150Project Life L 30 20 25 30Salvage Value at end of Life $1,500 $750 $1,000 $500

Now we have revenue, but need a way to evaluate the projects

In the next slide, we’ll choose the PW (present worth) method and see how to apply it to the problem.


Initial Cost C0 $8,000 $3,200 $3,500 $2,000

Fare $5 $5 $8 $10Daily Traffic (millions) 0.500 0.350 0.150 0.100

x325 Yearly traffic (millions) 163 114 49 33 <--- Because of public holidatys, etc. yearly traffic is 325x daily (not 365).Annual Revenue Ra (million hk$) $813 $569 $390 $325 <--- Annual Revenue = Yearly Traffic x Fare

Annual O&M Cost Ca $200 $400 $100 $150Project Life L 30 20 25 30Salvage Value at end of Life $1,500 $750 $1,000 $500

Present worth methodAlthough the tunnel project is simple, drawing a cash

flow diagram cam help to avoid mistakes.

Initial Cost C0

Year 1

Year L

Salvage

Value $S

Cash flow of $A/year for L years,

Where $A = Annual Revenue

– Annual O&M Cost

Find

$PW?

Present worth calculation

PW = -C0 + A * (P/A,i%,L) + S * (P/F,i%,L)

Initial Cost C0

Year 1

Year L

Salvage

Value $S

Cash flow of $A/year for L years,

Where $A = Annual Revenue

– Annual O&M Cost

Find

$PW?

Organizing the data

1. Arrange items according to life cycle (initial cost, annual O&M, salvage)

2. Calculate PW of each class of items

3. Add the PWs to get the total PW for each project

Organizing the data: 1 Arrange

Life cycle: Initial costs, then annual costs, then salvage


i% for calculations 6% 6.0% 6.0% 6.0% 6.0%Project Life L 30 20 25 30

Initial Cost C0 $8,000 $3,200 $3,500 $2,000

Fare $5 $5 $8 $10Daily Traffic (millions) 0.500 0.350 0.150 0.100Yearly traffic (millions) 163 114 49 33Annual Revenue Ra (million hk$) $813 $569 $390 $325Annual O&M Cost Ca $200 $400 $100 $150Annual cash flow = Ra-Ca $613 $169 $290 $175

Salvage Value at end of Life $1,500 $750 $1,000 $500

Except where noted, dollar figures are in millions

Organizing the data: 2 calculate each PW

Each PW is calculated with the standard formula. The initial cost is Year 0 and is thus a PW. The other costs need a conversion factor.



Initial Cost C0 $8,000 $3,200 $3,500 $2,000

Fare $5 $5 $8 $10Daily Traffic (millions) 0.500 0.350 0.150 0.100Yearly traffic (millions) 163 114 49 33Annual Revenue Ra (million hk$) $813 $569 $390 $325Annual O&M Cost Ca $200 $400 $100 $150Annual cash flow = Ra-Ca $613 $169 $290 $175(P/A,i%,L) factor for annuities 13.76483 11.46992 12.78336 13.76483PW of annual cash flow $8,431 $1,936 $3,707 $2,409

Salvage Value at end of Life $1,500 $750 $1,000 $500(P/F,i%,L) factor for Salvage Value 0.17411 0.31180 0.23300 0.17411PW of Salvage Value $261 $234 $233 $87

Organizing the data: 3 AddTunnel projectsSha Tin Tsuen WanLantau NWLantau SW


Initial Cost C0 $8,000 $3,200 $3,500 $2,000



Present Worth PW $692 -$1,031 $440 $496(sum of PW-C0) Except where noted, dollar figures are in millions

Conclusions from PW analysis

• The Sha Tin, and 2 Lantau projects are profitable because PW > 0 at MARR=6%.

• Of the profitable projects, the Sha Tin project is the MOST profitable. It produces the highest Present Worth.

• The Tsuen Wan project is not profitable at MARR=6%

Tunnel projectsSha Tin Tsuen Wan Lantau NW Lantau SW


Rates of Return

• The IRR is the most common rate of return

• The IRR is the i% where Present Worth = 0

• It is usually algebraically difficult to solve for IRR%, so– Use a spreadsheet and improved guesses to solve

– Or use a graph

Spreadsheet method for IRR% Tunnel projectsSha Tin Tsuen Wan Lantau NW Lantau SW


Initial Cost C0 $8,000 $3,200 $3,500 $2,000




Double click the spreadsheet, then play with i% until PW=0 for each project. The results are the Rates of Return IRR%.

IRR% results

• All except Tsuen Wan are above the 6% MARR.• A project is profitable when IRR>MARR• But the highest IRR% may not give the highest PW.



Initial Cost C0 $8,000 $3,200 $3,500 $2,000




Return on Investment 6.76% 2.18% 7.19% 8.10%(Find i% where PW=0)

Some problems with IRR

• Does not allow comparison of projects– Why not? Compare

– Can be ‘patched up’ with delta method.

• Easy to define, but Difficult to calculate– Trial and error, graphing, Newton’s method

– To solve f(I)=0, use I**=I*-(f(I)/f’(I))

• Assumes project revenues are reinvested at IRR%, which can be unrealistic if IRR% is very high.

We need a volunteer

The volunteer needs to have $100 in notes and $1 in coins.

There is no chance of losing money.

We need a volunteer

• Investment A costs $1 and earns a 100% rate of return during this class.

• Investment B costs $100 and earns a 20% rate of return during this class.

Choose which investment you prefer

Of course

$1 x 100% = $1 profit

$100 x 20% = $20 profit

The lower IRR% may give a higher profit, if the required dollar investment is higher.

Delta methodChoose the highest IRR investment, in this case A.Then look at cash flowInvestment A -$1, then +$2Investment B -$100, then +$120Delta = (B-A) has cash flows of

-$99, then + $118The IRR of Delta is about 20%. If 20% is over your MARR, then you should spend the

extra money (the $99) to go from investment A to investment B. You will then earn $19 more profit.

Other IRR complaints

• Too hard to calculate

• Unrealistic when high

ERR, or external rate of return, tries to solve these problems. It does NOT solve the project comparison problem.

ERR External Rate of Return

Procedure - Project life is L years.1. Determine which periods (years, months) have

net revenue and which periods have net costs2. Find FV (future value) of all net revenues, at

MARR%3. Find PV (present value) of all net costs, at MARR

%4. Find the E% that solves(FV of Revenues) = (PV of costs) * (P/F,E%,L) Or E% ={ [(FV of Rev)/(PV of Costs)](1/L)-1}



Initial Cost C0 $8,000 $3,200 $3,500 $2,000




IRR% return on investment 6.76% 2.18% 7.19% 8.10%(Find i% where PW=0)

FW of annual revenue $48,423.14 $6,207.57 $15,910.71 $13,835.18FW of Salvage Value $1,500 $750 $1,000 $500Total FW of revenues $49,923.14 $6,957.57 $16,910.71 $14,335.18

PW of Initial Cost $8,000 $3,200 $3,500 $2,000

Total PW of costs $8,000.00 $3,200.00 $3,500.00 $2,000.00

FW of revenues / PW of Costs 6.240392382 2.17424016 4.83163099 7.16759129to the (1/L) power 1.06293587 1.03959786 1.06503468 1.06785538ERR% 6.29% 3.96% 6.50% 6.79%

Odds and Ends

• MARR – where does it come from?

• Annual worth method – often worthwhile when projects repeat

MARR

The MARR is usually given to you in textbook problems. In real life, where does it come from?

Usually upper management decision, involving:• Rates of interest on borrowed money• Supply and demand for money in the company• Expectations of investors

– Return on companies stock– Return on stock index for the industry

• Project Risk (though adjusting MARR upwards for risk can be a bad move – since it weights present results over future)

Next Week: Annual Worth method

Useful for projects that repeat or when an annual value is desired for comparison to other projects with annual values.

[Our tunnel comparisons did not repeat – the company planned to sell the tunnel]

Main technical step is the annual capital recovery (CR) amount.

Annual CR = (Initial Cost) (A/P,i%,L) – (Salvage Value) (A/F,i%,L)

Summary

• To find the most profitable project or investment:– compare PW (or sometimes AW) using i%=MARR%

– Do not compare IRR of projects; misleading

• To find all the profitable projects or investments:– All of these methods produce equivalent answers

– Compare IRR to MARR

– Compare ERR to MARR

– Look for positive PW, using i%=MARR%

– Look for positive AW, using i%=MARR%

Documents

Applications of Money-Time Relationships And Methods For Evaluating Investments