Comparing Mutually Exclusive Alternatives

&Capital Budgetting

Ref: Chapter 5 & section 10.4

Capital Budgeting

• Trying to get the the most with what you’ve got.

• Look at all possible combinations of projects to create a capital plan

• Analyze each mutually exclusive alternative using one of the analysis methods

Why Capital Budgeting

• Capital Budgeting is critical in business and all organizations that have limited resources.

• Defence budget and discretionary spending - whose money is it anyway?

• Treasury board, the govt’s board of directors.

Evaluation of Multiple Investment Alternatives

• Basic Question: Should a project be included in the capital budget?

• For DND - APC’s, Submarines or Helicopters ?

Definitions

• PROJECT - a single engineering proposal being considered

• INVESTMENT ALTERNATIVE - a decision option.

Therefore one project represents two investment alternatives: accept or reject.

Definitions Continued

• INDEPENDENT PROJECT - can be accepted or rejected without influencing the accept/reject decision of another project.

• DEPENDENT PROJECTS - projects are related in such a way that the acceptance of one of them will influence the acceptance of others.

Types of Dependent Projects

• Mutually Exclusive - can only accept one alternative from the set.

• Contingent - acceptance of one requires the acceptance of another.

• Note - a fixed budget adds an external dependency if the cost of all projects exceeds the funds available.

Check

Mutually Exclusive alternatives are:

a. Independent

b. Dependent

Formulating Mutually Exclusive Alternatives

• Necessary to enumerate all feasible combinations of projects that could make up the capital project.

• Can then apply budget or other constraints and decision criteria to select the best capital project.

Approach 1 - Enumeration Method

• Create a matrix of all possible alternatives.

• For 2 independent projects, 4 alternatives - do nothing, P1 only, P2 only, or P1 and P2.

• A = 2n,

where A = the number of alternatives

and n = the number of projects.

Example

• Consider five investment projects A to E Suppose projects A and B are mutually exclusive. Project C is independent, but D is contingent on C. Also E is contingent on B.

• 25 = 32 Alternatives to consider

Approach 2: Graphical

RelationshipSymbol

A B

B

BA

A

A and B are independent.

A and B are mutuallyexclusive

B is contingent on A

Check

Suppose C is contingent on the acceptanceof both A and B, and B is contingent on theacceptance of A.

Which diagram is correct?

BA

C

BA

C

BA

C

A B C

Consider five investment projects with cash flows initial costs estimated above. Suppose projects A and B are mutually exclusive. Project C is independent, but D is contingent on C. Also E is contingent on B.

Formulate the total number of feasible investment alternatives and tabulate their Cash Flows.

EXAMPLE - Formulating Mutually Exclusive Alternatives

Project A -100Project B -200Project C -100Project D -300Project E -200

Graphical InterpretationA B

C D

E

Alternative A B C

1 0 0 0 0 0 2 1 0 0 0 0 3 0 1 0 0 0

4 0 0 1 0 05 1 0 1 0 0

D E

6 0 1 1 0 0 7 0 1 0 0 1

9 0 0 1 1 0 10 1 0 1 1 0 11 0 1 1 0 1

12 1 0 1 1 0 13 0 1 1 1 1

Total Number of Mutually Exclusive AlternativesAlternative Projects

1 -

2 A

3 C

4 B

5 A,C

6 B,C

7 B,E

8 C,D

9 A,C,D

10 B,C,E

11 B,C,D

12 B,C,D,E

Elements of Decision Criteria

• Basic problem: How do we select the best mutually exclusive alternative?

• Solution: depends on how you define best - your decision criteria

Three Important Factors

• Differences between alternatives

• The MARR

• The do nothing alternative

Differences Between Alternatives Fundamental Rule

• When comparing mutually exclusive alternatives, it is the difference between them that is relevant for determining the economic desirability of one over the other.

Fundamental Rule – General Process

• Step 1 - Put the mutually exclusive alternatives in order of increasing initial cost (investment at t = 0)

• Step 2 - The cheapest option becomes the defender

• Step 3 - The next cheapest becomes the challenger.

• Step 4 - Apply the appropriate decision criteria to the difference between the two options.

• If the increase in investment is economically desirable, the challenger becomes the defender otherwise, reject the challenger

• Step 5 - Continue until all alternatives are evaluated

• The last defender is the best choice

Fundamental Rule – General Process Cont’d

Minimum Acceptable Rate of Return (MARR)

• Decision criteria have as their objective the maximization of equivalent profit, given that all investment alternatives must yield a return that exceeds some MARR.

• MARR represents management’s cut-off rate - a policy decision.

The Do Nothing Alternative• Does not mean the money is buried in the back yard.• Means do nothing about the alternatives being

considered.• Money is still invested and expected to yield at least

the MARR• Therefore, for the do nothing alternative:

- PE (MARR) = 0

- AE (MARR) = 0

- FE (MARR) = 0

Decision Criterion - Analysis Methods

• PE/FE/AE on total investment

• PE on incremental investment

• IRR on incremental investment

Multiple-Alternative Comparison Based on Total Investment

• Step 1 - Remove non-profitable projects.

• Step 2 - Generate mutually exclusive alternatives.

• Step 3 - Order alternatives by increasing order of investment.

• Step 4 - Remove alternatives that exceed budget.

• Step 5 - Remove dominated alternatives.

• Step 6 - Calculate PE for remaining alternatives.

• Step 7 - Select alternative with highest PE.

Note: Step 6 can be replaced by FE or AE - same result

Multiple-Alternative Comparison Based on Total Investment Cont...

Multiple-Alternative Comparison Based on Total Investment - Example

Cash Flow

ProjectABCDE

0-100-200-100-300-200

1505030

100100

250

10080

100100

350

200120100150

Example - Steps 1-4

Alternatives Projects 0 1 2 31 - 0 0 0 02 A -100 50 50 503 C -100 30 80 1204 B -200 50 100 2005 AC -200 80 130 1706 B,C -300 80 180 3207 B,E -400 150 200 3508 C,D -400 130 180 2209 A,C,D -500 180 230 27010 B,C,E -500 180 280 47011 B,C,D -600 180 280 42012 B,C,D,E -800 280 380 570

Example - Steps 5 and 6Alternatives Projects 0 1 2 3 PE(10%)1 - 0 0 0 50 02 A -100 50 50 50 24.343 C -100 30 80 120 83.544 B -200 50 100 200 78.365 AC -200 80 130 170 107.886 B,C -300 80 180 320 161.97 B,E -400 150 200 350 164.618 C,D -400 130 180 220 32.239 A,C,D -500 180 230 270 56.5710 B,C,E -500 180 280 470 248.1511 B,C,D -600 180 280 420 110.5912 B,C,D,E -800 280 380 570 196.84

Present Worth on Incremental Investment Criteria

• Step 1 - Put the mutually exclusive options in increasing order of initial cost.

• Step 2 - Designate cheapest option as the first defender

• Step 3 - The next cheapest option becomes the challenger and calculate the present equivalent of the cash flow resulting from challenger - defender.

Present Worth on Incremental Investment Criteria Cont...

• Step 4 - If the result from step 3 is positive, challenger becomes new defender and continue, otherwise challenger is replaced with the next cheapest option and continue.

• Step 5 - Continue until all options are evaluated - the last defender is the best option.

(i.e., it is the largest possible investment that makes at least MARR. In other words, it is the option with the highest NPW that earns at least MARR.)

Example - PW on Incremental Investment

Additional data:- Fixed budget of $5,000- Projects 1 and 2 are mutually exclusive - meet the same requirement- Project 4 is contingent on project 1

Problem - Select the best capital program using both PE on total investment and IRR on incremental investment, MARR = 15%.

Consider the following projects:

Yr 0 1 2 3

Project 1 -1000 550 550 550Project 2 -2000 875 875 875Project 3 -3000 1400 1400 1400Project 4 -4000 1665 1665 1665

Solution

• A = 2n = 16 alternatives, where n = 4 projects.

• All projects appear profitable, none to remove.

• Generate mutually exclusive alternatives.

• Put options in increasing order of initial investment.

1

2

4

1

24

Graphical Approach

Mutually Exclusive Alternatives

Projects1 2 3 4

0 0 0 0 11 0 0 0 20 1 0 0 30 0 1 0 41 0 1 0 51 0 0 1 60 1 1 0 71 0 1 1 8

Option Initial Investment 0 1,000 2,000 3,000 4,000 5,000 5,000 8,000

Solution Cont...

• Remove any options that exceed budget.

• Remove dominated proposals.

• Designate cheapest option as the defender and next cheapest as the challenger and apply decision criteria to the difference between their respective cash flows.

• The last defender is the best option.

Alternative Cash Flows

Alt12345678

Projects-123

1,31,42,3

1,3,4

Yr 00

-1000-2000-3000-4000-5000-5000-8000

Yr 10

550875

14001950221522753615

Yr 20

550875

14001950221522753615

Yr 30

550875

14001950221522753615

Calculations - Present Worth on Incremental Investment

• Defender 1 Challenger 2 : Cash flow on difference:- 1000 550 550 550 PE (15%) = 255.8 > 0 therefore reject 1.

• Defender 2 Challenger 3 : Cash flow on the difference:-1000 325 325 325 PE (15%) = -258 < 0 therefore reject option 3.

• Defender 2 Challenger 4 : -2000 850 850 850 PE(15%) = -59 < 0 therefore reject option 4.

Calculations - Present Worth on Incremental Investment Cont...

• Defender 2 Challenger 5:

-3000 1400 1400 1400 PE(15%) = 196.5 >0 therefore reject 2.

• Defender 5 Challenger 7:

-1000 325 325 325 PE(15%) = -258 < 0 therefore option 5 is best

• Conclusion: invest in plan 1 and 3. The remaining $1000 to be invested at a min of MARR elsewhere.

Multiple-Alternative Comparison by IRR on Incremental Investment

• Step 1 - Order alternatives by increasing order of initial investment

• Step 2 - Compute cash flow difference between pairs and calculate IRR on the increment of investment: if greater than MARR accept the larger investment and continue. If less than MARR, reject larger investment and move to next alternative.

• Step 3 - Repeat step 2 until all alternatives have been compared - the last defender is the best alternative.

Why Use IRR ?

• PE - Absolute, IRR relative

• If you were CEO which would you rather hear?

- We made MARR plus a present value surplus of $268,000.

Or, - The investment yielded a 22% return.

IRR on Total Investment

Alt12345678

Yr 00

-1000-2000-3000-4000-5000-5000-8000

Yr 10

550875

14001950221522753615

Yr 20

550875

14001950221522753615

Yr 30

550875

14001950221522753615

IRR0

30%15%19%22%16%17%17%

Example - IRR on Increment

Alt12345678

Projects-123

1,31,42,3

1,3,4

Yr 00

-1000-2000-3000-4000-5000-5000-8000

Yr 10

550875

14001950221522753615

Yr 20

550875

14001950221522753615

Yr 30

550875

14001950221522753615

Calculations - IRR on Incremental Investment

• Defender 1 Challenger 2 : Cash flow on difference:

- 1000 550 550 550 IRR = 29.9% > MARR therefore reject 1.

• Defender 2 Challenger 3 : Cash flow on the difference:

-1000 325 325 325 IRR = -1.3% < MARR therefore reject option 3.

• Defender 2 Challenger 4 :

-2000 850 850 850 IRR = 13.2% < MARR therefore reject option 4.

Calculations - IRR on Incremental Investment Cont...• Defender 2 Challenger 5:

-3000 1400 1400 1400 IRR = 18.9% >MARR therefore reject 2

• Defender 5 Challenger 7:

-1000 325 325 325 IRR = -1.3% < MARR therefore option 5 is best

• Conclusion: invest in plan 1 and 3. The remaining $1000 to be invested at a min of MARR elsewhere.

General Conclusions

• The two methods provide consistent results. Option 5 has the highest PE evaluated at MARR. Option 5 is the largest investment that yields at least MARR on each increment of investment.

• Although not calculated, Option 5 also would have the highest FE and AE evaluated at MARR.

General Conclusions Cont...

• Option 5 is not however the largest investment that gains at least MARR. Option 7 represents a $1,000 increase in investment over option 5 and has an IRR of 17.3 %. Why is this not chosen?

• As demonstrated during the IRR on incremental investment analysis, the IRR on the $1,000 increment is -1.3% hence, it actually loses money. Therefore, the remaining $1,000 should be invested elsewhere.

SUMMARY – Capital Budgeting

• When dealing with multiple investment alternatives with various dependencies, we need to organize them into mutually exclusive projects that cover all feasible investment combinations.

• Independent Projects vs Dependent Projects

• PE/FE/AE on total investment

• PE and IRR on incremental investment.

Projects with unequal lives

• Two options– Study period– Least common multiple

Study Period

• Could be a company policy - say 5 yrs• A period for which accurate estimates of future

cash flows are available• A period that coincides with the life of one of the

projects• Any other period that the analyst thinks makes

sense

Example - Materials handlingA factory is considering three options for improving their materials handling system:

Option

Initial CostLabour/yrHydro /yr

Maintenance/yrTaxes &insurance/yr

service life(yrs)

A

$9 200

B

$15 0003 300400

2 40030010

C

$25 0001 450600

3 07550015

Which option is best? (MARR =9%)

Solution: Study period = 10yrs

PA = 9200(P/A,9,10) = 9200(6.4176) = $59 042

PB = 15000 + (3300+400+2400+300)(P/A,9,10) = 15000 + (6400)(6.4176)

= $56 073

For option 3 the service life is 5 yrs longer than the study period. We have to estimate a salvage value at the end of the study period.For example, if we assume a salvage value of $5000 at the end of 10 yrs:

PC =25000 + (1450+600+3075+500)(P/A,9,10)-5000(P/F,9,10) =25000 + (5625)(6.4176)-5000(.4224)

= $58 987

Least Common Multiple Method

For investments that have long periods of required service, it is preferable to set the study period equal to the least common multipleof the service lives of the contenders. For example if option A has 2 yr duration and option B has a 3 yr duration, we choose a study period of 6 yrs.

1 2

1 2 3

A

B

6

6

Example: Least common multiple

Option

Initial costLabour/ Yr.Hydro/ Yr.

Maintenance/ Yr.Taxes & Insur/ Yr.Service Life (Yrs.)

PE(Service life)

A

9 200$

n/a

B

15 000$3 300400

2 40030010

56 073

C

25 000$1 450600

3 07550015

70 341

Solution: Least Common Multiple

The least common multiple is 30Yrs. We can calculate PE or AEC but since A is already annual we calculate the AEC of Option B (3 times) and option C (2 times).

AECB = 56073(A/P,9,10) = 8 736$

AECC = (70341)(A/P,9,15) =6 844$

With this approach, C is preferred.

AECA = 9 200$

Recommended Problems Chapter 5 & 10

• Level 1: – Chapter 5: 5.1 – 5.6, 5.8 – 5.13 – Chapter 10: 10.7, 10.8

• Level 2:– Chapter 5: 5.17, 5.18, 5.21, 5.30, 5.33, 5.36– Chapter 10: 10.20, 10.22, 10.24,

• Level 3: 5.43, 10.32

Replacement Analysis Fundamentals (Section 6.6)

Reasons for Replacement

• Deterioration - machine is breaking too often or tolerances are no longer within acceptable limits.

• Obsolescence - caused by a change in the environment. Production rates increase or quality standards go up. Competitive advantage.

• A decision faced by all car owners.

Outline

• Allocation of costs to defender

• Sunk Costs

• Economic Life

• Cyclic replacement

• Finding the Best replacement

Overview

• Replacement is normally motivated by economic factors

• Other reasons include:– reliability– preferences– change for change sake

Defender Challenger Approach

• Defender - The existing asset being considered for replacement

• Challenger - The asset proposed to be the replacement

Two Primary Questions

• When should the item be replaced ?

• What should it be replaced with?

Allocation of Costs to the Defender

• Current market value– not necessarily trade-in value

• Sunk Costs– any past cost that will be unaffected by future

investment decisions

Sunk costs associated with asset disposal

$0

$5,000

$10,000

$15,000

$20,000Repair Cost

Book Loss

Sunk cost

Book Value

Market value

Example - Car Replacement

• Suppose you have a car you purchased 7 yrs ago for $10,000.

• On avg maint costs were between $500-$1,000 per yr.

• Salvage value and estimated maint costs are on the next slide.

• Assume that after 10 yrs the car will be scrapped for the salvage value indicated.

Old Car - Defender

Yr AGE SalvageValue ($)

MaintCosts ($)

0 7 1,000

1 8 500 2,000

2 9 400 2,500

3 10 300 3,000

Allocation of costs to the defender

• Sunk Cost - ARE NOT INCLUDED IN THE ANALYSIS.

• Initial Cost - For the purpose of the analysis, we need an initial cost for the defender. We use the salvage value since if we keep the car, we give up the opportunity to make $1,000 by selling (opportunity cost of keeping the defender).

Example continued

• Now consider the purchase of a new car to replace the defender.

• The dealership offers $1,250 trade in allowance on the old car.

• The new car retails for $10,250 and it is expected to last 7 years or more.

• Salvage values and maintenace costs for the new car over the next 7 years are estimated on the next slide.

Challenger CostsYr Salvage Value Maint Costs AE(15%) yr n

0 10,000

1 8,000 150 3,650

2 7,000 150 3,045

3 6,000 150 2,801

4 5,000 500 2,721

5 4,000 1,000 2,725

6 3,000 1,500 2,768

7 2,000 2,000 2,830

Cash Flows - Step 1Old Car

10002000 2500 3000

300

New Car

10000

150 500

500/yr

2000

2000

Study Period Approach

• One way to analyse the problem is to choose a study period equal to the service life of the defender.

• Given an interest rate of 15%:

• CR(i)= (P-S)(A/P,i,N) + iS

• where P = initial cost ans S = Salvage value

• Called the Capital Recovery Cost Factor

Capital Recovery Cost CR(i)

• CR(i) = P(A/P,i,N) - S(A/F,i,N)

• but, (A/F,i,N) = (A/P,i,N) - i

• substituting,

• CR(i)=P(A/P,i,N)-S[(A/P,i,N)-i]

• CR(i)=(P-S)(A/P,i,N)+iS

Solving AE(i) for both:

• AED=[(1000-300)+2000(P/F,15,1)+2500 (P/F,15,2) + 3000(P/F,15,3)](A/P,15,3) + 300(0.15) = $2,805

• AEC=(10,000-2,000)(A/P,15,7) + 150 + 2000(.15) = $2,373

• Conclusion - replace the old Car

Alternate Approach

• keep the three yr study period but assume we can get $6,000 salvage value for the new car at the end of yr three.

• AEC = [(10,000-6,000) + 150(P/A,15,3)] (A/P,15,3) + 6000 (0.15) = $2,801

• hence, it is still better to replace the old car.• However, is service life of the defender the

best study period to choose?

Economic Service Life

• Because assets O&M costs increase with age, we are more interested in knowing an assets practical vice physical service life.

• Economic service life is that period of useful life that minimizes the equivalent annual cost of an asset.

Old Car - Defender

Yr AGE SalvageValue ($)

MaintCosts ($)

AE(15%) for nyears

0 7 1,000

1 8 500 2,000 2,650*

2 9 400 2,500 2,662

3 10 300 3,000 2,805

* Economic Life Defender = 1 yr

Challenger CostsYr Salvage Value Maint Costs AE(15%) yr n

0 10,000

1 8,000 150 3,650

2 7,000 150 3,045

3 6,000 150 2,801

4 5,000 500 *2,721

5 4,000 1,000 2,725

6 3,000 1,500 2,768

7 2,000 2,000 2,830

* Economic service life challenger = 4 yrs.

Equ

ival

ent A

nnua

l Cos

t

1,000

2,000

3,000

4,000

Asset Life in years

1 2 3 4 5 6

Operating Costs

Capital Recovery Costs

Total Cost

Economic Life of an AssetEconomic Life of an Asset

Economic Life - special cases

• Case 1 - operating costs and salvage value are both constant - Economic life = ?

• The longer the asset is retained, the lower its total equivalent annual cost, hence keep for its service life.

• Case 2 - initial costs always equal salvage value and operating costs are always increasing - Economic Life = ?

• 1 period

Replacement Decisions and Assumptions

• Assume no replacement of current alternative.

• Assume replacement of each alternative by an identical alternative.

• Assume replacement of each alternative by the best challenger.

• Assume replacement of each alternative by dissimilar challengers.

Cyclic Replacement

• This approach assumes that an asset is retained for its economic life then succeeded by a sequence of identical replacements.

• To establish a study period, it is assumed that the cash flows are repeated until a common multiple of lives is reached.

• In some instances, it is assumed that the alternatives are repeated forever.

Cyclic replacement example

Old Car Options

New Car Options

1 2 3 4 5 6 7

D1 D1 D1 D1 D1D1 D1

D2 D2 D2 D2

D3D3D3

1 2 3 4 5 6 7C1

C1 C1 C1 C1 C1 C1

C2C2 C2 C2

C7

Cyclic replacement example 2 - no replacement of defender.

Old Car Options

New Car Options

1

D1

D2

D3

1 2 3 4 5 6 7C1

C1 C1 C1 C1 C1 C1

C2C2 C2 C2

C7

2

3

Replacement by the best Challenger

• The best challenger is the current challenger among all available current challengers that minimizes the AE based on economic life.

• We assume that the defender will be replaced indefinitely by a sequence of best challengers replaced after their economic life.

Example - Continued

• We can now decide whether it is better to replace our old car now or after 1,2 or 3 years (in general, to the end of the service life of the defender).

• Regardless, it will be replaced by a series of new cars replaced after their economic life(In general only the best challenger is considered, here their is only 1)

• Can then calculate NPW or AE for all combinations and pick the cheapest.

Special Cases

• If the AE for the defender for all possible durations are greater than the AE of the economic life of the challenger, replace immediately.

• If for only one option (say x yrs)the AE of the defender is less than the AE of the economic life of the challenger, replace at the end of yr x.

• Otherwise, calculate all combinations, for all values of n such that AE defender is less that AE of economic life of the challenger.

Replacement by the best Challenger.

Feasible Combinations

1

D1

2 3 4 5 6 7

C4 C4

D2

2

C4

6C4

D3

3C4

7

Solution

• In our example, their are 2 values of n such that the defender has a lower AE than the economic life of the challenger. Hence we must look at keeping the defender for 1 more year or two more years.

• Still must choose a study period to allow for comparison between combinations. Choosing 3 yrs calculate the following:

CalculationsKeeping the defender for 1 more yr:

NPWD1 = 1000 + [ (2000-500) + 2721 (P/A,15,2)](P/F,15,3) = $6,151

Keeping the defender for 2 yrs:

NPWD2 = 1000 + 2000 (P/F,15,1) + (2500-400)(P/F,15,2) + 2721(P/F,15,3) = $6,116

Therefore, keep the defender for two yrs then replace it with a series of challengers that are replaced every four yrs.

Conclusions• The minimum AE is obtained by keeping the old

car for two yrs then replacing it with the new car which is then replaced every 4 years (its economic life).

• NOTE: If we assume infinite replacement by the challenger, do not automatically assume you will only keep the defender for it’s economic life! In this example the economic life of the old car is 1 yr yet our solution is to keep it for two yrs. WHY?

Assume replacement of each alternative by dissimilar challengers

• The previous assumptions are restrictive considering all the forces at work in the economy.

• Technological changes make it likely that better challengers will come along before we replace our best challenger even once.

• This method employs network techniques that considers

all possible combinations of future replacements. • Alternatively, redo the analysis each yr or as factors

change.

Summary of Methods

1.Study period approach - good for finite problems or if there is a required service period. Compare the AE or PW of all options. If service life of asset is longer than study period, you must estimate salvage value.

2. Assume no replacement after service life of optimum choice - compare AE cost for each option’s economic life and choose the cheapest.

Summary of Methods Cont’d

3. Cyclic Replacement - Compare the AE cost for the economic life for each option and choose the cheapest.

4. Assume infinite cyclic replacement by the best challenger - Next slide ...

Infinite Replacement by best Challenger

1. Calculate AEC of the economic life of the best challenger

2. Calculate AEC for all possible durations of the defender.

3. If AECD > AECC for all values of AECD Replace now.

4. If AECD < AECC for only one value of n, replace then.

5.Otherwise, for all values of n where AECD < AECC

(economic life) calculate the PW or AEC for all combinations using either the service life of the defender or the economic life of the challenger as the study period (both yield the same result) . Select the minimum cost option.

Recommended Problems – Chapter 6

• Level 1: 6.5 – 6.7

• Level 2: 6.32, 6.38, 6.41, 6.49, 6.55