MIE 754 - Class #5 MIE 754 - Class #5 Manufacturing & Engineering Manufacturing & Engineering

EconomicsEconomics

• Concerns and QuestionsConcerns and Questions•Quick Recap of Previous ClassQuick Recap of Previous Class• Today’s Focus:Today’s Focus:

– Chap 3 Comparing Alternatives with Chap 3 Comparing Alternatives with Different Useful Lives Different Useful Lives

– Chap 4 Rate of Return MethodsChap 4 Rate of Return Methods• Hmwk #3 Due in 1 Week:Hmwk #3 Due in 1 Week:

– Chap 2 - 3, 4, 5, 6, 11, 12, 15, Chap 2 - 3, 4, 5, 6, 11, 12, 15, 16, 24, 30, 3716, 24, 30, 37

Concerns and Questions?Concerns and Questions?

Quick Recap of Previous ClassQuick Recap of Previous Class

Effective InterestEffective Interest Comparison of AlternativesComparison of Alternatives ProcedureProcedure Defining Investment AlternativesDefining Investment Alternatives Useful Life versus Study PeriodUseful Life versus Study Period

Useful Life versus Study PeriodUseful Life versus Study Period

Comparison must be over the Comparison must be over the samesame study period for ALL alternatives!study period for ALL alternatives!

Useful Lives = Study PeriodUseful Lives = Study Period Useful Lives are Different Among Useful Lives are Different Among

Alters.Alters.UL < SPUL < SPUL > SPUL > SP

Same Study Period Required!!Same Study Period Required!!

Use either:Use either:

Repeatability AssumptionRepeatability Assumption

Cotermination AssumptionCotermination Assumption

ExampleExample

AA BBCapital InvestmentCapital Investment -$3,500 -$3,500 -$5,000 -$5,000Annual Revenues Annual Revenues 1,900 1,900 2,500 2,500Annual Expenses Annual Expenses -645 -645 -1,020 -1,020Useful LifeUseful Life 4 4 6 6Market ValueMarket Value 0 0 0 0

Several In-class ExamplesSeveral In-class Examples

The Effect of CompoundingThe Effect of Compounding

Benjamin Franklin, according to the American Benjamin Franklin, according to the American Bankers Association, left $5,000 to the residents of Bankers Association, left $5,000 to the residents of Boston in 1791, with the understanding that it Boston in 1791, with the understanding that it should be allowed to accumulate for a hundred should be allowed to accumulate for a hundred years. By 1891 the $5,000 had grown to $322,000. years. By 1891 the $5,000 had grown to $322,000. A school was built, and $92,000 was set aside for a A school was built, and $92,000 was set aside for a second hundred years of growth. In 1960, this second hundred years of growth. In 1960, this second century fund had reached $1,400,000. As second century fund had reached $1,400,000. As Franklin put it, in anticipation: "Money makes Franklin put it, in anticipation: "Money makes money and the money that money makes makes money and the money that money makes makes more money."more money."

Question: What average interest rate per year was Question: What average interest rate per year was earned from earned from 1791 to 1891?1791 to 1891?

The “Ben Franklin” Problem SolutionThe “Ben Franklin” Problem Solution

Given: P=$5,000 N=100 Given: P=$5,000 N=100 F = $322,000F = $322,000Find: i'%Find: i'%F = P(F|P, i'%, 100)F = P(F|P, i'%, 100)

$322,000 = $5000(F|P, i'%, 100)$322,000 = $5000(F|P, i'%, 100)therefore, (F|P, i'%, 100) = 64.4therefore, (F|P, i'%, 100) = 64.4

From tables, From tables, (F|P, 4%, 100) = 50.5049(F|P, 4%, 100) = 50.5049(F|P, 5%, 100) = 131.501(F|P, 5%, 100) = 131.501

F = P(1+i')F = P(1+i')NN therefore, 322 = 5(1+i') therefore, 322 = 5(1+i')100 100

64.4 = (1+i')64.4 = (1+i')100 100 or (1+i') = 1.0425 so that i' = 4.25%or (1+i') = 1.0425 so that i' = 4.25%

Chapter 4 - Rate of Return MethodsChapter 4 - Rate of Return Methods

Compare against minimum Compare against minimum standard of desirability - minimum standard of desirability - minimum attractive rate of return (MARR)attractive rate of return (MARR)

Internal Rate of Return (IRR) Internal Rate of Return (IRR) MethodMethodSolves for the Solves for the interest rateinterest rate that that equates the equivalent worth of a equates the equivalent worth of a project's cash outflows (expenditures) project's cash outflows (expenditures) to the equivalent worth of cash inflows to the equivalent worth of cash inflows (receipts or savings). (receipts or savings).

IRR is Like a “Break-Even” ProblemIRR is Like a “Break-Even” Problem

Find i' such that Find i' such that FW(neg, i') = FW(pos, i')FW(neg, i') = FW(pos, i')

FW = 0 = FW(pos, i') - FW(neg, i')FW = 0 = FW(pos, i') - FW(neg, i')

Can use any one of the EW methods for IRR:Can use any one of the EW methods for IRR: PW(i' %) = 0PW(i' %) = 0 AW(i' %) = 0AW(i' %) = 0 FW(i' %) = 0FW(i' %) = 0

Why?Why?

Can Solve for Can Solve for ii'' by:by:

Trial and errorTrial and error Linear interpolationLinear interpolation An equation solverAn equation solver Computer programComputer program

You will need to know how to You will need to know how to interpolate in this course!interpolate in this course!

Evaluating Project with IRREvaluating Project with IRR

Compare IRR to MARR to Compare IRR to MARR to determine whether or not the determine whether or not the project is acceptable with respect project is acceptable with respect to profitability.to profitability.

IRR = IRR = i' i' MARR MARR acceptableacceptable

IRR = IRR = i' i' << MARR MARRunacceptableunacceptable

(reject)(reject)

Difficulties with the IRR MethodDifficulties with the IRR Method The IRR Method assumes that recovered The IRR Method assumes that recovered

funds are reinvested at the IRR rather than funds are reinvested at the IRR rather than the MARRthe MARR

Possible multiple IRRsPossible multiple IRRs

Why should you learn the RR Why should you learn the RR Methods?Methods?

The majority of companies favor the RR The majority of companies favor the RR methods for evaluating investment projectsmethods for evaluating investment projects

Example Problem Using IRRExample Problem Using IRR

Cost/Revenue Estimates Cost/Revenue Estimates

Initial Investment:Initial Investment: $50,000$50,000Annual Revenues:Annual Revenues: $20,000$20,000Annual Operating Costs:Annual Operating Costs: $2,500$2,500Salvage Value @ EOY 5:Salvage Value @ EOY 5: $10,000$10,000Study Period: Study Period: 5 years5 yearsMARRMARR 20%20%

Solution using IRR methodSolution using IRR method

Find i'% such that the PW(i'%) = 0Find i'% such that the PW(i'%) = 00=-50,000+17,500(P|A, i'%,5)+10,000(P|F, i'%,5)0=-50,000+17,500(P|A, i'%,5)+10,000(P|F, i'%,5)PW(20%) = 6354.50 tells us that i' > 20%PW(20%) = 6354.50 tells us that i' > 20%PW(25%) = 339.75 > 0, tells us that i'% > 25%PW(25%) = 339.75 > 0, tells us that i'% > 25%PW(30%) = -4,684.24 < 0, tells us that i'% < 30%PW(30%) = -4,684.24 < 0, tells us that i'% < 30%

25% < i' < 30%25% < i' < 30%Use linear interpolation to estimate i'% or Use linear interpolation to estimate i'% or Use “short-cut” to draw a conclusionUse “short-cut” to draw a conclusion

Linear Interpolation example Linear Interpolation example

i%i% PWPW2525 339.75339.75i'i' 003030 -4684.24-4684.24

i' = i' = 25.3% > MARR, accept25.3% > MARR, accept

Comparing Mutually Exclusive Comparing Mutually Exclusive Alternatives (MEAs) with RR MethodsAlternatives (MEAs) with RR Methods

Fundamental Purpose of Capital Fundamental Purpose of Capital Investment:Investment:• Obtain Obtain at leastat least the MARR for every dollar the MARR for every dollar

invested.invested.

Basic Rule:Basic Rule:• Spend the least amount of capital possible Spend the least amount of capital possible

unlessunless the extra capital can be justified by the the extra capital can be justified by the extra savings or benefits.extra savings or benefits.(i.e., any increment of capital spent above the (i.e., any increment of capital spent above the minimum must be able to pay its own way) minimum must be able to pay its own way)

Why not select the investment opportunity Why not select the investment opportunity that maximizes IRR? See example belowthat maximizes IRR? See example below

AA BB B-A(B-A())InvestmentInvestment -$100 -$10,000 -$100 -$10,000 --

$9,900$9,900Lump-Sum $1,000Lump-Sum $1,000 $15,000$15,000 $14,000$14,000

ReceiptReceiptNext YearNext Year

IRRIRR 900% 50%900% 50% 41.4%41.4%

If MARR = 20%, would you rather have A If MARR = 20%, would you rather have A or B if comparable risk is involved? or B if comparable risk is involved?

Comparing MEAs - using the IRR Comparing MEAs - using the IRR method cont'd.method cont'd.

If MARR = 20%, PWIf MARR = 20%, PWAA = $733 and PW = $733 and PWBB = $2,500 = $2,500 * Never simply maximize the IRR.* Never simply maximize the IRR. * Never compare the IRR to anything except * Never compare the IRR to anything except

the MARR. the MARR.

IRRIRRA->BA->B::

PWPWA->BA->B = 0 = -9,900 + 14,000(P|F, i'%, 1) = 0 = -9,900 + 14,000(P|F, i'%, 1)9,900/14,000 = (P|F, i'%, 1)9,900/14,000 = (P|F, i'%, 1)

i' = 41.4% > MARRi' = 41.4% > MARR

Rate of Return Methods for Comparing Rate of Return Methods for Comparing Alternatives Alternatives MUSTMUST use an Incremental use an Incremental Approach!Approach!

Step 1.Step 1. Rank order alternatives from least Rank order alternatives from least to greatest initial investment.to greatest initial investment.

Step 2.Step 2. Compare current feasible Compare current feasible alternative with next challenger in the alternative with next challenger in the listlist

Step 3.Step 3. Compute RR (IRR or ERR) and Compute RR (IRR or ERR) and compare with MARR. If RR < Marr compare with MARR. If RR < Marr choose the least initial investment choose the least initial investment alternative. If RR alternative. If RR MARR choose the MARR choose the greater initial investment alternativegreater initial investment alternative

Step 4.Step 4. Remove rejected alternative from Remove rejected alternative from list. Continue with next comparisonlist. Continue with next comparison

Example Problem: Example Problem: Given three Given three MEAs and MARR = 15%MEAs and MARR = 15%

11 22 33Investment (FC) -28,000 -16,000 -23,500Investment (FC) -28,000 -16,000 -23,500Net Cash Flow/yr 5,500Net Cash Flow/yr 5,500 3,300 4,800 3,300 4,800Salvage ValueSalvage Value 1,500 1,500 0 0 500500Useful LifeUseful Life 10 yrs 10 yrs 10 yrs 10 yrs10 yrs 10 yrsStudy PeriodStudy Period 10 yrs 10 yrs 10 yrs 10 yrs10 yrs 10 yrs

Use the IRR procedure to choose the best Use the IRR procedure to choose the best alternative.alternative.

Example Problem Cont.Example Problem Cont.

Step 1.Step 1. DN -> 2 -> 3 -> 1 DN -> 2 -> 3 -> 1Step 2.Step 2. Compare DN -> 2 Compare DN -> 2

cash flowscash flows InvestmentInvestment -16,000 - 0 = -16,000-16,000 - 0 = -16,000 Annual ReceiptsAnnual Receipts 3,300 - 0 = 3,300 3,300 - 0 = 3,300 Salvage ValueSalvage Value 0 - 0 = 0 - 0 = 0 0

Compute Compute IRR IRRDN->2DN->2

PW(PW(i') = 0 = -16,000 + 3,300(P|A, i') = 0 = -16,000 + 3,300(P|A, i'%, 10)i'%, 10)i'i'DN->2DN->2 15.9% 15.9%

Step 3.Step 3. Since Since i' > MARR, keep alt. 2 i' > MARR, keep alt. 2 (higher FC) as current best alternative. (higher FC) as current best alternative. Drop DN from further consideration. Drop DN from further consideration.

Step 4.Step 4. Next comparison: 2 -> 3 Next comparison: 2 -> 3 InvestmentInvestment -23,500 - (-16,000) = -23,500 - (-16,000) = --

7,5007,500 Annual ReceiptsAnnual Receipts 4,800 - 3,300 = 4,800 - 3,300 =

1,5001,500 Salvage ValueSalvage Value 500 - 0 = 500 - 0 =

500 500Computing Computing IRR IRR2->32->3 PW(PW(i') = 0 i') = 0 0= -7,500 + 1,500(P|A, 0= -7,500 + 1,500(P|A, i'%, 10) + i'%, 10) +

500(P|F, 500(P|F, i'%, 10)i'%, 10)

i'i'2->32->3 15.5% 15.5%Since Since i' > MARR, keep Alt. 3 (higher FC) i' > MARR, keep Alt. 3 (higher FC) as current best alternative. Drop Alt. 2 as current best alternative. Drop Alt. 2 from further consideration. from further consideration.

Next comparison: 3 -> 1Next comparison: 3 -> 1 cash flows cash flows InvestmentInvestment -28,000 - (-23,500) = -28,000 - (-23,500) = --

4,5004,500 Annual ReceiptsAnnual Receipts 5,500 - 4,800 = 5,500 - 4,800 = 700700 Salvage Value 1,500 - 500 = Salvage Value 1,500 - 500 =

1,0001,000

Compute Compute IRR IRR3->13->1 PW(PW(i') = 0 i') = 0 0= -4,500 + 700(P|A, 0= -4,500 + 700(P|A, i'%, 10) + 1,000(P|i'%, 10) + 1,000(P|

F, F, i'%, 10)i'%, 10)i'i'3->13->1 10.9% 10.9%

Since Since i' < MARR, keep alt. 3 (lower i' < MARR, keep alt. 3 (lower FC) as current best alternative. Drop FC) as current best alternative. Drop alt. 1 from further consideration. alt. 1 from further consideration.

Step 5.Step 5. All alternatives have been All alternatives have been considered.considered.

Recommend alternative 3 for Recommend alternative 3 for investment. investment.

Graphical Interpretation of ExampleGraphical Interpretation of Example