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**TM 661 Engineering Economics for Managers**

Investment Worth

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Investment Worth MARR Suppose a company can earn 12% / annum in U. S. Treasury bills No way would they ever invest in a project earning < 12% Def: The Investment Worth of all projects are measured at the Minimum Attractive Rate of Return (MARR) of a company.

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**MARR MARR is company specific MARR based on**

utilities - MARR = % mutuals - MARR = % new venture - MARR = % MARR based on firms cost of capital Price Index Treasury bills

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**Investment Worth Alternatives**

NPW(MARR) > 0 Good Investment

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**Investment Worth Alternatives**

NPW(MARR) > 0 Good Investment EUAW(MARR) > 0 Good Investment

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**Investment Worth Alternatives**

NPW(MARR) > 0 Good Investment EUAW(MARR) > 0 Good Investment IRR > MARR Good Investment

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Present Worth Example: Suppose you buy and sell a piece of equipment. Purchase Price $16,000 Sell Price (5 years) $ 4,000 Annual Maintenance $ 3,000 Net Profit Contribution $ 6,000 MARR % Is it worth it to the company to buy the machine?

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**Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5) 4,000 4,000 6,000**

16,000 6,000 3,000 5 4,000 3,000 5 4,000 16,000 NPW = (P/A,12,5) + 4(P/F,12,5)

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**Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)**

16,000 6,000 3,000 5 4,000 3,000 5 4,000 16,000 NPW = (P/A,12,5) + 4(P/F,12,5) = (3.6048) + 4(.5674)

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**Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)**

16,000 6,000 3,000 5 4,000 3,000 5 4,000 16,000 NPW = (P/A,12,5) + 4(P/F,12,5) = (3.6048) + 4(.5674) = = -$2,916

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**Annual Worth Annual Worth (AW or EUAW) AW(i) = PW(i) (A/P, i%, n)**

= [ At (P/F, i%, t)](A/P, i%, n) AW(i) = Annual Worth of Investment AW(i) > 0 **OK Investment**

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**Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)**

3,000 5 4,000 16,000 Repeating our PW example, we have AW(12) = -16(A/P,12,5) (A/F,12,5)

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**Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)**

3,000 5 4,000 16,000 Repeating our PW example, we have AW(12) = -16(A/P,12,5) (A/F,12,5) = -16(.2774) (.1574)

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**Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)**

3,000 5 4,000 16,000 Repeating our PW example, we have AW(12) = -16(A/P,12,5) (A/F,12,5) = -16(.2774) (.1574) = -.808 = -$808

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**Alternately AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774)**

3,000 5 4,000 16,000 AW(12) = PW(12) (A/P, 12%, 5) = (.2774) = - $810 < 0 NO GOOD

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**Internal Rate of Return**

IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment**

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**Internal Rate of Return**

IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t

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**Internal Rate of Return**

IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t PWrevenue(i*) = PWcosts(i*)

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**Internal Rate of Return**

Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) 3,000 5 4,000 16,000

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**Internal Rate of Return**

Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) 3,000 5 4,000 16,000

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**Internal Rate of Return**

Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) i* = 5 1/4 % i* < MARR 3,000 5 4,000 16,000

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Public School Funding

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Public School Funding 216% 16 yrs

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**School Funding F = P(F/P,i*,16) (F/P,i*,16) = F/P = 2.16**

3 16 100 216 F = P(F/P,i*,16) (F/P,i*,16) = F/P = 2.16 (1+i*)16 = 2.16

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**School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 1 2 3 16**

100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701

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**School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701**

3 16 100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481

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**School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701**

3 16 100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481 (1+i*) = e.0481 =

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**School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701**

3 16 100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481 (1+i*) = e.0481 = i* = = 4.93%

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**School Funding We know i = 4.93%, is that significant growth? 1 2 3 16**

100 216 We know i = 4.93%, is that significant growth?

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**School Funding We know i = 4.93%, is that significant growth?**

1 2 3 16 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period.

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**School Funding We know i = 4.93%, is that significant growth?**

1 2 3 16 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period. d i j 1 0493 0350 . d= 1.4%

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**School Funding We know that d, the real increase in school funding**

1 2 3 16 100 216 We know that d, the real increase in school funding after we discount for the effects of inflation is 1.4%. So schools have experienced a real increase in funding?

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**School Funding We know that d, the real increase in school funding**

1 2 3 16 100 216 We know that d, the real increase in school funding after we discount for the effects of inflation is 1.4%. So schools have experienced a real increase in funding? Rapid City growth rate 3% / yr.

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Summary NPW > Good Investment

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Summary NPW > Good Investment EUAW > Good Investment

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**Summary NPW > 0 Good Investment EUAW > 0 Good Investment**

IRR > MARR Good Investment

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**Summary NPW > 0 Good Investment EUAW > 0 Good Investment**

IRR > MARR Good Investment Note: If NPW > EUAW > 0 IRR > MARR

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**Internal Rate of Return**

IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t PWrevenue(i*) = PWcosts(i*)

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**Internal Rate of Return**

Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) 3,000 5 4,000 16,000

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**Internal Rate of Return**

Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) i* = 5 1/4 % i* < MARR 3,000 5 4,000 16,000

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Spreadsheet Example

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Spreadsheet Example

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Spreadsheet Example

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Spreadsheet Example

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Spreadsheet Example

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**IRR Problems Consider the following cash flow diagram. We**

1,000 4,100 5,580 2,520 n 1 2 3 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR).

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**IRR Problems Consider the following cash flow diagram. We**

1,000 4,100 5,580 2,520 n 1 2 3 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR). PWR(i*) = PWC(i*) 4,100(1+i*)-1 + 2,520(1+i*)-3 = 1, ,580(1+i*)-2

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**IRR Problems NPV vs. Interest ($5) $0 $5 $10 $15 $20 $25 0% 10% 20%**

30% 40% 50% 60% Interest Rate Net Present Value

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**External Rate of Return**

Purpose: to get around a problem of multiple roots in IRR method Notation: At = net cash flow of investment in period t At , At > 0 0 , else -At , At < 0 0 , else rt = reinvestment rate (+) cash flows (MARR) i’ = rate return (-) cash flows Rt = Ct =

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**External Rate of Return**

Method find i = ERR such that Rt (1 + rt) n - t = Ct (1 + i’) n - t Evaluation If i’ = ERR > MARR Investment is Good

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**External Rate of Return**

Example MARR = 15% Rt ( ) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 1 2 3 1,000 4,100 5,580 2,520

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**External Rate of Return**

Example MARR = 15% Rt ( ) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 ERR > MARR 1 2 3 1,000 4,100 5,580 2,520

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**External Rate of Return**

Example MARR = 15% Rt ( ) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 ERR > MARR Good Investment 1 2 3 1,000 4,100 5,580 2,520

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**Critical Thinking IRR < MARR a. IRR < MARR < ERR**

b. IRR < ERR < MARR c. ERR < IRR < MARR

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**Critical Thinking IRR > MARR a. IRR > MARR > ERR**

b. IRR > ERR > MARR c. ERR > IRR > MARR

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**Savings Investment Ratio**

Method 1 Let i = MARR SIR(i) = Rt (1 + i)-t Ct (1 + i)-t = PW (positive flows) - PW (negative flows)

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**Savings Investment Ratio**

Method #2 SIR(i) = At (1 + i) -t Ct (1 + i) -t SIR(i) = PW (all cash flows) PW (negative flows)

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**Savings Investment Ratio**

Method #2 SIR(i) = At (1 + i) -t Ct (1 + i) -t SIR(i) = PW (all cash flows) PW (negitive flows) Evaluation: Method 1: If SIR(t) > 1 Good Investment Method 2: If SIR(t) > 0 Good Investment

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**Savings Investment Ratio**

16 1 2 3 4 5 7 Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) = < 1.0

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**Savings Investment Ratio**

16 1 2 3 4 5 7 Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) = < 1.0

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**Payback Period Method Find smallest value m such that where**

Co = initial investment m = payback period investment m t = 1 R t C o

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Payback Period R t n å Example c n o 16 1 2 3 4 5 7 m = 5 years

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**Capitalized Costs Example: $10,000 into an account @ 20% / year**

A = P(A/P, i, ) P = A / i A 10,000

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Class Problem A flood control project has a construction cost of $10 million, an annual maintenance cost of $100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep.

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**Class Problem Pc = 10,000 + A/i = 10,000 + 100/.08 = 11,250**

10,000 A flood control project has a construction cost of $10 million, an annual maintenance cost of $100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep. Pc = 10,000 + A/i = 10, /.08 = 11,250 Capitalized Cost = $11.25 million

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Class Problem 2 Suppose that the flood control project has major repairs of $1 million scheduled every 5 years. We now wish to re-compute the capitalized cost.

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**Class Problem 2 Compute an annuity for the 1,000 every 5 years:**

10,000 1,000 Compute an annuity for the 1,000 every 5 years:

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**Class Problem 2 Compute an annuity for the 1,000 every 5 years:**

10,000 1,000 Compute an annuity for the 1,000 every 5 years: A = ,000(A/F,8,5) = ,000(.1705) = 270.5

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Class Problem 2 10,000 1,000 10,000 Pc = 10, /.08 = 13,381

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**How Many to Change a Bulb?**

How does Bill Gates change a light bulb?

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**How Many to Change a Bulb?**

How does Bill Gates change a light bulb? He doesn’t, he declares darkness a new industry standard!!!

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**How Many to Change a Bulb?**

How many Industrial Engineers does it take to change a light bulb?

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**How Many to Change a Bulb?**

How many Industrial Engineers does it take to change a light bulb? None, IE’s only change dark bulbs!!!!!

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