TM 661 Engineering Economics for Managers InvestmentWorth Investment Worth.

TM 661 Engineering Economics for Managers InvestmentWorth Investment Worth

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.

MARR MARR is company specific u utilities - MARR = 10 - 15% u mutuals - MARR = 12 - 18% u new venture - MARR = 20 - 30% MARR based on u firms cost of capital u Price Index u Treasury bills

Investment Worth Alternatives u NPW( MARR ) > 0Good Investment

Investment Worth Alternatives u NPW( MARR ) > 0Good Investment u EUAW( MARR ) > 0Good Investment

Investment Worth Alternatives u NPW( MARR ) > 0Good Investment u EUAW( MARR ) > 0Good Investment u IRR > MARRGood Investment

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 12% Is it worth it to the company to buy the machine?

Present Worth 16,000 6,000 3,000 5 0 4,000 NPW= -16 + 3(P/A,12,5) + 4(P/F,12,5) 16,000 3,000 5 0 4,000

Present Worth 16,000 6,000 3,000 5 0 4,000 NPW= -16 + 3(P/A,12,5) + 4(P/F,12,5) = -16 +3(3.6048) + 4(.5674) 16,000 3,000 5 0 4,000

Present Worth 16,000 6,000 3,000 5 0 4,000 NPW= -16 + 3(P/A,12,5) + 4(P/F,12,5) = -16 +3(3.6048) + 4(.5674) = -2.916 = -$2,916 16,000 3,000 5 0 4,000

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

Annual Worth; Example Repeating our PW example, we have AW(12)= -16(A/P,12,5) + 3 + 4(A/F,12,5) 3,000 5 0 4,000 16,000

Annual Worth; Example Repeating our PW example, we have AW(12)= -16(A/P,12,5) + 3 + 4(A/F,12,5) = -16(.2774) + 3 + 4(.1574) 3,000 5 0 4,000 16,000

Annual Worth; Example Repeating our PW example, we have AW(12)= -16(A/P,12,5) + 3 + 4(A/F,12,5) = -16(.2774) + 3 + 4(.1574) = -.808 = -$808 3,000 5 0 4,000 16,000

Alternately AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774) = - $810 < 0 NO GOOD 3,000 5 0 4,000 16,000

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

Internal Rate of Return 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 = A t (1 + i*) n - t

Internal Rate of Return 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 = A t (1 + i*) n - t PW revenue (i*) = PW costs (i*)

Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) 3,000 5 0 4,000 16,000

Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) i* = 5 1 / 4 % i* < MARR 3,000 5 0 4,000 16,000

Public School Funding

216% 16 yrs

School Funding 12316 100 216 F = P(F/P,i *,16) (F/P,i *,16) = F/P = 2.16 (1+i * ) 16 = 2.16

School Funding 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701

School Funding 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481

School Funding 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481 (1+i * ) = e.0481 = 1.0493

School Funding 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481 (1+i * ) = e.0481 = 1.0493 i* =.0493 = 4.93%

School Funding 12316 100 216 We know i = 4.93%, is that significant growth?

School Funding 12316 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period.

School Funding 12316 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period. d ij j 1 04930350 1... d= 1.4%

School Funding 12316 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?

School Funding 12316 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.

Summary u NPW > 0 Good Investment

Summary u NPW > 0 Good Investment u EUAW > 0 Good Investment

Summary u NPW > 0 Good Investment u EUAW > 0 Good Investment u IRR > MARR Good Investment

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

Internal Rate of Return 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 = A t (1 + i*) n - t PW revenue (i*) = PW costs (i*)

Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) 3,000 5 0 4,000 16,000

Internal Rate of Return Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) i* = 5 1 / 4 % i* < MARR 3,000 5 0 4,000 16,000

Spreadsheet Example

IRR Problems 1,000 4,100 5,580 2,520 n 0123 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR).

IRR Problems 1,000 4,100 5,580 2,520 n 0123 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR). PW R (i * ) = PW C (i * ) 4,100(1+i * ) -1 + 2,520(1+i * ) -3 = 1,000 + 5,580(1+i * ) -2

IRR Problems NPV vs. Interest ($5) $0 $5 $10 $15 $20 $25 0%10%20%30%40%50%60% Interest Rate Net Present Value

External Rate of Return Purpose: to get around a problem of multiple roots in IRR method Notation: A t = net cash flow of investment in period t A t, A t > 0 0, else -A t, A t < 0 0, else r t = reinvestment rate (+) cash flows (MARR) i = rate return (-) cash flows R t = C t =

External Rate of Return Method find i = ERR such that R t (1 + r t ) n - t = C t (1 + i ) n - t Evaluation If i = ERR > MARR Investment is Good

Example MARR = 15% R t (1 +.15) n - t = C t (1 + i ) n - t 4,100(1.15) 2 + 2,520 = 1,000(1 + i ) 3 + 5,580(1 + i ) 1 i =.1505 External Rate of Return 0 1 2 3 1,000 4,100 5,580 2,520

Example MARR = 15% R t (1 +.15) n - t = C t (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 External Rate of Return 0 1 2 3 1,000 4,100 5,580 2,520

Example MARR = 15% R t (1 +.15) n - t = C t (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 External Rate of Return 0 1 2 3 1,000 4,100 5,580 2,520

Critical Thinking IRR < MARR a.IRR < MARR < ERR b.IRR < ERR < MARR c.ERR < IRR < MARR

Critical Thinking IRR > MARR a.IRR > MARR > ERR b.IRR > ERR > MARR c.ERR > IRR > MARR

Savings Investment Ratio Method 1 Let i = MARR SIR(i) = R t (1 + i) -t C t (1 + i) -t = PW (positive flows) - PW (negative flows)

Savings Investment Ratio Method #2 SIR(i) = A t (1 + i) -t C t (1 + i) -t SIR(i) = PW (all cash flows) PW (negative flows)

Savings Investment Ratio Method #2 SIR(i) = A t (1 + i) -t C t (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

Savings Investment Ratio Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) 16 =.818 < 1.0 16 0 123 4 5 3 3 33 7

Payback Period Method Find smallest value m such that where C o = initial investment m = payback period investment m t = 1 R t C o

Payback Period Example m = 5 years n c o 1 16 3 3 16 9 4 16 12 5 16 19 2 16 6 16 0 123 4 5 3 3 33 7 R t n 0

Capitalized Costs Example: $10,000 into an account @ 20% / year A = P(A/P, i, ) P = A / i.. AAA 10,000

Class Problem u 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.

Class Problem 1 2 3 4... 100 100 10,000 P c = 10,000 + A/i = 10,000 + 100/.08 = 11,250 u 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. Capitalized Cost = $11.25 million

Class Problem 2 u 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.

Class Problem 2 Compute an annuity for the 1,000 every 5 years: 1 2 3 4 5... 100 100 100 100 100 10,000 1,000

Class Problem 2 u Compute an annuity for the 1,000 every 5 years: 1 2 3 4 5... 100 100 100 100 100 10,000 1,000 A = 100 + 1,000(A/F,8,5) = 100 + 1,000(.1705) = 270.5

Class Problem 2 1 2 3 4 5... 170.5 170.5 10,000 P c = 10,000 + 270.5/.08 = 13,381 1 2 3 4 5... 100 100 100 100 100 10,000 1,000

How Many to Change a Bulb? How does Bill Gates change a light bulb?

How Many to Change a Bulb? How does Bill Gates change a light bulb? He doesnt, he declares darkness a new industry standard!!!

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TM 661 Engineering Economics for Managers InvestmentWorth Investment Worth.

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