TM 661 Engineering Economics for Managers InvestmentWorth Investment Worth
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- Slide 1
- TM 661 Engineering Economics for Managers InvestmentWorth
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 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
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- Investment Worth Alternatives u NPW( MARR ) > 0Good
Investment
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- Investment Worth Alternatives u NPW( MARR ) > 0Good
Investment u EUAW( MARR ) > 0Good Investment
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- Investment Worth Alternatives u NPW( MARR ) > 0Good
Investment u EUAW( MARR ) > 0Good Investment u IRR > MARRGood
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 12%
Is it worth it to the company to buy the machine?
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- 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
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- 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
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- 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
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- 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**
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- 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
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- 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
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- 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
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- Alternately AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774) = -
$810 < 0 NO GOOD 3,000 5 0 4,000 16,000
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- 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**
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- 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
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- 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*)
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- 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
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- 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
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- 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
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- Public School Funding
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- 216% 16 yrs
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- School Funding 12316 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 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) =
ln(2.16) =.7701
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- School Funding 12316 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 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
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- 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%
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- School Funding 12316 100 216 We know i = 4.93%, is that
significant growth?
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- School Funding 12316 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 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%
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- 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?
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- 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.
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- Summary u NPW > 0 Good Investment
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- Summary u NPW > 0 Good Investment u EUAW > 0 Good
Investment
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- Summary u NPW > 0 Good Investment u EUAW > 0 Good
Investment u IRR > MARR Good Investment
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- Summary u NPW > 0 Good Investment u EUAW > 0 Good
Investment u IRR > MARR Good Investment Note: If NPW > 0 EUAW
> 0 IRR > MARR
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- 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*)
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- 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
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- 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
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- Spreadsheet Example
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- 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).
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- 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
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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: 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 =
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- 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
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- 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
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- 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
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- 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
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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
- Slide 56
- 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)
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- 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)
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- 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
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- 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
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- 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
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- 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
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- 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
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- Capitalized Costs Example: $10,000 into an account @ 20% / year
A = P(A/P, i, ) P = A / i.. AAA 10,000
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- 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.
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- 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
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- 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.
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- 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
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- 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
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- 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
- Slide 70
- 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 doesnt, 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?
- Slide 73
- How Many to Change a Bulb? How many Industrial Engineers does
it take to change a light bulb? None, IEs only change dark
bulbs!!!!!