TM 661 Engineering Economics for Managers

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TM 661 Engineering Economics for Managers

• Investment Worth

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Topics

• Minimum Attractive Rate of Return (MARR)
• Present Worth Analysis (PW)
• Annual Worth Analysis (AW)
• Future Worth Analysis (FW)
• Internal Rate of Return Analysis (IRR)
• External Rate of Return Analysis (ERR)
• Payback Period (PP)
• Capitalized Cost (CC)
• Examples

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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 lt 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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What is MARR?

• MARR is the minimum interest rate that encourages
  the investor to invest in financial projects. It
  is not the rate of return on the project. MARR
  shows how much an investor is interested in
  investment.

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Question

• Who is more interested in investment A person
  with high MARR or a person with low MARR?
• Remember to invest in a project, ROR should be
  greater or equal to MARR.

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MARR

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

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Investment Worth Alternatives

• NPW(MARR) gt 0 Good Investment

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Investment Worth Alternatives

• NPW(MARR) gt 0 Good Investment
• EUAW(MARR) gt 0 Good Investment

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Investment Worth Alternatives

• NPW(MARR) gt 0 Good Investment
• EUAW(MARR) gt 0 Good Investment
• IRR gt 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 12
• Is it worth it to the company to buy the machine?

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Present Worth
16,000

• NPW -16 3(P/A,12,5) 4(P/F,12,5)

12
Present Worth
16,000

• NPW -16 3(P/A,12,5) 4(P/F,12,5)
• -16 3(3.6048) 4(.5674)

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Present Worth
16,000

• NPW -16 3(P/A,12,5) 4(P/F,12,5)
• -16 3(3.6048) 4(.5674)
• -2.916
• -2,916

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Future Worth Analysis

• It is an extension of PW method.
• FW PW (F/P, i, n)
• In previous example,
• FW -2,916 (F/P, 12,5)
• -2,916(1.7623)
• - 5138.87

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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) gt 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)

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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)

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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

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Alternately

• AW(12) PW(12) (A/P, 12, 5)
• -2.92 (.2774)
• - 810 lt 0 NO GOOD

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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 gt 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 gt MARR OK Investment
• Alt
• FW(i) 0 At(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 gt 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) -16 3(P/A, i, 5) 4(P/F, i, 5)

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Internal Rate of Return

• Example
• PW(i) -16 3(P/A, i, 5) 4(P/F, i, 5)

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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 lt MARR

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Public School Funding
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Public School Funding
216
16 yrs
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School Funding
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School Funding
(1i)16 2.16 16 ln(1i) ln(2.16) .7701
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School Funding
(1i)16 2.16 16 ln(1i) ln(2.16)
.7701 ln(1i) .0481
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School Funding
(1i)16 2.16 16 ln(1i) ln(2.16)
.7701 ln(1i) .0481 (1i) e.0481 1.0493
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School Funding
(1i)16 2.16 16 ln(1i) ln(2.16)
.7701 ln(1i) .0481 (1i) e.0481 1.0493
i .0493 4.93
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School Funding
We know i 4.93, is that significant growth?
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School Funding
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? Suppose inflation 3.5 over that same
period.
d 1.4
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School Funding
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 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 gt 0 Good Investment

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Summary

• NPW gt 0 Good Investment
• EUAW gt 0 Good Investment

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Summary

• NPW gt 0 Good Investment
• EUAW gt 0 Good Investment
• IRR gt MARR Good Investment

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Summary

• NPW gt 0 Good Investment
• EUAW gt 0 Good Investment
• IRR gt MARR Good Investment
• Note If NPW gt 0 EUAW gt 0
• IRR gt MARR

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IRR Problems
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
wish to find the Internal Rate-of-Return (IRR).
PWR(i) PWC(i) 4,100(1i)-1 2,520(1i)-3
1,000 5,580(1i)-2
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IRR Problems
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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 gt 0
• 0 , else
• -At , At lt 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 gt MARR
• Investment is Good

?
?
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External Rate of Return

• Example MARR 15
• Rt (1 .15) 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

?
?
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External Rate of Return

• Example MARR 15
• Rt (1 .15) 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 gt MARR

?
?
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External Rate of Return

• Example MARR 15
• Rt (1 .15) 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 gt MARR Good Investment

?
?
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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 (negative flows)
• Evaluation
• Method 1 If SIR(t) gt 1 Good Investment
• Method 2 If SIR(t) gt 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 lt 1.0

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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 lt 1.0

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Payback Period

• Is the length of time required to recover the
  initial investment.
• Method
• Find smallest value m such that
• where
• Co initial investment
• m payback period investment

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Payback Period

• Example

c
n
o
1 16 3

2 16 6
3 16 9
4 16 12
5 16 19
m 5 years
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Capitalized Cost (CC)

• Capitalized cost (CC) is the present worth of an
  alternative that will last forever.
• Example Operating cost of a dam which last for
  100
• CC A/i
• A Annual Cost
• i Interest Rate

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Capitalized Costs

• Example 10,000 into an account _at_ 20 / year
• A P(A/P, i, )
• P A / i

?
?
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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

• 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 10,000 100/.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

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Class Problem 2
A 100 1,000(A/F,8,5) 100
1,000(.1705) 270.5

• Compute an annuity for the 1,000 every 5 years

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Class Problem 2
Pc 10,000 270.5/.08 13,381
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• BREAK

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Example (2.30, Page 74)

• David borrows 25,000 at 8 compounded
• quarterly. He wishes to repay the money with
• 10 equal semiannual installments. What must
• be the size of payment if the first payment is
• made 1 year after obtaining the 25,000?

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Solution

• r 8
• m4
• i (quarter) 8 / 4
• 2
• ieff (1r/m)m - 1
• Isemi-annual (12)2 - 1
• 4.04

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Solution

• A25,000(F/P,4.04 ,1)(A/P, 4.04, 10)
• 25,000 (1.0404)(0.1235)
• 3213.19

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Example (3.38, Page 75)

• Lynne borrows 15,000 at 1.5 /month. She
• desires to repay the money using equal monthly
• payments over 36 months. Lynne makes for of
• such payments and decides to pay off the
• remaining debt with one lump sum payment at
• the time for the fifth payment. What should be
• the size of the payment if interest rate is truly
• compounded at a rate of 1.5 /month?

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Solution

• A 15,000 (A/P, 1.5, 36)
• 15,000(0.0362)
• 543
• 15,000 543(P/A, 1.5, 4) X (P/F, 1.5, 5)
• 15,000 543 (3.8544) X (0.9283)
• X(0.9283) 12907.06
• X13,903.98

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Example (3.9, Page 106)

• The inflation rates for 4 years are forecast to
  be
• 6, 5, 4, and 5 over the same period. If
• labor is projected to be 1000, 1500, 2000, and
• 1000 in then-current dollars, during these year,
• determine the present worth equivalent for
• labor cost.

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Solution

• First calculate combined rate for each year.
• Year j d i
• 3 6 9.18
• 3 5 8.15
• 4 4 8.16
• 5 5 10.25
• ij d jd
• I (for year 1) 3 6 (3)(6)
• 9.18

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Solution

• Next, return back the all cash flow to year 0.
• Year CF i1 i2 i3
  i4 PW
• 1 1000 1.0918
  915.92
• 2 1500 1.0815 1.0918
  270.34
• 2000 1.0816 1.0815 1.0918
  1566.01
• 1000 1.1025 1.0816 1.0815 1.0918 710.21
  
• Total
  4462.48

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Example (3.18, Page 107)

• Dr. Ramirez wishes to purchase a bond having
• a face value of 10,000 and a bond rate of 15
• payable annually. The bond has a remaining life
• of 8 years. In order to earn a 20 rate of return
• on the investment, what amount should be paid
• for the bond?

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Solution

• I v. b / c
• I (10,000) (15) / 1
• I 1500
• P 1500(P/A, 20, 8) 10,000(P/F, 20,8)
• 1500(3.8372) 10,000(0.2326)
• 8081.80