Evaluation Alternatives Present Worth Analysis

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Evaluation AlternativesPresent Worth Analysis

• Companies constantly evaluate whether or not to
  pursue projects.
• Mutually Exclusive Projects several projects
  proposed to address the same need. Only one of
  the projects can be selected.
• Independent Projects projects that do not
  compete, and are selected merely on their
  economic value.
• Do Nothing (DN) Option projects are often
  compared to the option of taking no action.

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Present Worth Analysis

• Project Types
• Revenue each alternative project being
  evaluated generate costs and revenues. These
  alternatives usually involve the purchase of new
  systems and equipment in order to increase
  revenue. Both the cost streams and revenue
  streams vary by alternative.
• Service Each alternative has only cost cash
  flow estimates. These projects are typically for
  safety, or are government mandated projects.

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Present Worth Analysis Equal Life Alternatives

• One alternative Calculate the present worth
  (PW) at the MARR. If PW gt 0, the requested MARR
  is met or exceeded and the alternative is
  financially viable.
• Two or more alternatives calculate the PW of
  each alternative at the MARR. Select the
  alternative with the PW value that is numerically
  largest. (If all PW are negative, and do nothing
  is an alternative, then do nothing.)

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Present Worth Analysis Equal Life Alternatives

• Revenue Example You are evaluating the purchase
  of a two income properties. You expect a MARR of
  15. You have enough funds for both purchases.
• 75K Home 37.5K Home
• Purchase Price 15,000 7,500
• Annual Maint. 6,000 4,000
• Annual Income 7,500 5,000
• Resale (after
• Expenses) 90,000 40,000
• Life, years 15 15

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Present Worth Analysis Equal Life Alternatives

• 75K Home
• PW -15,000 1500(P/A,15,15)
  90,000(P/F,15,15)
• PW 4832
• 37.5K Home
• PW -7,500 1000(P/A, 15,15)
  40,000(P/F,15,15)
• PW 3263
• What if 8 MARR was used?
• PW (75K Home) 26,207
• PW (37.5K Home) 13,667

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Present Worth Analysis Equal Life Alternatives

• Service Example You are evaluating the purchase
  of a new or used car that needs to last you for
  only 5 years.
• New Car Used Car
• Purchase Price 20,000 10,000
• Annual Maint. 500 1,000
• Resale Value 8,000 4,000
• Life, years 5 5

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Present Worth Analysis Equal Life Alternatives

• Use 8 for i since 8 is the expected rate of
  return if money invested in stock market rather
  than purchasing a vehicle.
• New Car
• PW -20,000 - 500(P/A,8,5) 8000(P/F,8,5)
• PW -16,552
• Used Car
• PW -10,000 -1000(P/A,8,5) 4,000(P/F,8,5)
• PW -11,270

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Present Worth Analysis Different Life
Alternatives

• The Present Worth of alternatives must be
  compared over the same number of years.
• If project alternative have different service
  lives, the equal service requirement can be
  satisfied by
• Compare the alternative over a period of time
  equal to the least common multiple (LCM) of their
  lives.
• Compare the alternative using a study period of
  length n, which does not necessarily take into
  consideration the useful lives of the
  alternatives. This period n is called the
  planning horizon.

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Present Worth Analysis Different Life
Alternatives

• Revenue Example You are evaluating the upgrade
  of some production equipment to increase
  productivity. You are considering two
  alternatives. Company policy dictates a MARR of
  20.
• Alternative 1 Alternative 2
• Purchase Price 200,000 100,000
• Annual Maint. 5,000 2,000
• Productivity Imp.
• (per year) 40,000 20,000
• Life, years 10 12

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Present Worth Analysis Different Life
Alternatives

• Revenue Example Because life alternatives are
  different, you decide to use a 10 year planning
  horizon and estimate a resale value of
  alternative of 5000 for the remaining 2 year of
  life.
• Alternative 1
• PW -200,000 35,000(P/A,20,10)
• PW -53,262
• Alternative 2
• PW -100,000 18,000(P/A,20,10)
  5000(P/F,20,10)
• PW -23,727
• Which alternative should you choose?
• Do nothing.

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Present Worth Analysis Different Life
Alternatives

• Service Example You are evaluating the purchase
  of a new or used cars that needs to last you for
  10 years.
• New Car Used Car
• Purchase Price 20,000 10,000
• Annual Maint. 750 1,000
• Resale Value 4,000 4,000
• Life, years 10 5

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Present Worth Analysis Different Life
Alternatives

• Use 8 for i since 8 is the expected rate of
  return if money invested in stock market rather
  than purchasing a vehicle.
• New Car
• PW -20,000 - 750(P/A,8,10)
  4000(P/F,8,10)
• PW -23,180
• Used Car
• Buy used car in year 0 Buy
  used car in year 5
• Sell car in year 5
  
• PW -10,000 4,000(P/F,8,5)
  -10,000(P/F,8,5) 4000(P/F,8,10)
  -1000(P/A,8,10)
• PW -18,941

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Future Worth Analysis

• Future worth analysis is similar to present worth
  analysis, expect that all cash flows are
  normalized to some future point in time.
• Future worth is often used if an asset is to be
  sold at some future point in time, but before its
  expected life is reached. The future worth would
  be an indicator of how much the asset could be
  sold for at that future point in time (of course
  this assumes the buyer expects the same cash
  flows you anticipate).

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Future Worth Analysis

• Example A company is considering selling off its
  power generation plants in 5 years. The cash
  flow projection over the next 5 years for this
  power generation operation unit is depicted
  below. What sales price in year 5 (future
  worth)must be received to achieve the companys
  ROR of 15 per year.
• FW -60(F/P,15,4) 25(F/P,15,3)
  50(F/P,15,2) 75(F/P,15,1) 50

FW
75
50
50
25
0 1 2 3 4 5 6 7
8
60
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Capitalized Cost

• Capitalized cost (CC) is the present worth of an
  alternative that will last forever. Examples
  include University endowments, and large public
  sector projects (dams, bridges, tollroads, etc.).
• CC is related to P/A formula
• P A(P/A,i,inf.)
• As n approaches infinity, CC A/i.
• This result should make common sense. For
  example if an endowment was invested at 10
  interest, and 1000 was to be withdrawn every
  year indefinitely, then 1000/.1 10,000 must
  be the present amount in the endowment.

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Capitalized Cost

• The following examples demonstrates how to obtain
  the capitalized cost of an asset which contains
  both recurrent and non-recurrent cash flows.
• A toll-road was just completed at a cost of 1.5
  billion, with major maintenance expenditures of
  500 million forecast every 10 years. Annual
  receipts minus maintenance results in a positive
  cash flow of 150 million. What is the present
  worth, assuming i 5?

150
150
0 1 2 3 4 5 6 7
8 9 10 11 20
500
1500
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Capitalized Cost
150
150
150


0 1 2 3 4 5 6 7
8 9 10 11 20 ..
500
1500

• 1) Distribute the 500million every 10 years to
  an annual cost.
• A1 -500(A/F, 5,10) -39.75 million.
• Therefore AT 110.25 million.
• 2) CC -1,500 110.25/.05 705 million

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Payback Period Analysis

• The payback period, np, is the estimated time,
  usually in years, it will take for the estimated
  revenues and other economic benefits to recover
  the initial investment and a stated rate of
  return i.
• In other words, find np that satisfies the
  following equation
• Or if all end of year cash flows are equal,
• where NCF is the net cash flow in period t, and
  DP is the initial downpayment or cash flow at
  time 0

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Payback Period Analysis

• Example A pharmaceutical company anticipates
  RD cost of 1.5 Billion for the development of a
  new drug. In addition, production startup costs
  are estimated at 1.0 Billion. Annual marketing
  costs are expected to be 50 million, annual
  production costs are 100 million, and annual
  sales are expected to be 500 million. What is
  the payback period for an ROR of 10?
• DP 2,500 Million
• A 350 million
• 2,500 350(P/A, 10,n)
• (P/A,10,n) 7.14
• n 13.1

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Payback Period Analysis

• Caution Payback period does not necessarily
  indicate one alternative being preferable to
  another alternative.
• Example Using ROR of 15
• Machine 1 Machine 2
• DP 12,000 8,000
• Annual
• NCF 3000 1000 (year 1-5)
• 3000 (year 6-14)
• Max Life 7 14
• (years)

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Payback Period Analysis

• Example
• Machine 1 0 -12,000 3000(P/A,15,np)
• np 6.57
• Machine 2 0 -8,000 1000(P/A,15,5)
• 3,000(P/A,15,np-5)(P/F,15,5)
• np 9.52
• Using LCM of 14 years.
• PW 1 -12,000 - 12,000(P/F,15,7)
  3000(P/A,15,14) 663
• PW 2 -8,000 1000(P/A,15,5)
  3000(P/A,15,9)(P/F,15,5)
• 2470

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Life Cycle Cost (LCC)

• LCC analysis is the present worth (PW) of an
  alternative at a stated MARR, covering all costs
  from the early stages of design through the final
  stages of phase-out and disposal.
• Example A pharmaceutical company anticipates
  RD cost of 1.5 Billion for the development of a
  new drug. In addition, production startup costs
  are estimated at 1.0 Billion. Annual marketing
  costs are expected to be 50 million, annual
  production costs are 100 million, and annual
  sales are expected to be 500 million. At the
  end of production, the plant and equipment can be
  sold for 100 million. This drug is to be phased
  out after 20 years. What is the LCC assuming a
  10 MARR.

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Life Cycle Cost (LCC)

• Example A pharmaceutical company anticipates
  RD cost of 1.5 Billion for the development of a
  new drug (in years 12). In addition, production
  startup costs are estimated at 1.0 Billion (in
  year 34). Annual marketing costs are expected
  to be 50 million, annual production costs are
  100 million (both starting in year 5). At the
  end of production, the plant and equipment can be
  sold for 100 million (year 24). This drug is to
  be phased out after 20 years of production. What
  is the LCC assuming a 10 MARR.
• (in millions)
• PW 750(P/F,10,1) 750(P/F,10,2)
  500(P/F,10,3) 500(P/F,10,4)
  150(P/A,10,20)(P/F,10,4) - 100(P/F,10,24)

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Present Worth of Bonds

• Bonds are financial instruments for raising
  capital. In other words, in order to finance
  major projects, the government or corporations
  issue bonds in return for cash.
• The borrower (corporation) promises to pay the
  face value of the bond upon maturity (V), and
  agrees to pay interest or dividends (I) at
  periodic times. Expected dividend payments are
  quarterly or semi-annually (c 4 or 2). The
  Interest is determined using the stated bond
  coupon rate (b).

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Present Worth of Bonds

• Example To find the interest payment on a 1000
  US treasury bond stated as paying 5 quarterly.
• V 1000
• b 5
• c 4
• Therefore I 1000(.05)/4 12.5 per quarter

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Present Worth of Bonds

• How much should you purchase a bond for (PW)?
• To find the present worth (to you) of a bond,
  perform the following
• Determine I, the interest per payment period.
• Construct the cash flow diagram, including the
  dividend payments and the face value payment upon
  maturity.
• Establish a MARR.
• Calculate PW.

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Present Worth of Bonds

• Example Find the PW on a 1000 US treasury bond
  stated as paying 5 quarterly with a maturity
  date 10 years from now. Assume a MARR of 12,
  compounded quarterly.
• I 1000(.05)/4 12.5 per quarter.
• n 104 40 quarters.
• i 12/4 3
• PW 12.5(P/A,3,40) 1000(P/F,3,40)
  595.54