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True or False Questions

1. Life Cycle Cost Analysis (LCCA) does not include "User benefits resulting from the project" or "Externalities resulting from the project"

2. Unlike analysis based on net annual worth, net present worth analysis does NOT require equal analysis periods for different alternatives?

3. Life-cycle cost analysis (LCCA) uses a common period of time called "Analysis Period" to assess cost differences between alternatives so that the results can be fairly compared.

4. Whether Present Value (PV) or equivalent uniform annual cost (EUAC) is used, the decision supported by the analysis will be the

5. Remaining Service Life (RSL) value exists only if the alternative will continue in operation after the end of the analysis period, whereas salvage value requires termination.

6. Remaining service life (RSL) is service life exceeding the analysis period.

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Darryn D'Souza
Darryn D'SouzaLv10
28 Sep 2019

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Related questions

  1. 1. What is the capitalized value of an infinite series of $1,000 biennial (every two years) payments? MARR = 10%

    $4,950

    $4,025
    .

    $4,350


    $4,762

    1. 2. Frank West Enterpises would like to choose one of the following plans for expansion. Using present worth analysis, find the best alternative among the three alternatives and the Do-Nothing.

      Alt. A

      Alt. B

      Alt. C

      Initial Cost

      $10,000

      $21,000

      $8,000

      EUAB

      $6,000

      $7,000

      $4,000

      EUAC

      $1,500

      $1,000

      $2,000

      Salvage Value

      $2,000

      $8,000

      $3,000

      Life

      2 Years

      3 Years

      2 Years

      Use a MARR of 10%.

      .

      Do-Nothing
      .

      Alt. C
      .

      Alt. B
      .

      Alt. A

      3.

      1. U. S. Engineering is investigating the possibility of acquiring new automated packaging equipment at a cost of $12,000. It is expected that the equipment will have a salvage value of $1,000 at the end of its useful life of 10 years. It is determined by the plant engineering department at the company that the operation and maintenance cost will be $500 in the first year and will gradually increase every year starting year 2 at the rate of $50 until the equipment is retired. Determine the equivalent uniform annual cost (EUAC) if MARR for the company is 10%.

        $2,575.95


        $2,150.50
        .

        $2,176.60
        .

        $2,184.40

        4.

        1. The cash flows given in table below are for two different alternatives. MARR =10%

          Data

          C

          D

          Initial Cost

          $20,000

          $80,000

          Uniform Annual Benefits

          $6,000

          $10,000

          Salvage Value

          $5,000

          $20,000

          Useful Life in years

          5

          infinity

          The equivalent uniform annual worth (EUAW) of alternative N is ______________.

          Hint: Assume the Salvage Value is never realized since N is infinite.

          $2,000

          $1,920

          $1,762


          $2,476

          5.

          1. Compare the following plans using a MARR of 6%.

            Data

            Plan X

            Plan Y

            Equipment First Cost

            $50,000

            $75,000

            Annual Operation & Maintenance Cost

            $3,000

            $2,500

            Salvage Value

            $10,000

            $0

            Service Life, Years)

            25

            50


            $6,728; $7,255


            $7,255; $9,000
            .

            $6,728; $7,728
            .

            $6,728; $8,200

            6.

            1. LAEP, Inc. wants to evaluate two methods of shipping their products. The following cash flows are associated with each alternative:

              Data

              Method

              A

              B

              Life (Years)

              10

              10

              First Cost

              $700,000

              $1,512,000

              M&O Cost

              $18,000

              $9,000

              M&O Cost Gradient

              $900

              $775

              Annual Benefit

              $154,000

              $303,000

              Salvage Value

              $142,000

              $210,000

              Annual profits are based on amount of products which can ordinarily be shipped each year as a function of the amount of vehicles or service purchased with the first cost and the M&O costs.

              Using a MARR of 15%, calculate the equivalent uniform annual cash flow (EUAB - EUAC) for each alternative.

              Determine the most desirable alternative based on the results.

              $389.57, $445.90; Method A


              $445.90, $389.57; Method A
              .

              $445.90, $389.57; Method B
              .

              $389.57, $445.90; Method B
            7.

            1. For the cash flow diagram shown, which of the following equations properly calculates the uniform equivalent (A)?

              n

              0

              3

              6

              9

              12

              15

              Dollars

              $100

              $100

              $100

              $100

              $100

              $100


              A=100(A/P,i,15)+ 100(A/F,i,3)


              A=100(A/P,i,15)


              A=100(A/F,i,3) + 100(A/F,i,15)


              A=100(A/P,i,3) + 100(A/F,i,3)
              8.

              1. Three different alternatives shown in table below are being considered by U. S. Engineering systems.

                Assume that alternatives X and Z are replaced at the end of their lives.

                Data

                Alternative X

                Alternative Y

                Alternative Z

                Initial Cost

                $6,000

                $1,000

                $1,500

                Uniform Annual Benefits

                $810

                $125

                $ 230

                Useful Life in Years

                20

                infinite

                10

                MARR

                12%

                The NPW for alternative

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