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Engineering Economy class. Please, when answering this problem, it will be more than appreciated if you can present the whole process (equations) before solving it. Just to understand better everything. Example: Present Value = 20,000(P/F, i, n)+45,000(A/F, i, n)(P/F, i, n)... Thanks in advance!

There are two manufacturing processes that can be used by a company to reduce energy losses in the equipment. The following cost information for these two processes is known:

  Process A Process B
Initial Cost $170,000 $220,000
Operation and Maintenance Cost $7,000 quarterly $5,000 quarterly
Useful Life (Life Cycle) 2 years 4 years
Residual Value $30,000 $42,000

By doing the following analysis, select which manufacturing process must be selected (A or B):

a) Find the Present Value with an interest rate of 12% annually computed quarterly.

b) Quarterly equivalent uniform value with an interest rate of 8% annually computed quarterly

c) Capitalized Cost with an interest rate of 14% annually computed quarterly

d) Which manufacturing process must be selected

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Darryn D'Souza
Darryn D'SouzaLv10
28 Sep 2019

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Related questions

Compare the alternatives shown below on the basis of their capitalized costs, using an interest rate of 12% per year, compounded quarterly. (10 points)

 

Alternative

Alternative

Alternative
 

A

B

C

First

-200,000

-300,000

-700,000

cost, $

      

Quarterly

36,000

32,000

40,000

income,

      

$/quarter

      

Salvage

50,000

120,000

100,000

value, $

      

Life, years

3

6

∞

It was my answer but I got points off because of the incorrect interest rate. Please find out where I went wrong.

Effective interest rate = (1+0.12/4)4-1 = 12.55%

AWA = -200,000(A/P, 12.55%, 3) + 36,000 + 50,000(A/F, 12.55%,3)

= -84,057.92+36,000+14,7439.47 = -33,318.45

Pcap A = AWA / i

= -33,318.45 / 0.1255 = -265,485.65

AWB = -300,000(A/P, 12.55%, 6) +32,000 + 120,000(A/F, 12.55%, 6)

= -74,100+32,000+14,583.15 = -27,516.85

PcapB = -27,516.85 / 0.1255 = -219,257.77

PcapC = -700,000+40,000 / 0.1255 = -381,275

1. The first cost of a bridge that is expected to have an infinite life is $10 million. The maintenance cost of the bridge will be $150,000 per year. At an interest rate of 10% per year, the annual worth of the bridge is equal to:

a. $3,200,000

b. $1,000,000

c. $1,150,000

d. $850,000

2. When calculating the i* value, all the net positive cash flow are assumed to be reinvested at:

a. The current market interest rate

b. The i* rate

c. The company's MARR

d. The company's cost of capital

3. If you have the present worth of an alternative with a 5-year life, you can obtain its annual worth by:

a. Multiplying the PW by i

b. Multiplying the PW by (A/F, i, 5)

c. Multiplying the PW by (P/A, i, 5)

d. Multiplying the PW by (A/P, i, 5)

4. If you have the capitalized cost of an alternative that has an infinite life, you can get its annual cost over a very long number of years by:

a. Multiplying the capitalized cost by i

b. Multiplying the capitalized cost by (A/F,i, n)

c. Dividing the capitalized cost by (P/A, i,n)

d. Dividing the capitalized cost by i

5. An investment of $85,000 resulted in uniforms income of $15,000 per year for 15 years. The rate of return of the investment is equal to:

a. 10.6%

b. 15.6%

c. 13.7%

d. 23.5%

  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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