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    Concepts in Engineering Economics

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    1. The value of P (The present worth) is most nearly
    $ 367
    $ 371
    $ 377
    $ 386

    2. The value of P (the present worth) is most nearly:
    $80
    $83
    $85
    $89

    3. The value of Q is most nearly
    $193
    $243
    $258
    $272

    4. The present value, P is most nearly:
    $265
    $275
    $290
    $305

    5. (Be careful with the time line.) The value of P is most nearly
    $ 1640
    $ 1670
    $ 1690
    $ 1710

    6. Consider the following Cash Flow table for questions 6 - 8...
    Year Cash Flow
    0 - 30,000
    1 + 10,000
    2 + 10,000
    3 + 20,000
    4 + 20,000
    5 - 5,000
    The interest rate is 10 %

    The Present Worth of costs is most nearly
    $ 29,100
    $ 31,400
    $ 33,100
    $ 34,400

    7. The Present Worth of Benefits is most nearly
    $44,200
    $44,600
    $45,100
    $46,000

    8. The Net Present Worth is most nearly
    $ 12,300
    $ 12,600
    $ 12,900
    $ 13,300

    9. The correct Analysis Period is:
    10 years
    15 years
    30 years
    150 years

    10. The Present Worth (PW) of the cost (- Installed cost and operating cost + Salvage value) of Westinghome is most nearly
    - $ 87,400
    - $ 92,700
    - $ 93,600
    - $ 95,100

    11. The Present Worth (PW) of the cost ( -Installed cost and operating cost + Salvage Value) of Itis is most nearly:
    - $ 92,500
    - $ 92,700
    - $ 94,600
    - $ 95,100

    12. The recommendation should be to accept the bid of (Points : 1)
    Westinghome
    Itis

    13. The annual income from a rental house is $ 14,000. The annual expenses are $ 4,000. The house is bought for $ 90,000. At the end of eight years the house is sold for $ 140,000. For an interest rate of 8 %, the Net Present Worth of this investment is most nearly
    $ 42,400
    $ 42,900
    $ 43,100
    $ 45,400

    14. A machine is purchased for $ 10,000. Its life is six years and there is no salvage value. Annual maintenance is paid at the end of each year. The first year maintenance is $ 2,000 and maintenance increases by 15 % each year. The interest rate is 8 %. The Present Value of the cost is most nearly
    $ 21,900
    $ 22,400
    $ 23,100
    $ 23,800

    15. The present worth of the cost for alternative B is most nearly
    - $ 1,105
    - $ 1,215
    - $ 1,295
    - $ 1,305

    16. Which course of action should be selected?
    Alternative A
    Alternative B

    17. The present worth of costs is most nearly:
    $ 6,000
    $ 6,025
    $ 6,150
    $ 6,200

    18. The present worth of benefits is most nearly:
    $ 6,000
    $ 6,025
    $ 6,100
    $ 6,150

    19. The appropriate recommendation should be:
    Invest in the equipment
    Don't invest in the equipment

    20. The NPW of alternative B is most nearly
    $3.50
    $6.80
    $7.70
    $9.20

    21. The NPW of alternative A is most nearly
    $3.50
    $6.80
    $7.70
    $9.20

    22. The NPW of alternative C is most nearly
    $3.50
    $6.80
    $7.70
    $12.40

    23. The NPW of alternative D is most nearly
    $3.50
    $7.70
    $9.20
    $32.40

    24. The NPW of alternative E is most nearly:
    $7.70
    $9.20
    $12.40
    $32.40

    25. The NPW of alternative F is most nearly:
    $3.50
    $7.70
    $12.40
    $32.40

    26. Which alternative should be selected?
    A
    B
    C
    D
    E
    F

    27. The amount of money deposited 50 years ago is most nearly:
    $2,600
    $5,300
    $7,100
    $20,400

    28. The capitalized cost of the dam, including maintenance is most nearly
    $ 2.2 million
    $ 2.3 million
    $ 2.4 million
    $ 2.5 million

    29. The maximum equal annual withdrawal is most nearly
    $ 1150
    $ 1180
    $ 1270
    $ 1390

    © BrainMass Inc. brainmass.com October 10, 2019, 5:26 am ad1c9bdddf
    https://brainmass.com/economics/macroeconomics/concepts-engineering-economics-504966

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    1.
    P=50/(1+10%)+100/(1+10%)^2+150/(1+10%)^3+200/(1+10%)^4=$377.399
    Option C i.e. $377 is correct.

    2.
    P=30/(1+15%)^0+20/(1+15%)^1+20/(1+15%)^2+30/(1+15%)^3=82.2397
    Option B i.e. $83 is correct.

    3.
    Periodic Cash inflows=C=$50
    Number of periods=n=5
    PV of cash inflows=Co+ C/r*(1-1/(1+r)^n)=50+50/12%*(1-1/(1+12%)^5)=230.2388
    PV of cash outflows=Q/(1+12%)^1
    So, Q/1.12=230.2388
    Q=257.8675
    Option C i.e. $258 is correct

    4.
    P=50/(1+10%)^4+120/(1+10^)^5+50/(1+10%)^6+120/(1+10%)^7+50/(1+10%)^8+120/(1+10%)^9=$272.68
    Correct option is $275

    5.
    (P/G,I,n)=((1+i)^n-(i*n)-1)/(i^2*(1+i)^n)=((1+6%)^7-(6%*7)-1)/(6%^2*(1+6%)^7)= 15.44969
    PV of cash inflows=100*15.44969=1544.969
    PV of cash outflows=P/(1+6%)^1
    P/(1+6%)^1=1544.969
    P=1544.969*1.06=1637.667
    Option A i.e. $1640 is correct.

    6.
    The interest rate is 10 %

    PV of costs=-30000/(1+10%)^0-5000/(1+10%)^5=-33104.61
    Option C i.e. $33100 is correct

    7.
    PV of ...

    Solution Summary

    There are 29 basic problems related to time value of money concepts. Solutions to given problems illustrate the methodology to find out the present value and NPV of cash flows in different cases. Solutions are derived with the help of suitable formulas.

    $2.19