Engineering Economy (Reviewer) Flashcards

1
Q

Analysis and evaluation of the factors that will affect the economic success of engineering projects to the end that a recommendation can be made which will ensure the best use of capital

A

Engineering Economy

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2
Q

the amount of money paid for the use of borrowed capital

The income earned by money which has been loaned

A

Interest

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3
Q

Calculated using the principal only, ignoring any interest that
had been accrued in preceding interest periods

A

Simple Interest

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4
Q

computed on the basis of 12 months of 30 days each or 360 days a year (1 interest period = 360 days)

A

Ordinary Simple Interest

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5
Q

computed on the basis of exact number of days in a year: 365 days for ordinary year or 366 days for leap year (1 interest period = 365 or 366 days)

A

Exact Simple Interest

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6
Q

Simple graphical representation of cash flows drawn on a time scale

A

Cash Flow Diagram

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7
Q

Interest on top of interest

A

Compound Interest

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8
Q

specifies the rate of interest and the number of interest period/s in one year

A

Nominal Rate of Interest

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9
Q

actual rate of interest on the principal for one year

A

Effective Rate of Interest

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10
Q

of several future cash payments maybe defined as the sum
of the values of the future payments discounted at a given rate for the corresponding periods to the present.

A

Present Worth

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11
Q

is obtained by setting the sum of the values on a certain comparison or focal date of one set of obligations equal to the sum of the values on the same date of another set of obligations

A

Equation of Values

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12
Q

It is assumed that cash payments occur once per year, but, compounding is continuous throughout the year.

A

Continuous Compounding

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13
Q

the difference between the present worth and the worth at some time in the future.

A

Discount

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14
Q

the discount on one unit principal for one unit of time.

A

Rate of Discount

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15
Q

It is a series of equal payments made at equal intervals of time.

A

Annuity

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16
Q

equal payments are made at the end of each period

A

Ordinary Annuity

17
Q

start of equal payments is deferred (delayed) for a certain number of periods

A

Deferred Annuity

18
Q

equal payments that go on forever

A

Perpetuity

19
Q

One of the most important applications of Perpetuity

of any property is the sum of the first cost and the present worth of all costs of replacement, operation and maintenance for a long time or forever.

A

Capitalized Cost

20
Q

No replacement, only maintenance and/or operation every period

Capitalized Cost = First Cost + Present Worth of Cost of Perpetual Maintenance and/or Operation

A

Case 1

21
Q

Replacement only, no maintenance and operation

Capitalized Cost = First Cost + Present Worth of Cost of Perpetual Replacement

A

Case 2

22
Q

Replacement, maintenance and/or operation every period

Capitalized Cost = First Cost + Present Worth of Cost of Perpetual Maintenance and/or Operation + Present Worth of Cost of Perpetual Replacement

A

Case 3