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Engineering Economics II > Engineering Economics II > Flashcards

Engineering Economics II Flashcards

1
Q

Amount of money paid for the use of borrowed capital or the income produced by money which has been loaned.

A

Interest

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2
Q

The percentage of money charged as interest.

A

interest rate

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3
Q

Calculated using the principal only, ignoring any interest that has been accrued in preceding 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.

A

Ordinary simple interest

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5
Q

Based on the exact number of days in a year, 365 days for an ordinary year and 366 days for a leap year.

A

Exact simple interest

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6
Q

The interest for an interest period is calculated on the principal plus total amount of interest accumulated in previous periods. Thus it means “interest on top of interest.”

A

Compound Interest

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7
Q

A graphical representation of cash flows drawn on a time scale. Cash-flow diagram for economic analysis problems is analogous to that of free body diagram for mechanics problems.

A

Cash Flow Diagram

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8
Q

positive cash flow or cash inflow

A

Receipt

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9
Q

negative cash flow or cash outflow

A

Disbursement

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10
Q

Difference between total cash receipts (inflows) and total cash disbursements (outflows) expenditures for a given period of time.

A

Cash flow

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11
Q

the interest is compounded at the end of each finite- length period, such as a month, a quarter, or a year.

A

discrete compounding

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12
Q

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

A

continuous compounding

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13
Q

It is often said that money makes money.

A

Cash flow diagram

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14
Q

Changes in the amount of money over a given period of time.

A

Time Value of Money

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15
Q

different sums of money at different times can be equal in economic value.

A

equivalence

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16
Q

the interest is compounded at the end of each finite length period, such as a month, a quarter year.

A

Discrete compounding

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17
Q

Single payment compound amount factor.

A

(1+i)^n

18
Q

single payment present worth factor.

A

(1+i)^-n

19
Q

Specifies the rate of interest and a number of interest periods in one year.

A

Nominal rate of interest

20
Q

The actual or exact rate of interest on the principal during one year.

A

Effective rate of interest

21
Q

Assumed that cash payment occur one a year, but the compounding is continuous throughout the year.

A

continuous compounding

22
Q

the difference between Future Worth and Present Worth

A

Discount

23
Q

difference between the present worth (the amount received for the paper in cash) and the worth of the paper at some time in the future (the face value of the paper or principal)

A

Discount

24
Q

interest paid in advance.

A

Discount

25
Q

The discount on one unit of principal per unit of time.

A

rate of discount

26
Q

Is the increase in the prices for goods and services from one year to another, thus decreasing the purchasing power of money.

A

Inflation

27
Q

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 Value

28
Q

a series of equal payments occurring at equal periods of time.

A

annuity

29
Q

one where the payments are made at the end of each period.

A

Ordinary Annuity

30
Q

one where the first payment is made several periods after the beginning of the annuity.

A

Deferred Annuity

31
Q

one where the payments are made at the beginning of each period.

A

Annuity Due

32
Q

an annuity in which the payments continue indefinitely.

A

Perpetuity

33
Q

One of the most important applications of perpetuity.

A

Capitalized Cost

34
Q

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

35
Q

Capitalized cost=

1st cost + Present worth of perpetual operation and or maintenance.

A

Case 1: no replacement, only maintenance and or operation every period.

36
Q

Capitalized cost=

1st cost + Present worth of perpetual replacement.

A

Case 2- replacement only, no maintenance and or operation.

37
Q

Capitalized cost=

1st cost + Present worth of perpetual operation and or maintenance + Present worth of cost of perpetual replacement.

A

Case 3- Replacement, maintenance and or operation every period.

38
Q

method of repaying a debt, the principal and interest included, usually by a series of equal payments at equal interval of time.

A

Amortization

39
Q

Involves the analysis of problems for manufacturing a product or rendering a service based on present or immediate cost

A

present Economy

40
Q

The decrease in the vale of physical property with the passage of time

A

depreciation

Engineering Economics II (2 decks)

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