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Systems Eng. Lecture 2 Begin Reading Chapter 2 35- 45, Problems 1, 3, 5 by Wednesday, January 24, 2001.

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Presentation on theme: "Systems Eng. Lecture 2 Begin Reading Chapter 2 35- 45, Problems 1, 3, 5 by Wednesday, January 24, 2001."— Presentation transcript:

1 Systems Eng. Lecture 2 Begin Reading Chapter 2 35- 45, Problems 1, 3, 5 by Wednesday, January 24, 2001

2 Homework Format Name, Date, Assignment number, should be written in the upper right hand corner Typed or Neatly Written Assignments that are calculation problems should be organized 1.Given values 2.Unknowns 3.Associated Equations 4.Calculations 5.A boxed answer

3 Homework If this format is not followed your homework will be returned to you without a grade.

4 Time Value of Money Money today is not the same as money in the Future! Example Which would you prefer, $100 today or $100 in one year? Today!!!!!!!! If you had it today you could use it today, or lend it to someone that would pay you interest for the privilege of using your money

5 Earning power - money can be put to work! $1 today is worth more than $1 at some future time. This is because money Costs, and is measured by an interest rate. i = interest rate = the cost or price of money, expressed as a percentage per period of time.

6 Earning Power Plan for receipts or disbursements - yields a cash flow pattern over a specified period of time Future amount of money - result of interest accruing on principal over a certain number of interest periods

7 Many transactions involve interest: School loans, car loans, savings accounts, credit cards, etc. You could say that we all have an interest in interest :).

8 There are other elements involving transactions involving interest rates: Principal - an initial amount of money Interest period - determines how frequently interest is calculated (monthly, quarterly, yearly) Number of interest periods - length of time that marks the duration of the transaction (30 year mortgage, 5 year car loan

9 Variables are associated with these concepts in order to perform calculations: A n = a discrete payment or receipt occurring at the end of some interest period i = interest rate per period N = total number of interest periods

10 Variables P = present value or present worth of a sum of money (designated as time zero for analysis) F = future sum of money A = end of period receipt or payment V n = equivalent sum of money at the end of a specified period n that consider time value of money

11 More Variables G = uniform period-by period increase or decrease in cash flow; arithmetic gradient g = uniform rate of period-by period increase in cash flow; geometric gradient

12 Cash Flow Diagram Engineering economics uses mathematical and economic techniques to systematically analyze situations with alternative courses of action. Initially alternative courses are compared by their possible outcomes in terms of money, or consequences.

13 Money consequences (which we all prefer to be positive) can be shown by a cash flow table.

14 Ex. year by year consequences of buying and owning a used car YearCash Flow Begin. of first yr.-$4500 Car purchased now, minus sign indicates a disbursement. End of year 1-350 End of year 2-350 Maintenance costs End of year 3-350 End of year 4-350 +2000 Car sold at the end of 4 th year

15 Cash flow may be represented graphically 4500 4 2000 really 1650 321 350 each yr

16 Methods of Calculating Interest

17 Simple Interest Interest earned on only the principal amount during each period. Thus the total amount of interest earned from P with a simple interest i for N periods = I = (iP)N A future amount F = P + I = P(1 + iN) (not used often)

18 Compound interest Suppose a present sum is invested P for one year at i. At the end of the year you would receive P +Pi, the principal and whatever interest you have earned on the principal. Factoring P from P + Pi, we get P (1+i). What about the next years

19 Amount at beginning +Interest for period = Amount at End of period of the period 1st yr. PPi P(1+i) 2nd yr.P(1+i)Pi(1+i) P(1+i) 2 3rd yr.P(1+i) 2 Pi(1+i) 2 P(1+i) 3 4th yr.P(1+i) n-1 Pi(1+i) n-1 P(1+i) n

20 Example Earlier example, P = $500, i = 10%, n = 3 years, what is the future amount in three years? F = 500(1 +.01) 3 = 665.5 Notation for this equation reads, Find Future given P, i, n (F/P,i,n)

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