IE 492 Engineering Economics PRESENT WORTH CONT Additionally you can also use

Ie 492 engineering economics present worth cont

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IE 492 Engineering Economics
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PRESENT WORTH (CONT.) Additionally, you can also use Present Worth method to get an equivalent of different cash flows in today’s dollar / at the end of year 0. PW is represented by: Example: You decide to buy $1,000 worth of technology stock a year form now. You also expect to increase purchases by $200 per year for the next 9 years. What would the present worth of all stocks be if they yield, a uniform dividend rate of 10% throughout entire investment period, if the price per share remains constant? In above example there are two cash flows. First: $1,000 that occurs at the end of year 1. Second: A gradient series ($200) that occurs for the future years. 5 Summation of present worth’s of all cash flows. IE 492 Engineering Economics
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PRESENT WORTH (CONT.) In order to get the present worth of these two cash flows, we need to obtain present worth of cash flow #1, and add present worth of cash flow #2 to it. PW of cash flow #1 (base amount) = 1,000 x (P/A,10%,10) = 1,000 x (A/P,10%,10) -1 = 1,000 x 6.1444 = $6,144.40 PW of cash flow #2 (gradient) = 200 x (P/G,10%,10) = 200 x (3.7255)x (0.16275) -1 = 200 x 3.7255 x 6.1444 = $4,578.19 Total Present Worth (PW) of Cash Flows is $6,144.40 + $4,578.19 = $10,722.59 6 IE 492 Engineering Economics
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FUTURE WORTH (FW) Future worth (FW) is determined by calculating the future value of the Present worth (PW). Basically, multiplying PW by (F/P) factor at the interest rate. Interest rate is the MARR. FW = PW x (F/P,i%,n) where i% = MARR. In case PW is used to calculate FW, ‘n’ value in F/P depends on which time period was used to determine PW. Additionally, we can also calculate FW by calculating future value of each cash flow, similar to the approach we took in PW. In this case, FW is represented by: 7 Summation of future worth’s of all cash flows.
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