mm 314 engineering economy--- 2. interest rate and economic equivalence

8/12/2019 Mm 314 Engineering Economy--- 2. Interest Rate and Economic Equivalence

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MM 314 Engineering Economy

INTEREST RATE ANDECONOMIC EQUIVALENCE

SPRING SEMESTER2011-2012MM 314 ENGINEERING ECONOMY

DR. MUSTAFA YURDAKUL



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WHAT DOES ’Money has a time value’ MEAN?

EVERY COMPANY DOES ITS JOB (SELLS ITS PRODUCTS;

PROVIDES SERVICES) TO INCREASE ITS WEALTH. TOINCREASE ITS WEALTH A COMPANY MAKES

INVESTMENTS, DEVELOPS PROJECTS SO THAT MONEY IS

NOT HIDDEN. INSTEAD, MONEY IS USED TO EARN MORE

MONEY.

Capital refers to wealth in the form of money or propertythat can be used to produce more wealth.

Engineering economy studies involve the commitment ofcapital for extended periods of time. Commitment ofcapital for extended periods of time (more than one year)increases the importance of time value of money.



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A dollar today is worth more than a dollar one or more yearsfrom now. We prefer to receive a dollar NOW instead ofreceiving it one year later or two years later even we assume

there is no inflation.

WHY?

WHEN WE RECEIVE ONE DOLLAR NOW, WE CAN USE

ONE DOLLAR IN PROJECTS AND MAKE ONE DOLLAR 1.10

DOLLAR ONE YEAR LATER. SO, ONE DOLLAR BECOMES

1.10 DOLLAR ONE YEAR LATER, 1.21 DOLLAR TWO YEARS

LATER SO ON. SO THAT WE PREFER TO RECEIVE MONEY

EARLIER THAN LATER.



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$1,000 becomes $1,331 at 10% earning rate in three years



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If you have no current use for one dollar (no projects or

investments to use money to EARN more money (no earningrate– 1 dollar will remain as 1 dollar in the coming years)), you

can let someone else use it to earn more money.

WHY SHOULD A FIRM LET SOMEONE ELSE USE ITSMONEY?---Because the lender receives a portion ot the borrower’searnings. The borrower pays INTEREST to the lender let himuse its money.

The charge for using money is called INTEREST. Cost of using

the borrowed money is measured by interest rate a percentagethat is periodically applied and added to an amount of moneyover specified length of time



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THEN WE CAN SAY THAT-----

IF OUR EARNING RATE FROM PROJECTS IS 10%, 1 DOLLAR NOW IS EQUIVALENT TO 1.10 DOLLAR ONEYEAR LATER.

OR

WE MAY LEND 1 DOLLAR NOW AT AN INTEREST RATEOF 10% AND RECEIVE 1.10 DOLLAR ONE YEAR LATER.

THEN 1 DOLLAR NOW IS EQUIVALENT TO 1.10 DOLLAR

ONE YEAR LATER.

SO WHAT?



TIME VALUE OF MONEY (1.10 DOLLAR IS EQUIVALENTTO 1 DOLLAR NOW ) LEADS US THE ABILITY TO REPLACE

1.10 DOLLAR ONE YEAR LATER WITH ONE DOLLAR NOW .

FOR EXAMPLE, AT 10% INTEREST RATE (=EARNING RATE)

WE CAN REPLACE OPTION 1 WITH OPTION 2. IT MEANS

THEY ARE ECONOMICALLY EQUIVALENT TO EACH

OTHER. WE DO NOT PREFER ONE OPTION OVER THEOTHER OPTION.



NOW WE CAN START TO COMPARE DIFFERENTCASH FLOWS AND DETERMINE WHICH ONE IS

ECONOMICALLY PREFERRED USING ECONOMICEQUIVALENCE CONCEPT.



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BEFORE GOING ANY FURTHER WE SHOULD MAKE SOME

POINTS CLEAR SUCH AS CASH FLOWS OCCUR AT

DICRETE POINTS



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SHOW THAT P= $ 2000 IS EQUIVALENT TO F= $4,287.18EIGHT YEARS LATER AT THE INTEREST RATE OF 10%

USING THE FORMULATIONS



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SHOW THAT P= $ 10 IS EQUIVALENT TO F= $20 FIVE

YEARS LATER AT THE INTEREST RATE OF i=14.87%.



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SHOW THAT P= $ 6000 IS EQUIVALENT TO F= $1200

N=3.80 YEARS LATER AT THE INTEREST RATE OF 20%.



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Number of years required to double an initial investment atvarious interest rates?



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Decomposition of uneven cash flow series

HOW DO YOU CALCULATE P (= THE EQUIVALENT VALUEOF THE FOLLOWING CASH FLOW)?



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WAIT A MINUTE! WHAT DOES IT MEAN (P/F,10%,1)?WE CAN EITHER USE FORMULATIONS

OR TABLES.



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AN EXAMPLE--- Foreman family decided to start investing totheir son Eric’s college fund. They started it by depositing$1,000 on January 1, 2008 to the Washington Mutual Bank.

They plan to deposit $2,000 on June 31, 2010 and $3,000 on June31, 2013. If they manage to invest according to that plan, howmuch money will they have on January 1, 2018 if the interest istaken as 8% per each year.



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UP TO NOW WE HAVE ONLY SINGLE PAYMENT CASH FLOWS. BUT INREAL LIFE WE HAVE OTHER TYPES OF CASH FLOWS. Five types of cashflows: (a) Single cash flows, (b) equal (uniform) payment series, (c) linear-gradient series, (d) geometric-gradient series, and (e) irregular payment series.



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EQUAL (UNIFORM) PAYMENT SERIES



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EXAMPLE C l l t A t k t i l t t



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EXAMPLE--- Calculate A to make payments eqıivalent toreceipts at the interest rate of 10%.



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EXAMPLE--- Calculate C to make cash flow 1 equivalent tocash flow 2 at the interest rate of 10%.



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EXAMPLE--- How much money has to deposited at he end ofyear 1 through year 18 to receive $40,000 in years 18 through 21at the interest rate of ….%.

EXAMPLES OF UNIFORM (EQUAL) PAYMENT SERIES



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EXAMPLES OF UNIFORM (EQUAL) PAYMENT SERIESX = ?; C = ?; P (t = 0) = ?

LINEAR (ARITHMETIC) GRADIENT SERIES



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LINEAR (ARITHMETIC) GRADIENT SERIES



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A TYPICAL EXAMPLE FOR LINEAR (ARITHMETIC) GRADIENT SERIES



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A TYPICAL EXAMPLE FOR LINEAR (ARITHMETIC) GRADIENT SERIES

CALCULATE P (t = 0) OF THE FOLLOWING CASH FLOW



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ANOTHER EXAMPLE



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ANOTHER EXAMPLE



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TWO EXAMPLES FOR LINEAR (ARITHMETIC) GRADIENT SERIES



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1

1

1 (1 ) (1 ), eğer

/(1 ), eğer

N N g i A i g

P i g

NA i i g



Yaz 2008END 321

Mühendislik Ekonomisi



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mm 314 engineering economy--- 2. interest rate and economic equivalence

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