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## ISYE 3025 - Week 4

by: Moriah Mattson

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# ISYE 3025 - Week 4 ISYE 3025

Moriah Mattson
Georgia Tech
GPA 3.67

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These notes cover the first two videos on Learnding Cycle 2, Fundamentals of an Economic Decision and Future Worth, Present Worth, Future Worth.
COURSE
Engineering Economy
PROF.
Kelly Bartlett
TYPE
Class Notes
PAGES
7
WORDS
CONCEPTS
Economics, Engineering
KARMA
25 ?

## Popular in Industrial Engineering

This 7 page Class Notes was uploaded by Moriah Mattson on Saturday September 24, 2016. The Class Notes belongs to ISYE 3025 at Georgia Institute of Technology taught by Kelly Bartlett in Fall 2016. Since its upload, it has received 8 views. For similar materials see Engineering Economy in Industrial Engineering at Georgia Institute of Technology.

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Date Created: 09/24/16
ISYE 3025 Engineering Economics ­ Week 4    Video ­ Fundamentals of an Economics Decision   Goals:  ­ To illustrate the fundamental choice between Benefits and Costs  ­ Develop a formal approach and notation to determining the most economical  decision  ­ Introduce the basic decision criterions of Prospective Benefit and Prospective  Cost    Introducing Economics Decisions  Assume you can either get a 1­year magazine subscription for \$15 or a 2­year  magazine subscription for \$25. And, as a special offer these prices are guaranteed for  the next four years. Which option would you prefer?    In order to determine this, we should base this decision on ​the differences​!    Then, we must break down the differences into two components.   (Benefit and cost series)  Difference = Benefit ­ Cost    The benefit and cost series represent two vectors.   These funds (F​ anB ​​ ) caC​be invested. And, we can determine which series would  accumulate more money by an equivalence calculation.       Formal Representation  ­ And economic decision criterion is hot we make economics decisions  ­ Measure of Worth:​ expression of monetary value associated with an investment  alternative  ­ Decision Rule:​ judges if the monetary value is an economical option or not    Minimum Attractive Rate of Return (MARR)  ­ We can look at the special interest rate (i) known as the MARR to determine if  the opportunity is economical or not  ­ The MARR looks at those opportunities readily available to us to invest our funds  ­ This is the minimum standard to compare alternatives to  ­ Also known as, marginal growth rate, discount rate, cutoff rate, hurdle rate, and  yield    Example  Suppose we invest \$100 at an MARR, i = 5%.     Is this an economical decision?  We cannot say if it is an economical decision without a standard to compare it to.  The MARR sets this standard...  If MARR < 5%, it would be a good investment  If MARR > 5%, it would not be a good investment         Values of the MARR  1) For individuals, the MARR typically represents the minimum attractive  opportunities to invest in money markets  2) For corporations, the MARR typically represents the minimum attractive  opportunities to invest in the company    Notation  A​jt​ net cash flow  j = index on opportunities  A​  > 0 is a net receipt  jt​ A​jt​ 0 is a net expense  A​jt​ 0 for t<0 and t>N  A​  = B​  ­ C​   jt​ jt​ jt B​jt​ A​ jt​ A​  jt​, else B​  = 0  jt​ C​jt​ A​  jt​A​  jt​, else C​  = 0  jt​   Measure of Worth  N N B  (i) =   ∑ B (1+i)  T−t and C  (i) =   ∑ C (1+i)  T−t , where i = MARR  j t = 0 j j t = 0 j     Typically, T = 0 or T = N    Decision Rule  Accept j if B​ (i) jt​​ (i), ojt​rwise reject j    Example  A manufacturer is considering buying 10 new machines, which cost \$100,000 a piece  and have a life expectancy of 5 years. The cost to install all machines is \$25,000. Each  machine will reduce labor costs by \$30,000, but will increase energy costs by \$8000.   If the MARR = 8%/year, are these machines economical?    Break down into A​  = B​  ­ C​   Rt​ Rt​ Rt   Now, we determine which series would accumulate more money.    By accepting the machines, you would be poorer by relative comparison.    Every decision has benefits and costs, and the question is   “Do the Benefits exceed the Cost?”    Video ­ Future Worth, Present Worth, Annual Worth     There are three classic economic criteria based on worth:  ­ Future Worth, FW​(i)  j​ ­ Present Worth, PW​(i)  j​ ­ Future Worth, FW​(i)  j​   Future Worth  Measure of Worth:  FW​(ij​= B​(i)j​ C​(i)j​where i = MARR  N N N =  ∑ B  (1+i)  T−t − ∑ C  (1+i)  T−t =  ∑ [B  −C  ](1+i)  T−t   t = 0 jt t = 0 jt t = 0 jt jt N =  ∑ A  (1+i)  T−t , typically T = N  t = 0 jt Decision Rule:  Accept j if FW​(i) >j​, otherwise reject    Future Worth Example  Referring back to the machine example from the previous videos. We can use these  equations for Future Worth to determine whether it is a good investment.    FW​(ij​= B​(i) j​C​(i) j​\$1,290,652 ­ \$1,506,061 = ­\$215,409 < 0, OR  FW​(0.08) = ­\$1,025,000(F/P, 0.08, 5) + \$220,000(F/A,0.08,5) = ­\$215,409 < 0  j​   Present Worth  Measure of Worth:  ­N​ PW​(ij​= FW​(i)(1+j​​ , where i = MARR  N N−t −N = [ ∑ A  (1+jt  ](1+i)    t = 0 N −t =  ∑ A  (1jt)    t = 0 Decision Rule:  Accept j if PW​(i) > j​ otherwise reject    Present Worth Example  Consider an engineer that is evaluating 3 alternative routes for a power line. Use an  i = 0.05/year and N = 15 years. Below is the relevant data:              Then, identify the cash flows.      We will compare PW​ (i) to PWR1​i).  R2​ PW​ (0.05) = ­72K + 24.9K(P/A, 0.05, 14) + 54.9K(P/F, 0.05, 15) = +200.8K  R1​ PW​ (R2​5) = ­110K + 28.6K(P/A, 0.05, 14) + 48.6K(P/F, 0.05, 15) = +196.5K    ­> Can we say Route 1 is the best choice because it is the highest number?  No, we cannot right now. We do not know if we can claim this yet, we must evaluate the  differences   Since PW​ (i) R2​W​ (i) = ­R1​K < 0, Route 1 is the more economical option.    Now, we will compare PW​ (i) to PW​ (i).  R1​ R3​ PW​ (R1​5) = ­72K + 24.9K(P/A, 0.05, 14) + 54.9K(P/F, 0.05, 15) = +200.8K  PW​ (R3​5) = ­64K + 27.3K(P/A, 0.05, 14) + 51.3K(P/F, 0.05, 15) = +230.9K    Since PW​ (i) R3​W​ (i) = +R1​1K > 0, ​Route 3 is the most economical option​.    An Equivalent Decision Rule  For any two alternatives j = A1, A2:  PW​ (i) = PW​ (i) ­ PW​ (i)  A1­A2​ A1​ A2​     Thus, equivalent decision rule is:   prefer A1 over A2 when PW​ (i) > PW​ (i) A1​en there is A2​capital rationing        (which will be talked about later)    Annual Worth  Measure of Worth:  AW​(ij​= PW​(i)(Aj​, i, N), where i = MARR  = FW​(i)j​/F, i, N)    Decision Rule:  Accept j if AW​(i) j​0, otherwise reject     AW​ (R1​5) = +19.3K > 0   AW​ (R2​5) = +18.9K > 0   AW​ (0.05) = +22.2K > 0  R3​  AW​ R2­R1​(0.05) = ­0.4K < 0   ​AW​ (0.05) = +2.9K > 0  R3­R1​

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