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ISYE 3025 Engineering Economy - Week 5

by: Moriah Mattson

ISYE 3025 Engineering Economy - Week 5 ISYE 3025

Marketplace > Georgia Institute of Technology > Industrial Engineering > ISYE 3025 > ISYE 3025 Engineering Economy Week 5
Moriah Mattson
Georgia Tech
GPA 3.67

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About this Document

These notes cover the second two videos in Learning Cycle 2. The notes cover Internal Rate of Return, Benefit/Cost Ratio, and Payback Period.
Engineering Economy
Kelly Bartlett
Class Notes
Economics, Engineering, economy, industrial
25 ?




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Popular in Industrial Engineering

This 6 page Class Notes was uploaded by Moriah Mattson on Monday September 26, 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 9 views. For similar materials see Engineering Economy in Industrial Engineering at Georgia Institute of Technology.


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Date Created: 09/26/16
ISYE 3025 - Week 5 Video - Internal Rate of Return - IRR is the most popular decision making criteria in practice - However, it is also the most misunderstood and most misused - Notice that there are TWO Decision Rules for this decision criteria Measure of Worth ij​= {i | “PW​(i)”j​ 0} or *​ ij​= {i | “FW​(i)”j​ 0} - You must solve for i​ , which can be done with guess-and-check j​ Decision Rules Accept ​investment​ j if i​ > MARR, otherwise reject j​ Accept ​borrowing​ j if i​ < MARR, j​herwise reject Example- An Investment Say you can buy a machine for $20,000 that has an estimated life and salvage value of 5 years and $5000. You will save $7000 per year with this machine, but it will cost you $1000 per year in maintenance and upkeep. Assume the MARR = 9%. What is the IRR? -> Make a cash flow table *​ * -> Using the equations i​ = {i | “PW​(ij​ = 0}, we can gj​ss-and-check to find i​ j​ PW​(​0.15​) = 2598.81 <- this number is greater than zero, therefore we need to j​ pick a larger IRR PW​(​0j​5​) = -2225.92 <- this number is less than zero, therefore we need to pick a smaller IRR, and that the IRR will be between 0.15 and 0.25 PW​(​0j​0​) = -20K + 6K(P/A, 0.20, 4) + 11K(P/F, 0.20, 5) PW​(​0.20​) = -20K + 15,534.41 + 4420.65 = ​-46.94 j​ ij​≈ 0.20 ​<- we can approximate IRR to be 20% 0.20 > 0.09 (the MARR)​, therefore we can say that we ​Accept the investment and buying the machine is an economical course of action -> This means that by investing the $20K in the machine, you will get a return of (capital growth) 20% each year for the 5 years in addition to the return of the $20K. This is more economical in comparison to not buying the machine and investing the $20K at a MARR of 9%. Example - A Borrowing Now, assume you can only buy the above machine if you finance a loan from the supplier. For a down payment of $5000, you can finance the remainder to be repaid in 5 years with annual payments of $3561. What is the IRR? -> Make a net cash flow table -> Using the equations i​ = {i | “PW​(i)” = 0}, we can guess-and-check to find i​ * j​ j​ j​ -> We find: PW​(i) = 15K - 3561(P/A, i, 5) j​ ij​≈ 0.06 < 0.09 (the MARR), ​ therefore we can say that we ​Accept​ the investment. Notice that since this was a ​borrowing​ example, we want the IRR to be l​ ess​ than the MARR. Now let us go back to the magazine subscription example from last notes: -> Let us find the IRR for A​ 2YR-1YR *​ *​ -1​ *​ -2​ *​ -3​ PW​ (2-1​ = -10 + 15(1 + i​ )​ - 10(1 + i​ )​ + 15(1 + i​ )​ = 0 -> We find: *​ j​ = 0.50 > 0.03 (the MARR) *​ **NOTICE: if we found PW​ (i​ ) instea1-2​ 2-1, we would get the same answer Do we ​Accept or Reject?​ Is it ​Investing or Borrowing? We must make a better definition of Investing and Borrowing -> Let’s go back to the machine example of Investing. -> We will make graphs depicting each step/change in capital -> We can see that the capital remains invested at ​all times​ until year 5, where nothing remains. This classifies this transaction as an i ​ nvestment​. -> We can see that the capital remains borrowed at ​all times​ until year 5, where nothing remains. This classifies this transaction as a b ​ orrow​. **Not all graphs will be so uniform. A transaction can still be an investment or borrowing if it has sign changes. However! If it has more than one sign change, it is possible for it to have more than one “root” (IRR). If you are solving for IRR and you find more than one IRR, you cannot classify the transaction into investment or borrowing and therefore cannot determine if you should accept or reject. Therefore, if you find more than one root, simply change to using a ​different decision criteria​ ​(FW, PW, AW, etc.), which do not rely on this classification. **Also note that IRR is ​not​ additive/have an equivalent decision rule. Therefore, when comparing alternatives you CANNOT just compare the IRR. You must take the incremental cash flow calculation (e.g. A​ not (A​ - A​ )) 2YR-1YR​ 2YR​ 1YR​ Video - Benefit/Cost Ratio, and Payback Period Benefit/Cost Ratio:​ The ratio of the prospective benefits and the prospective costs or the Present Worth of the benefits and the Present Worth of the costs. Measure of Worth​: N B ji) = ∑ Bjt1+i)−t BCR (i)j= N=0 , where i = MARR and T = 0 or N (the planning horizon) C ji) = ∑ Cjt1+i)−t t=0 Decision Rule Accept j if BCR​(i) j​1.0, otherwise reject j. This means that the benefits exceed the costs since the ratio is greater than 1.0 Example This appears to suggest that the BCR and PW are indicating opposite choices since BCR(A​ ) 1​BCR(A​ ) ​bu2​​PW(A​ ) < PW1​​ ). ​We n2​d to do an incremental cash flow stream calculation​. **Notice: BCR (just like IRR) has ​no equivalent decision rule/is not additive Payback Period -> Often used in conjunction with other methods we have learned -> This criterion is designed to determine what point in time when one “just” earns back the initial investment. Everything after the Payback Period would be profit. Measure of “Worth” (actually, liquidity) * N N−t N =j{N|[∑ A (1+i)jt = 0]}, where i is either the MARR or sometimes simply t=0 zero, in which case it is called the “Undiscounted Payback Period” Decision Rule Accept investment j if N​ < N​j​ max​ otherwise reject (for multiple alternatives, prefer the shortest payback period) -> This decision may not necessarily lead you to the same decision on economic attractiveness that the other methods would lead you to. Example 1 -> We find: *​ ij​= 0.43 PW​ (MARR = 0.25) = 3686 <- This investment is economical 2-1​ N​A1​*​= 3​ <- We ​accept​ this if our maximum Payback Period is greater than 3 years, and ​reject​ if our maximum Payback Period is less than 3 years. **This calculation is different from all of the others that we have done so far because it is not based on the same type of calculation. It only takes into account cash flow up to the Payback Period. In this example, you can see that it does not take into account year 4 and year 5. Example 2 *​ 3 < N​ A2​​< 4 -> In this example, we can see that the Payback Period occurs between years 3 and 4.


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