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## ISYE 3025 Engineering Economy - Week 6

by: Moriah Mattson

11

0

8

# ISYE 3025 Engineering Economy - Week 6 ISYE 3025

Marketplace > Georgia Institute of Technology > Industrial Engineering > ISYE 3025 > ISYE 3025 Engineering Economy Week 6
Moriah Mattson
Georgia Tech
GPA 3.67

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These notes cover the last two required videos for Learning Cycle Two, which are on Multiple Alternatives.
COURSE
Engineering Economy
PROF.
Kelly Bartlett
TYPE
Class Notes
PAGES
8
WORDS
CONCEPTS
Economics, economy, industrial, Engineering
KARMA
25 ?

## Popular in Industrial Engineering

This 8 page Class Notes was uploaded by Moriah Mattson on Tuesday September 27, 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 11 views. For similar materials see Engineering Economy in Industrial Engineering at Georgia Institute of Technology.

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Date Created: 09/27/16
ISYE 3025 Engineering Economy - Week 6 Video - Multiple Alternatives, Part 1 Overview: - Define the possible relationships between investment alternatives and give examples - Demonstrate the selection process for investment alternatives Independent Alternatives -> Investment alternatives where the selection of one does not restrict the selection of another Example: Alternative A - Replace a gas-fired boiler with a more fuel efficient unit Alternative B - Replace plant electric control system with “smart” system that reduces the peak load from 450 to 350 kW - There is very little interaction between these two alternatives technically - Therefore, we can say they are independent We can consider our options to be: Φ = Do Nothing A = Replace gas-fired boiler B = Replace plant electric control systems C = (A+B) = Replace Both Mutually Exclusive Alternatives -> Investment alternatives where the selection of one precludes the selection of the other -> You can only pick one alternative, and they are not independent Example 1: Alternative A - Replace a gas-fired boiler with a more fuel efficient unit from Manufacturer X Alternative B - Replace a gas-fired boiler with a more fuel efficient unit from Manufacturer Y - It does not make sense to do both of these options, you want to pick one We can consider our options to be: Φ = Do Nothing A = Replace gas-fired boiler with one from Manufacturer X B = Replace gas-fired boiler with one from Manufacturer Y Contingent Alternatives -> Investment alternatives where the selection of one depends on the selection of another Example: Alternative A - Install stand-by 350kW diesel generator for refrigeration system Alternative B - Install “smart” electric control system that reduces peak load from 450 to 350kW - It does not make sense to pick A by itself because without the addition of B, Alternative A will not work. We can consider our option to be: Φ = Do Nothing B = Install “smart” electric control system that reduces peak load from 450 to 350kW C = (A+B) = Install generator plus the “smart” control system Complementary Alternatives -> Two or more alternatives whose relative worth is larger as a combination than individually Example: Alternative A - Company Z gains the right to distribute its products in zone A Alternative B - Company Z gains the right to distribute its products in zone B Alternative (A+B) - Company Z gains the right to distribute its products in zones A and B - Zones A and B are close to each other, so we can also expect some cost saving in transportation if both zones are gained We can consider our options to be: Φ = Do Nothing A = Distribute in Zone A only B = Distribute in Zone B only C = (A+B) = Distribute in Zones A and B with cost savings reflecting the combined transportation cost Competitive Alternatives -> Two or more alternatives whose relative worth is less as a combination than individually Example: Alternative A - Replace a gas-fired boiler for a heating system with a more fuel-efficient unit Alternative B - Install a system of sensors and closers at doors, windows, and other openings to reduce heat loss during the winter season Alternative (A+B) - This is technically feasible, but the savings are not additive We can consider our options to be: Φ = Do Nothing A = Replace a gas-fired boiler for a heating system with a more fuel-efficient unit B = Install a system of sensors and closers at doors, windows, and other openings to reduce heat loss during the winter season C = (A+B) = This will reflect a loss of savings Numerical Example - Present Worth Consider the two ​independent ​alternatives A and B below: We will use an MARR = 15%. Assume no budget constraint, and we will just pick to use Present Worth to analyze the alternatives. Consider: Φ, A, B, (A+B) PW​ ɸ​.15) = 0 <- B​ y definition PW​ (0.15) = -300K + 115K(P/A, 0.15, 4) A​ PW​ A​.15) = \$28,323 ​> 0, so A is acceptable PW​ B​.15) = -400K + 125K(P/A, 0.15, 5) PW​ B​.15) = \$19,019 ​> 0, so B is acceptable PW​ (A+B)​0.15) = 28,323 + 19,019 PW​ (0.15) = \$47,342 ​> 0, so (A+B) is acceptable A​ Because Present Worth is ​additive​, we will choose (A+B). Likewise, IF these two alternatives were mutually exclusive (A+B was not an option), you would select A. Numerical Example - Internal Rate of Return Consider the two ​mutually exclusive a ​ lternatives below: We will use an MARR = 15%. Assume no budget constraint, and we will use IRR to analyze these alternatives. Consider: Φ, C, D PW​ ɸ​.15) = 0 <- B ​ y definition PW​ C​) = -200K + 85K(P/A, i, 6) (P/A, i, 6) = 200K/85K = 2.35294 We find, i​ = 35.7% ​> 0, so we accept C for now -> Now compare D-C PW​ D-C​i) = -(260K - 200K) + (125K - 85K)(P/A, i, 6) (P/A, i, 6) = 60K/40K = 1.5 *​ We find, i​ = 63.1% ​> 15%, so accept D as the ​most economical​ choice. Video - Multiple Alternatives, Part 2 -> In these notes, we will present a method for defining combinations of independent and mutually exclusive investment alternatives Combination Alternatives Method (“Brute force method”) This first method is very literal and time consuming, but demonstrates the concept very well. - First, construct logical combinations of alternatives that are technically feasible - Then, eliminate those combinations that do not meet the applied constraints - Lastly, apply Present Worth to select the best combination Example 1: A manufacturing company is trying to expand its market into areas E, F, and G. And, there are a number of potential distributors to serve in each of these new areas: Area E: distributors W and X Area F: distributors W and Y Area G: distributors X and Z The above alternatives with their associated investment costs can be seen in the table below: Alternative Investment Alternative Investment Alternative Investment EΦ 0 FΦ 0 GΦ 0 EW 80,000 FW 70,000 GX 45,000 EX 90,000 FY 50,000 GZ 75,000 We have the given constraints for these options: - There can only be one distributor in each area, so there is no competition - No distributer can serve in more than one area to prevent stretching the sales staff - There is a budget constraint of \$150,000 With these constraints, we can say that within an area, the distributor choice is ​mutually exclusive​. However, the choice of what areas to expand to is ​independent​. There are 3 alternatives for each area, and 3 areas, so there are 27 combinations. Below is the table that show each combination, the total investment for that combination, and whether or not it complies to the given constraints. Combination Investment Valid Option? EΦ,FΦ,GΦ 0 YES EΦ,FΦ,GX 45,000 YES EΦ,FΦ,GZ 75,000 YES EΦ,FW,GΦ 70,000 YES EΦ,FY,GΦ 50,000 YES EW,FΦ,GΦ 80,000 YES EX,FΦ,GΦ 90,000 YES EΦ,FW,GX 115,000 YES EΦ,FY,GX 95,000 YES EΦ,FW,GZ 145,000 YES EΦ,FY,GZ 125,000 YES EW,FΦ,GX 125,000 YES EW,FΦ,GZ 155,000 NO EX,FΦ,GX --- NO EX,FΦ,GZ 165,000 NO EW,FW,GΦ --- NO EW,FY,GΦ 130,000 YES EX,FW,GΦ 160,000 NO EX,FY,GΦ 140,000 YES EW,FW,GX --- NO EW,FW,GZ --- NO EW,FY,GX 175,000 NO EW,FY,GZ 205,000 NO EX,FW,GX --- NO EX,FW,GZ 235,000 NO EX,FY,GX --- NO EX,FY,GZ 215,000 NO We would then apply Present Worth to the remaining combinations. We will not be working out this math here, since there are so many equations. Example 2: We will now consider three ​mutually exclusive​ options shown below using IRR and a MARR = 15%. Alternatives t A B C Net Cash Flows 0 -30K -15K -20K 1 11.175 6.159 7.546 2 11.175 6.159 7.546 3 11.175 6.159 7.546 4 11.175 6.159 7.546 5 11.175 6.159 7.546 We will sequence these from least to greatest to analyze using PW: ​B, C, A PW​ B​) = -15K + 6159(P/A, i, 5) = 0 (P/A, i, 5) = 15K/6159 = 2.43546 *​ We find, i​ = 30% ​> 15%, so we accept B for now PW​ (i) = -(20K-15K) + (7546 - 6159)(P/A, i, 5) = 0 C-B​ (P/A, i, 5) = 5000/1357 = 3.60490 *​ We find, i​ = 12% ​< 15%, so we reject C, and B remains the best option so far PW​ A-B​= -(30K-15K) + (11,175 - 6159)(P/A, i, 5) = 0 (P/A, i, 5) = 15K/5016 = 2.99043 We find, i​ = 20% ​> 15%, so we accept A. Which is our final selection. Example 3: We will consider two ​mutually exclusive​ options A and B, which have cash flows shown on the table below. We will use Benefit/Cost Ratio to select the best option, and use a MARR = 15%. Alternatives t A B Outflows 0 20K 35K 1 40K 57K 2 60K 56K 3 80K 42K Inflows 0 0 0 1 90K 80K 2 95K 120K 3 125K 140K **Notice: B and C are defined slightly differently in this table compared to the initial way we defined them. Initially, we had B or C equal to zero in each line. However, both definition are used in practice and are both correct. Consider B​ A​5%) = 232,284 Now C​ A​5%) = 152,753 So, BCR​ A​5%) = 232,284/152,753 = 1.52 > ​ 1, so accept A for now Now consider B​ B-A​) = 252,355 - 232,284 = 20,071 And, C​B-A​%) = 154,525 - 152,753 = 1772 So, BCR​ (B-A​ = 20,071/1772 = 11.35 > ​ 1, so we accept B as our final choice

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