Ch8 ror of_multiple_alternatives_rev1
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This document provides instructions for performing an incremental rate of return (ROR) analysis between two alternatives: 1. Order the alternatives by initial investment cost, with the lower-cost option labeled A and the higher-cost option labeled B. Develop the cash flow and incremental cash flow series over the common time period. 2. Draw an incremental cash flow diagram if needed. Count sign changes in the incremental cash flows to determine if multiple RORs are possible. 3. Set up the present worth equation for the incremental cash flows and use trial and error to determine the incremental rate of return, Δi*B-A. Select the alternative with the higher ROR if it exceeds the minimum acceptable rate of return
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This document contains engineering economics practice problems involving calculating present and equivalent worth of cash flows with varying interest rates. It includes questions about determining the present worth of income from a machine purchased to bend beams, calculating equivalent worth of cash flows in a diagram at 12% interest, and finding the value of x that makes a cash flow diagram equal $15,000 at 15% interest.
Hw1
This document contains an engineering economics homework assignment from Zarqa University. It includes 14 questions about time value of money concepts like simple vs. compound interest, calculating rates of return, and using the rule of 72. The questions cover cash flows, loans, certificates of deposit, and investment growth over various time periods at given interest rates. Students are asked to define terms, calculate amounts, draw cash flow diagrams, and estimate doubling/quadrupling times using the rule of 72.
Ch8 ror of_multiple_alternatives_examples_rev1
This document discusses return on investment (ROI) analysis for multiple alternatives. It provides examples of using ROI to evaluate different options. The chapter focuses on calculating and comparing the ROI of various potential choices or scenarios.
Ch7 ror analysis_for_single_project_rev3
This document discusses how to evaluate the rate of return (ROR) of a single project alternative to determine if it is financially viable. The ROR must meet or exceed the minimum attractive rate of return (MARR), which is the hurdle rate that is set upfront. To evaluate a project, its internal rate of return (IRR) is calculated and compared to the MARR. If the IRR is greater than or equal to the MARR, then the project is accepted, and if it is less than the MARR then it is rejected. Examples are provided to demonstrate how to calculate the IRR and evaluate if a project meets the MARR.
Ch5 pw analysis_part1_rev4
This document provides an overview of present worth analysis and cash flow analysis methods for evaluating investment alternatives. It discusses measures of profitability like present worth, annual worth, and future worth. It covers evaluating single alternatives as well as comparing alternatives of equal lives and different lives. It also introduces the concepts of planning horizon, coterminated assumptions, and capitalized worth for comparing long-term alternatives. Examples are provided to illustrate how to apply these concepts and calculate present worth, annual worth, capitalized worth, and to compare investment alternatives.
Ch4 nom&effective ir_rev2
This document discusses nominal and effective interest rates, including continuous compounding. It defines nominal and effective interest rates, and how to calculate effective rates for different compounding periods. It provides examples of calculating effective rates for annual, semi-annual, quarterly, and continuous compounding. It also discusses how to handle calculations when the payment period is equal to, longer than, or shorter than the compounding period. This includes using the appropriate effective rate and number of periods in calculations.
Ch3 hw2 soln
This document contains 10 engineering economics problems involving calculations of present worth, future worth, uniform annual worth, and equivalent cash flows using formulas for present and future value. The problems involve cash flows from equipment purchases and sales, loan repayment schedules, and production from oil wells over several years. Formulas are provided and calculations are shown for finding the present worth, future worth, or equivalent uniform annual worth that satisfy the given cash flow diagrams and requirements for rates of return.
Ch3 combinig factors_rev2
This document discusses methods for combining cash flow factors to solve more complex engineering economy problems involving shifted uniform series and gradients. It explains how to combine factors to find the present, future, or annual worth for cash flows that do not begin immediately or that change over time. Examples are provided to illustrate combining factors for shifted uniform series, non-standard shifted series, shifted gradients, and decreasing gradients. The use of spreadsheet functions to simplify the calculations is also discussed.
Ch2 (part2)arithmetic gradient
This document discusses arithmetic and geometric gradients for cash flow analysis. It provides examples of how to calculate the present worth of: 1) a base amount with an arithmetic gradient added, and 2) the total annual cash flow from a base amount and arithmetic gradient. It also defines a geometric gradient as a cash flow that increases or decreases by a constant percentage each period, and states that additional formulas are needed to calculate the present worth of a geometric gradient series.
Ch2 (part1)econ factors_rev2
This document discusses various time value of money factors used to solve problems involving interest rates over time. It covers:
1) Single payment factors (F/P and P/F) which determine the future or present value of a single payment.
2) Uniform series factors (P/A and A/P) which determine the present or equivalent annual value of a uniform series of payments.
3) Sinking fund factors (A/F and F/A) which determine the annual payment or future value for accumulating to a target amount.
4) Arithmetic gradient factors (P/G and G/P) which handle cash flows that increase or decrease by a constant amount each period.
Ch1 hw rev1_solution
This document contains solutions to homework problems from an engineering economics course. It addresses topics like the time value of money, simple vs compound interest, calculating rates of return, and cash flows. Example calculations include determining the rate of increase in company profits, the rate of return on an investment, and the amount originally borrowed given the loan payoff amount. The document also covers the time it would take investments to double or quadruple at given compound interest rates using the Rule of 72.
Ch1 why engecon_rev2
Engineering economy is important for engineers to incorporate economic analysis into their design decisions. It allows engineers to evaluate multiple alternatives and select the best option. Key concepts in engineering economy include time value of money, equivalence, cost estimation, and analyzing decisions like whether a company should buy or rent a cement mixer based on measures of worth, present/future worth, payback period, and rate of return. Cash flows are also important to engineering decisions and involve constructing timelines that show the inflows and outflows of money over time in order to calculate net cash flow for each period.
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Height and depth gauge linear metrology.pdf
Height gauges may also be used to measure the height of an object by using the underside of the scriber as the datum. The datum may be permanently fixed or the height gauge may have provision to adjust the scale, this is done by sliding the scale vertically along the body of the height gauge by turning a fine feed screw at the top of the gauge; then with the scriber set to the same level as the base, the scale can be matched to it. This adjustment allows different scribers or probes to be used, as well as adjusting for any errors in a damaged or resharpened probe.
AI-Based Home Security System : Home security
Home security is of paramount importance in today's world, where we rely more on technology, home
security is crucial. Using technology to make homes safer and easier to control from anywhere is
important. Home security is important for the occupant’s safety. In this paper, we came up with a low cost,
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Mechatronics deals with robotics, control systems, and electro-mechanical systems.
一比一原版(爱大毕业证书)爱荷华大学毕业证如何办理
原版一模一样【微信：741003700 】【(爱大毕业证书)爱荷华大学毕业证成绩单】【微信：741003700 】学位证，留信认证（真实可查，永久存档）原件一模一样纸张工艺/offer、雅思、外壳等材料/诚信可靠,可直接看成品样本，帮您解决无法毕业带来的各种难题！外壳，原版制作，诚信可靠，可直接看成品样本。行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备。十五年致力于帮助留学生解决难题，包您满意。
本公司拥有海外各大学样板无数，能完美还原。
1:1完美还原海外各大学毕业材料上的工艺：水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠。文字图案浮雕、激光镭射、紫外荧光、温感、复印防伪等防伪工艺。材料咨询办理、认证咨询办理请加学历顾问Q/微741003700
【主营项目】
一.毕业证【q微741003700】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【q/微741003700】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
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留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分
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办理(爱大毕业证书)爱荷华大学毕业证【微信：741003700 】外观非常简单，由纸质材料制成，上面印有校徽、校名、毕业生姓名、专业等信息。
办理(爱大毕业证书)爱荷华大学毕业证【微信：741003700 】格式相对统一，各专业都有相应的模板。通常包括以下部分：
校徽：象征着学校的荣誉和传承。
校名:学校英文全称
授予学位：本部分将注明获得的具体学位名称。
毕业生姓名：这是最重要的信息之一，标志着该证书是由特定人员获得的。
颁发日期：这是毕业正式生效的时间，也代表着毕业生学业的结束。
其他信息：根据不同的专业和学位，可能会有一些特定的信息或章节。
办理(爱大毕业证书)爱荷华大学毕业证【微信：741003700 】价值很高，需要妥善保管。一般来说，应放置在安全、干燥、防潮的地方，避免长时间暴露在阳光下。如需使用，最好使用复印件而不是原件，以免丢失。
综上所述，办理(爱大毕业证书)爱荷华大学毕业证【微信：741003700 】是证明身份和学历的高价值文件。外观简单庄重，格式统一，包括重要的个人信息和发布日期。对持有人来说，妥善保管是非常重要的。
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Supermarket Management System Project Report.pdf
Supermarket management is a stand-alone J2EE using Eclipse Juno program.
This project contains all the necessary required information about maintaining
the supermarket billing system.
The core idea of this project to minimize the paper work and centralize the
data. Here all the communication is taken in secure manner. That is, in this
application the information will be stored in client itself. For further security the
data base is stored in the back-end oracle and so no intruders can access it.
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Mechanical Engineering PROJECT REPORT ON SUMMER VOCATIONAL TRAINING
AT MBB AIRPORT
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一比一原版(CalArts毕业证)加利福尼亚艺术学院毕业证如何办理
CalArts毕业证学历书【微信95270640】CalArts毕业证’圣力嘉学院毕业证《Q微信95270640》办理CalArts毕业证√文凭学历制作{CalArts文凭}购买学历学位证书本科硕士,CalArts毕业证学历学位证【实体公司】办毕业证、成绩单、学历认证、学位证、文凭认证、办留信网认证、（网上可查，实体公司，专业可靠）
(诚招代理)办理国外高校毕业证成绩单文凭学位证,真实使馆公证（留学回国人员证明）真实留信网认证国外学历学位认证雅思代考国外学校代申请名校保录开请假条改GPA改成绩ID卡
1.高仿业务:【本科硕士】毕业证,成绩单（GPA修改）,学历认证（教育部认证）,大学Offer,,ID,留信认证,使馆认证,雅思,语言证书等高仿类证书；
2.认证服务: 学历认证（教育部认证）,大使馆认证（回国人员证明）,留信认证（可查有编号证书）,大学保录取,雅思保分成绩单。
3.技术服务：钢印水印烫金激光防伪凹凸版设计印刷激凸温感光标底纹镭射速度快。
办理加利福尼亚艺术学院加利福尼亚艺术学院毕业证文凭证书流程：
1客户提供办理信息：姓名生日专业学位毕业时间等（如信息不确定可以咨询顾问：我们有专业老师帮你查询）；
2开始安排制作毕业证成绩单电子图；
3毕业证成绩单电子版做好以后发送给您确认；
4毕业证成绩单电子版您确认信息无误之后安排制作成品；
5成品做好拍照或者视频给您确认；
6快递给客户（国内顺丰国外DHLUPS等快读邮寄）
-办理真实使馆公证（即留学回国人员证明）
-办理各国各大学文凭（世界名校一对一专业服务,可全程监控跟踪进度）
-全套服务：毕业证成绩单真实使馆公证真实教育部认证。让您回国发展信心十足！
（详情请加一下 文凭顾问+微信:95270640）欢迎咨询！子小伍玩小伍比山娃小一岁虎头虎脑的很霸气父亲让山娃跟小伍去夏令营听课山娃很高兴夏令营就设在附近一所小学山娃发现那所小学比自己的学校更大更美操场上还铺有塑胶跑道呢里面很多小朋友一班一班的快快乐乐原来城里娃都藏这儿来了怪不得平时见不到他们山娃恍然大悟起来吹拉弹唱琴棋书画山娃都不懂却什么都想学山娃怨自己太笨什么都不会斟酌再三山娃终于选定了学美术当听说每月要交元时父亲犹豫了山娃也说爸算了吧咱学校一学期才转
一比一原版(uofo毕业证书)美国俄勒冈大学毕业证如何办理
原版一模一样【微信：741003700 】【(uofo毕业证书)美国俄勒冈大学毕业证成绩单】【微信：741003700 】学位证，留信认证（真实可查，永久存档）原件一模一样纸张工艺/offer、雅思、外壳等材料/诚信可靠,可直接看成品样本，帮您解决无法毕业带来的各种难题！外壳，原版制作，诚信可靠，可直接看成品样本。行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备。十五年致力于帮助留学生解决难题，包您满意。
本公司拥有海外各大学样板无数，能完美还原。
1:1完美还原海外各大学毕业材料上的工艺：水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠。文字图案浮雕、激光镭射、紫外荧光、温感、复印防伪等防伪工艺。材料咨询办理、认证咨询办理请加学历顾问Q/微741003700
【主营项目】
一.毕业证【q微741003700】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【q/微741003700】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
◇企事业单位必须要求办理的
◇需要报考公务员、购买免税车、落转户口
◇申请留学生创业基金
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才
办理(uofo毕业证书)美国俄勒冈大学毕业证【微信：741003700 】外观非常简单，由纸质材料制成，上面印有校徽、校名、毕业生姓名、专业等信息。
办理(uofo毕业证书)美国俄勒冈大学毕业证【微信：741003700 】格式相对统一，各专业都有相应的模板。通常包括以下部分：
校徽：象征着学校的荣誉和传承。
校名:学校英文全称
授予学位：本部分将注明获得的具体学位名称。
毕业生姓名：这是最重要的信息之一，标志着该证书是由特定人员获得的。
颁发日期：这是毕业正式生效的时间，也代表着毕业生学业的结束。
其他信息：根据不同的专业和学位，可能会有一些特定的信息或章节。
办理(uofo毕业证书)美国俄勒冈大学毕业证【微信：741003700 】价值很高，需要妥善保管。一般来说，应放置在安全、干燥、防潮的地方，避免长时间暴露在阳光下。如需使用，最好使用复印件而不是原件，以免丢失。
综上所述，办理(uofo毕业证书)美国俄勒冈大学毕业证【微信：741003700 】是证明身份和学历的高价值文件。外观简单庄重，格式统一，包括重要的个人信息和发布日期。对持有人来说，妥善保管是非常重要的。
Blood finder application project report (1).pdf
Blood Finder is an emergency time app where a user can search for the blood banks as
well as the registered blood donors around Mumbai. This application also provide an
opportunity for the user of this application to become a registered donor for this user have
to enroll for the donor request from the application itself. If the admin wish to make user
a registered donor, with some of the formalities with the organization it can be done.
Specialization of this application is that the user will not have to register on sign-in for
searching the blood banks and blood donors it can be just done by installing the
application to the mobile.
The purpose of making this application is to save the user’s time for searching blood of
needed blood group during the time of the emergency.
This is an android application developed in Java and XML with the connectivity of
SQLite database. This application will provide most of basic functionality required for an
emergency time application. All the details of Blood banks and Blood donors are stored
in the database i.e. SQLite.
This application allowed the user to get all the information regarding blood banks and
blood donors such as Name, Number, Address, Blood Group, rather than searching it on
the different websites and wasting the precious time. This application is effective and
user friendly.
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Ch8 ror of_multiple_alternatives_rev1
- 1. ROR Analysis of multiple alternatives Chapter 8 1
- 2. ROR Incremental Analysis • The complete procedure by hand or computer for an incremental ROR analysis for two alternatives is as follows: • 1. Order the alternatives by initial investment or cost, starting with the smaller one, called A. The one with the larger initial investment is in the column labeled B. • 2. Develop the cash flow and incremental cash flow series using the LCM of years, assuming reinvestment in alternatives. • 3. Draw an incremental cash flow diagram , if needed. 2
- 3. ROR Incremental Analysis • 4. Count the number of sign changes in the incremental cash flow series to determine if multiple rates of return may be present. If necessary, use Norstrom's criterion on the cumulative incremental cash flow series to determine if a single positive root exists. • 5. Set up the PW equation for the incremental cash flows, and determine Δi*B-A using trial and error by hand or spreadsheet functions. • 6. Select the economically better alternative as follows: • If Δi*B-A < MARR, select alternative A. • If Δi*B-A >= MARR, the extra investment is justified; select alternative B. 3
- 4. ROR Incremental Analysis (EX 8.3) • 1. Alternatives A and B are correctly ordered with the higher first-cost alternative in coJumn (2). • 2. The cash flows for the LCM of 10 years are tabulated in TabJe 8-4. • 3. The incremental cash flow diagram is shown in Figure 8-l. • 4. There are three sign changes in the incremental cash flow series, indicating as many as three roots. There are also three sign changes in the cumulative incremental series, which starts negatively at So = $- 5000 and continues to SI0 = $+5000, indicating that more than one positive root may exist. 4
- 5. ROR Incremental Analysis (EX 8.3) 5
- 6. ROR Incremental Analysis (EX 8.3) • Assume that the reinvestment rate is equal to the resulting Δi*B-A (or Δi* for a shortened symbol). Solution of Equation [8.2] for the first root Discovered results in Δi*: between 12 and 15%. By interpolation Δi* = 12.65%. • 6. Since the rate of return of 12.65% on the extra investment is less than the 15% MARR, the lower- cost vendor A is selected. The extra investment of $5000 is not economically justified by the lower annual cost and higher salvage estimates. 6
- 7. Breakeven Example (Ex 8.4) 7
- 8. Ex 8.4 8
- 9. Ex 8.4 9
- 10. Ex 8.4 (plot) 10
- 11. Ex 8.4 (plot) 11
- 12. Example 8.7