Present Worth and Annual Worth Explained Engineering Economics Live Class Recording 

Thanks for the nice comment. Good luck in your course!

Glad it was helpful! Good luck in your course!

Good luck on the exam!

@@EngineeringEconomicsGuy I actually wrote it very well. Your lectures have been nothing but extremely helpful. Thank you very much. Good teachers like you are a gem in this world🙏🏾❤️

haha here i am, in 2 days

Here we go😂

👋 👋

Thank you for the kind words! Glad I could help you.

You are so welcome!

You are most welcome!

Just spit-balling here, but maybe it's because the *earth fill* for the wood bleachers lasts 90 years and the wood only lasts 30 years? If the earth fill only lasted 30 years like the wood, then would it be fair to say that the AW method would work at 90 years for the concrete bleachers and 30 years for the wood bleachers? Maybe it's more about making the wood bleacher project internally consistent so we're not paying for 90 years worth of earth fill but only getting 30 years of benefit from it?

You're doing some good thinking. You should never use AW if you're using the repeated-lives-approach...well, you CAN, but you get the same answer. In the example you mentioned, the value of the AW for the 30-year wood bleachers will be the same if you use 1-instance of 30-years or 3 consecutive instances of 30-years to make 90-years-of-bleachers. If you use Present Worth (PW) - the you MUST use the repeated-lives approach and you must assume that the projects automatically repeat. For AW, you need to make the assumption that there will be a "90-year need for bleachers", but you don't need to go through the hassle of the repeated-lives approach. AW is really a much better approach as long as the assumption of an ongoing need is valid. SO...you CAN use AW to directly compare the 90-year concrete bleachers and the 30-year wood bleachers - you are correct!!

Ha! Your second comment falls into the category of 'over-thinking'. Not bad thinking...just over-thinking. Remember your course is an Engineering Economics course, not a Geotechnical Engineering course (which is what I am actually!).

Thanks for the comment! I'm using the textbook: "Engineering Economics: financial decision making for engineers" 6th edition. Authors are Fraser, Jewkes, and Pirnia. Published by Pearson. There are many other texts available and they are all similar. Any text with 'Engineering Economics' in the title will have content similar to my videos.

Hello in Indonesia! Thank you for viewing my Channel! I am a Canadian Civil Engineer, now teaching at a University. I worked for a short time in Surabaya, Indonesia (1996-7) - I thought you might be interested to know! I'm happy to be helpful to you. Other viewers have requested a 'suggested order' for watching the videos/playlists. My suggestion is to follow the playlists as follows (you will need to go to my main Channel and click on 'Playlists': 1. Introduction 2. Interest 3. Cash Flow Analysis 4. Applications of Cash Flow Analysis 5. Comparison Methods 6. Replacement Decisions 7. Depreciation 8. Tax 9. Inflation The above order would most closely follow the structure of my Engineering Economics course. I hope you find this useful. Please tell you classmates and even your instructor about my Channel! All the best!

@@EngineeringEconomicsGuy Wow, you've been in Indonesia!? When that happened I wasn't even born yet hahaha. Hey you know, I hope one day you can visit Indonesia once again and I want to attend your classes or seminars. Thanks to technology, I'm grateful to be able to find your channel, Sir. Next week I'll present the equivalent annual cost and present worth method to my classmates. I hope I can give my best. And of course I'll tell all my friends about your amazing channel. May you always be healthy to educate your followers, Sir. Thank you for your reply!

I wish you good luck with your presentation and with your Engineering Economics course! I'm glad you discovered my Channel!

@@EngineeringEconomicsGuy thanks man lectures are the best explanation on the internet

I think the video: Compound Interest Factors Reminders Live Class Recording ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-21P7VCw1DwI.html also answers this question for you. Thanks for watching my Channel!

i tried to compute this problem but its giving me another answer,used this A((1.8)⁶-1))/(0.8(1.8)6).

@@albertmungamelo2936 careful... 8% is 0.08 so (1 + i) is 1.08 NOT 1.8.

Nice of you to say! Thank you!

Yes, in fact, when done correctly all three methods always lead to the same conclusion.

@@EngineeringEconomicsGuy Thanks a bunch!!!

My pleasure!

Thanks for watching my video. I'm sorry but you will need to clarify your question. It is important to note that this video assumes students are familiar with compound interest factors such as A/P and P/A - please look in the 'Cash Flow Analysis' Playlist on my main Channel page for some other videos that might help your understanding.

Or inflow = outflow?

Yes and no. Yes, for these types of problems we are equating inflow and outflow, BUT, we do it using the principles of the time value of money... fancy way of saying we account for the effects of compound interest associated with the timing of each cash flow.

@@EngineeringEconomicsGuy Thank you for clarification!

In the case you describe you need to separate the problem into two parts. 1. Use the P/A factor to find the "PW" of the annuity using n =7. 2. Recognize that the P you calculated in part 1 is NOT positioned at t =0...it will be at t=2. Then use P/F to move the value to t =0. Note, if the first payment starts 3 years from now and goes to year 9...that is 7 pmts! Draw the cash flow diagram, it helps!

Or, if the annuity itself only has 6 pmts, then n=6 in the first part of the analysis.

@@EngineeringEconomicsGuy Thank you a lot, got it....

RU-vid is the only place to find my videos, unless you want to enroll in my class!

It is definitely an option. How do I find your class?

I'm sorry for the delay. I don't think it would be possible for you to take my course unless you are enrolled in an Engineering program at the University where I teach! However, I suggest you go to my Playlists on my Main Channel page and follow the guidelines below. www.youtube.com/@EngineeringEconomicsGuy/playlists Some viewers have requested a 'suggested order' for watching the videos/playlists. There is some flexibility within each playlist in terms of order, however, I suggest using the video titles as a way to determine the difficulty of the topic. I've tried to 'list' the videos within each Playlist in the most logical order. My suggestion is to work through the Playlists as follows: 1. Introduction 2. Interest 3. Cash Flow Analysis 4. Applications of Cash Flow Analysis 5. Comparison Methods 6. Replacement Decisions 7. Depreciation 8. Tax 9. Inflation The above order would most closely follow the structure of my Engineering Economics course. I hope you find this useful. Best Regards, Engineering Economics Guy

Good question! The first 3-minutes of this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-QGJOkgcmdMo.html answer your question! We often have a present value (usually a purchase price 'P') as well as a salvage value 'S' in problems where we are trying to decide when to replace an asset. To make these decisions we convert the P and the S into an equivalent ANNUAL cost (EAC). There are also 2 sample problems linked at the end of the video I just gave the link to. I hope this helps! (pay close attention to the + and the - signs for the P and the S!)

@@EngineeringEconomicsGuy Thanks let me check it out

No, sorry. I guess I should make one!

@@EngineeringEconomicsGuy yes please am waiting for it

Hello! I have many other videos you can explore on my Channel. I hope you find them useful.

Kristine the numbers 4.6299 and 0.21632 come from a compound interest TABLE (usually found in the Appendix of an Engineering Economics or Finance textbook -or you can find them online). The notation (P/A, 8%, 6) means you look in the 8% TABLE in the column labeled P/A, and then look in the row with a value of 6 (n). You can also use the formula [(1+i)^n - 1] / [i * (1+i)^n] . For values of n = 6 and i = 0.08 you should get 4.6299. This is the factor that converts an A (annuity) into a P (present value). The "opposite" is true for the 0.21632 value (it is an A/P factor). I hope this helps. I have other videos that describe the various compound interest factors and the associated 'patterns' of cash flows. Please explore my Cash Flow Playlist: ru-vid.com/group/PLcfz9wmNxKqgciRucJOr8VdEQQT_cicOT or, the best video for you to watch is probably: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-nQuUjDuilAw.html Good Luck!

@@EngineeringEconomicsGuy thank you sir

Happy to help! Good luck in your course.

I think I know what you mean... if the PW or AW equals ZERO, then the project exactly meets the MARR, therefore the company would be 'indifferent' (meaning they could do the project OR not do the project).

The book is called Engineering Economics: financial decision making for engineers, by Fraser and Jewkes

Actually, the video is mirrored...I'm writing with my right hand.

You are correct about never accepting a project with a negative present worth. I have several other videos on my Channel that discuss this. But, please be careful! The MARR is not exactly like the bank (rate), or, what you think you can earn elsewhere. Using the MARR in the context of an individual rather than a company is not really appropriate but we sometimes construct problems with phrases like "Jim has a MARR of 8%" just to give students something to work on. What you've described in your comment is more like something called the 'risk-free-rate'. The MARR comes from a company's 'Weighted-Average-Cost-of-Capital' (WACC). This term refers to the "cost" (expressed as an interest rate) of a company's sources of money (capital); things like loans that are owing, bonds they have issued, and the expectation of their shareholders/owners. A company should only take-on projects that earn at-least their MARR otherwise they will not meet their financial obligations and satisfy their owners. Some people refer to the MARR as the 'hurdle-rate'. This is not explained very well in most Engineering Economics textbooks - it is more of a Business-Finance topic. In the text I use to teach my EE course the WACC is discussed in the Appendix of the Chapter where the MARR is introduced as a concept.

@@EngineeringEconomicsGuy Thanks.

P/A is the compound interest factor that converts an Annuity to a Present value. A/P is the opposite. I invite you to explore the other videos on my channel, for a complete explanation.

Thank You Sir

You're welcome! And that you for the positive energy!

Thanks for watching!

I'm using the textbook "Engineering Economics: financial decision making for engineers" by Fraser, et al. but almost every finance or engineering economics text will have the tables and formulas you need.

OK - yes, for i = 8% that would be true. Thanks!