Annual Worth Method of Analysis - Engineering Economics Lightboard
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- Опубликовано: 29 мар 2020
- Engineering Economics, Annual worth method of analysis; annuity with a gradient; arithmetic gradient; equivalent annual worth; annual value; mutually exclusive alternatives; comparison methods
so so so much better than pearson online & a teacher who only quizzes you on what you comprehend out of the book. I now have 7 weeks of video to catch up on cause fake it till you make it ran out last week.
LOL is overused but I really did laugh out loud when I read your comment! Thank you, and good luck!
Your videos literally hit right on the dot for the stuff I've got to learn, this channel is the reason why I prevented my grade average from dropping. Thanks!!
Great news! Glad I could help.
i really am thankful i came across your videos, you are such a big help. thank you very much and please keep on making these great and helpful videos. stay healthy, sir!
Thank you for the kind comment - I really appreciate it!!
I wish that I had professor like you.. thank you
Thank you so much for the great complement! It means a lot to me.
Really helpful thank you.
Glad it was helpful!
you earned a subscriber, ty!
You're welcome! And, thank you for subscribing!
Thank you so much !! I was reviewing my knowledge of annual worth analysis and came across your video which lead me to another great thing will cost my professor some points and so much of his ego 😂.. we debated with him if you can prounounce "marr" as one word .. he insisted that western professors only say (Minimum Attractive Rate of Return) and never say "mar" .. now guess who is gonna get a 5 bonus points 😂😂😂
Well, I guess this is good news for you, but please extend my apologies to your professor. If your professor would like to include my videos as supplementary content to your course I would gladly invite him to do so. If it helps lend credibility, a leading textbook publisher also reached out to me wanting to purchase the rights to my Engineering Economics videos. I teach this content at a University in Canada (ranked in the top 100 in the world). I hope the 5 bonus points gets you an 'A'. I can be reached at eeconomicsguy@gmail.com
@@EngineeringEconomicsGuy I'll definitely pass this to him and see what happens.
Much respect.
Nour
Amman - Jordan
OK - great! Thanks. And, good luck with your course!
شكرا لك
You're very welcome!
So you're able to compare projects with *unequal lives* directly using the *annual worth* method?
(No need to use the repeated lives or study period approaches?)
My textbook never made it clear whether or not this was possible.
Yes! Annual worth is absolutely the best, and easiest, way to compare projects of unequal lives...but, be aware of the underlying assumption that the projects "continue" - i.e. - repeat - forever (or at least as long as practical for the purpose of the analysis). Using Present-Worth and the repeated-lives-with-least-common-multiple-of-years-approach (or study-period) is a bit of an academic exercise to help students practice their calculation skills but there are certain real situations where it is appropriate. You should learn and understand both Present Worth and Annual Worth analysis.
@@EngineeringEconomicsGuy My textbook (Engineering Economics by Fraser and Jewkes) describes the difference between the PW and AW methods as being one of convenience, meaning do you prefer looking at the result as a lump sum or as an annual "salary" type of thing.
The textbook emphasizes that the projects must have equal lives to be compared using the PW method, and that if they don't, then there are two options:
1.) Repeated lives. (In your example I think you had 7 and 10 years so that would mean evaluating those projects at a common time period of 70 years.)
2.) Study period. (In your example that would mean capping the analysis after 7 years and messing around with the salvage value of the device that lasts 10 years to figure out what it's worth after 7 years.
But the textbook then does the same thing with the AW method (using the repeated lives approach)... and then doesn't do either. It kind of threw me for a loop.
I'm glad that it's unnecessary because it makes a lot of additional and tedious calculations. :)
What is the process you would do in the (A/P, 10%,7) in numbers?
A/P is a compound interest factor. It's what we use to convert a present value (P) to and annuity (A). Use the formula: [i(1+i)^n]/[(1+i)^n−1] to calculate the value of the factor, where n is the number of periods (often years - i.e. 7) and i is the interest rate (10% = 0.10). You can also use a compound interest table to find the value.
Thank you for a great video and explanation. This explanation makes sense, but I'm wondering about the book examples. They have a positive sign in front of the FC and a negative sign for savings (which doesn't make sense), but if I use your method to solve for many examples in the textbook, I will get a (negative) AW. Will that still be ok? Will my conclusion be to go with the alternative that got the lower (negative_ AW, if that makes sense?
Good question! I don't agree with a FC being a positive number - UNLESS, your book is treating ALL 'COSTS' as positive - but this is pretty unconventional. If your book views savings as negative, then I would assume that a negative AW is a good thing, AND, the more negative the better! So, my answer to your question is YES! But, again, this is opposite to the sign conventions that I use.
@@EngineeringEconomicsGuy I use Fraser, N., Jewkes, E. & Pirnia, M., Schmitt, K. (2022). Engineering Economics: Financial Decision Making for Engineers, 7th edition. and yes, it is so confusing to think about FC as costs as positives. Thank you for your reply. It is much appreciated.
My pleasure!
8:23 is there a by-hand method or analytical way of deriving the value 0.16275. Thank you for the upload!
Yes! Thanks for the question. The formula is (i * (1+i)^n) / ((1+i)^n - 1), where i = the interest rate, and n = number of periods. This is referred to as the A/P compound interest factor. A derivation of this formula can be found in most engineering economics textbooks. Hope that helps. You can also watch my other videos on compound interest factors and patterns of cash flows.
@@EngineeringEconomicsGuy Thank you!
You're welcome! Good luck in your course!
Can you do Future Worth Method, Sir? It would really be helpful 😊💛 You're the best teacher I never had 😢
Thank you for the complement! I love hearing from viewers. I will try to make a Future Worth video soon!
I need an annualized return on investment calculation for the capital being injected in project during 16weeks in 3trenches
I recommend you draw a cash flow diagram with 'units' of 'weeks'. Also, pay close attention to the quoted interest-rate; is it compounded weekly, monthly, yearly, etc.? You will need the content of 4 of my videos (perhaps):
Nominal and Effective Interest Rates:
ruclips.net/video/aT1n_bbQQbM/видео.html
Drawing Cash Flow Diagrams:
ruclips.net/video/KnZdHPs04EI/видео.html
Compounding More Frequent than Payments:
ruclips.net/video/kEcGEXlK5oY/видео.html
Compounding Less Frequent than Payments:
ruclips.net/video/plotBZ-kvpc/видео.html
This is in addition to this video on Annual Worth!
Good luck! Let me know how it goes!
EE Guy
Hi, May I ask what is the formula for A/G?
Here's a good reference page with all of the EE formulas - A/G is on there - enjoy! staff.emu.edu.tr/gokhanizbirak/Documents/courses/ieng323-mane323/assignments-homeworks/Formulas-Engineering-Economy.pdf
@@EngineeringEconomicsGuy Thank you! your videos are a big help.
Glad you like them! I'm happy to help.
Greetings! I came across a problem I regard as highly...complex. Would anyone happen to know the general approach to this? Machine A, Machine B. Purchase value of machine A is 180.000 and purchase value of machine B is 205.000. Annual interest rate is 10%. The annual operational costs are given by this function: 20.000+4.000t (t being umber of years) for machine A and 15.000+3000t for machine B. Tires need to be changed every 3 years, and they cost 2.000 for machine A and 2.500 for machine B. The salvage value is also given as a function, (180-2t) for A and (205-1,5t) for B. The question is, how do we determine the economic service life of a machine with such data? Whats the most viable option for purchase?
There are 4 Videos in my "Replacement Decisions" Playlist. I suggest you watch them all...you will need to become familiar with them to answer this complex question: ruclips.net/p/PLcfz9wmNxKqhSDF7C1FkR3B92tdo-Nzeh
You will need to use the concept of 'economic service life' or 'economic life' to solve the problem - NOTE - this refers to the number of years whereby the Equivalent Annual Cost of ownership is MINIMIZED...these terms will make more sense after you watch the videos. You will need to work through the options of owning each machine for 1 year, 2 years, 3 years, etc. (each as separate, independent scenarios) to find the number of years-of-ownership that minimizes the EAC (this is the variable "t" in the formulas you've been given). GOOD LUCK!!
@@EngineeringEconomicsGuy I watched all die of them! Great job explaining the EAC. The thing is, we use the factor formulas to calculate different cost in my school. For example, i(i+1)^n/(1+i)^n-1 instead of (A/P,i,N). My issue with this particular problem is its seemingly slow depreciation rate and slow repair/maintenance cost increase. I always get that the best time to sell is the first year, but doesn’t really make sense for its ESL to be 1 year. Anyway, thanks for the content, you’ve helped me grasp things I’d otherwise not be able to!
Thank you. Why dont you explain (A/P, 10%,10)??
ruclips.net/video/21P7VCw1DwI/видео.html
ruclips.net/video/t3i7OFWxIN8/видео.html
Johnny, I hope these videos help!