Capital Cost Comparison: Present Worth Analysis

[@ThatsRed]

Oh my god, thank you for making this video. I'm an engineering student and we didn't have to take economics at all in school but still will be expected to understand economic concepts when we take out FE Exam after graduation. This video helped soooooooooo much. Thank you! Thank you! Thank you!

[@sahanagowda9422]

Nice description of what P/A is, Thank you so much

[@Hawxye]

Man this has save me on my exam, bless u and thanks.

[@subrataghosh320]

Nice explanation of what P/A and P/F is. Thank you.

[@mateussaar4071]

This saved me. All other videos on this really lack

[@weikeazeh602]

I completely understand it, thanks.

[@StenTarvo]

great video!

[@nsg2425]

This video has benefited me greatly.....🙏👌

[@hajerjm]

Very helpful, thank you

[@olusolaoniyelu6901]

Thank you Sir!

[@danframpton12]

what does this interest rate mean?

[@tloumokobodi]

Thank you for a great video, helpful, Question, Overhaul 3.5k for B, should it not be accounted for at the end of year 6 and 12?

[@LearnChemE]

Good question. You would not be spending $3,500 on an overhaul at the end of the reactor's lifetime, so it is not counted.

[@TurtleboyTom]

If lifetimes are different, isn’t it better to do an annual worth analysis?

[@muhamadrazwan1061]

for case B why for overhaul, just only at 3&9 year only?

[@LearnChemE]

+Muhamad Razwan At years 3 and 9, there is an "overhaul" cost of $3,500 that is different from the annual maintenance cost of $4,000.

[@olajuwonkosoko9117]

+LearnChemE But the overhaul is at the end of the third year, shouldn't that be years 3, 6 and 9??

[@fatinquzazi3767]

thanks bro

[@nativuks]

Thank you very much for the video , which software do you use?

[@LearnChemE]

We use Camtasia Studio to create the videos.

[@eMAjooD]

isn't p/f take positive value ? why -22,000 (p/f,8%,8( and 4) ) ?

[@LearnChemE]

Costs are negative values. The cost of the new reactor is $25,000, but it was partially offset by the salvage value (positive) of $3000. The difference between those leaves -$22,000.

[@eMAjooD]

in case A, why did you put 12 yrs ? isn't 4 yrs ?

[@LearnChemE]

We are comparing the 2 cases so we need to take them out to a common multiple of their service lifetime to make the Present value comparison. In this case, that is 12 years.

[@eMAjooD]

Thank you..that is helpful

[@velozdrums]

Thank you for asking this question, it confused me as well

[@meteorush]

THANK YOU!

[@JV-iw8my]

Weren't you supposed to account for three salvages in case A? i.e. PW_A=-25000-2000(P/A,8,12)-22000[(P/F,8,4)+(P/F,8,8)]+3000[(P/F,8,4)+(P/F,8,8)+(P/F,8,12)]

[@LearnChemE]

In the case of the first two salvages, the salvage value was subtracted from the cost of the new replacement equipment for those years. That is why the cost is -$22,000 in those years instead of -$25,000, which would have been the cost if the old equipment had no salvage value. .

[@reynaldocaay]

How to input (P/A, (interest rate), years) in calculator?

[@tahasrori]

you should find the value throw a table

[@slicric7416]

where does the 1.08 come from in this equation

[@LearnChemE]

+Eric Iacaboni The interest rate is 8%, so the (1+i) in the equation becomes 1.08.

[@Zeupiter_MCG]

Subscried

[@reynaldocaay]

I see it now hehehe

[@tahasrori]

good lol