Oh that's a very interesting viewpoint.The stack exchange post says that over a few years, the replacement batteries become the dominant cost, so buying a high end UPS at say twice the costs, kind of almost doesn't matter after several years, the total costs approach each other - so buy a very good one.
I'd have to figure out if that is the case for my stuff too. It makes sense though, because if we buy a 1000 watt unit sine and 1000 watt unit crap sine (ha ha) then when we replace the batteries it will be the same cost for both. So let's see...
Purchases both types: $200, $100 [totals 200 and 100] cost ratio: 2
Batteries after 2 years: $50, $50 [totals 250 and 150] cost ratio: 1.67
Batteries after 4 years: $60, $60, [totals 310 and 210] cost ratio: 1.48
Batteries after 6 years: $70, $70, [totals 380 and 280] cost ratio: 1.36
Batteries after 8 years: $80, $80, [totals 460 and 360] cost ratio: 1.28
Batteries after 10 years: $90, $90, [totals 550 and 450] cost ratio: 1.22
So we can see that the cost ratio does start to become the same after some years, but it takes a long time.
First set we see it drop by about 33 percent, then about 19 percent, then about 12 percent, then about 8 percent then about 6 percent.
So they do start to even out, but it takes a long time.
It takes 20 years to get down to 10 percent difference.
It takes about 30 years to get down to 5 percent difference.
After 40 years about 3.3 percent difference.
I guess we can say though that 200 and 100 are widely different, while 550 and 450 are not that much different in the long term.
This is based on a battery or battery set cost rising by $10 USD every 2 years.
This is the function I used:
f(x):=[%[1]+%[3],%[2]+%[3],%[3]+10,%[4]+2,float((%[1]+%[3])/(%[2]+%[3]))]
with initial values:
[200,100,50,0,0]
which is the two costs then the $50 then, then the number of years, then the ratio of the higher cost unit over the lower cost unit.
The first update becomes:
[250,150, 60,2,1.6667]
where we can see the cost increased from 50 to 60, the years increased from 0 to 2, and the ratio is 1.6667.
The 60 is applied after the next 2 years but it's updated at the present so it can be used later. The cost applied was actual 50.
BTW the notation "%[n]" just means the nth element of that vector of 5 numbers. So %[3] is 50 to start then 60 after 2 years.
the function f(x) is just applied over and over again, which updates everything every 2 years.
One thing to note here though is that not everyone will have to buy new batteries after 2 years. It depends on how much the power goes out and how long you have to run the UPS before you can shut down manually.
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