A friend asked me what size disk drive to buy.  Is there an optimal size?
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It turns out the answer is yes and the optimal size is e = 2.72.  The base of the natural logarithm  times the size of the drive you have now.  Isn’t this marvelous?

Here is why.

Unless you have very special needs you don’t need to worry about RPM, cache size  and the like when you buy a disk drive.  The reason is that, at any given time, the performance of the disk drives available in the market is within a very narrow band.

This is because at the less performant end you can not substitute slowness for sufficient cost savings. It has been tried a few times but with little success. At the high performance end the component cost rises dramatically* and the cost of drives accordingly. This leaves high performance drives to a small special requirement niches. The trend toward RAID and JBOD for even big storage farms like Google and Amazon result in the disk market being very homogeneous performance wise.

This leaves the size of your new disk as the only real decision parameter. Furthermore Cost per GB is almost constant over the range of the sizes. If you think otherwise look at Pricewatch.com and check yourselves.

So what is a good replacement strategy?

Think about your investment in a disk drive as a cycle that keeps repeating.

Your drive gets full and you need to buy a new one. The only decision parameter you have is what size to buy,. namely S

Each cycle is identical because I assumed that the storage growth rate is equal to the technology driven price decline of storage. (Moore’s law etc)

This is an excellent assumption if you think it through, and you can look at historical data to verify the assumption.

What I mean by thinking it through is that most of the increase in storage needs is externally driven. (This goes for companies and institutions as well)

You start to encode mp3’s at higher bitrate as storage prices falls, You store your Video at better quality and buy more when storage is cheap. Etc.

Most importantly for this assumption to hold is that there is a feedback loop. If storage cost gets ahead of the curve meaning being cheaper than expected we change our behavior. Same if it falls behind.

This assumption leads to the equality:

Storage cost decline = storage requirement growth. The drive St (t is any given time) will always cost the same. The one you buy next time at optimum size S will always be the same dollar amount as the one you bought last time. The increase in S is offset by the decline of C, Both S and C should have a little subscript t like this St and Ct.

Obvious if the assumption does not hold and storage increases higher or lower than prices declines this shifts the optimal S = e, but not very dramatic and you can adjust for it but the math gets quite complicated.

For any economic optimization rule the important thing to understand and articulate is the two forces that pulls in opposite directions thereby creating a cost minimum.

For disk drives it is the cost of unused space over time versus the declining cost of storage plus cost of scrapping the drive. Very large drive and the cost of unused space is high. Very small drive and the cost of ditching the drive is high as you amortize the cost over a short period.

Here is the Math. It look more complicated than it really is.

S = Renewal size relative to current capacity
g = storage growth / time unit
c = cost / storage unit
n = Life of unit
How many time periods n does it take before your new drive of s is full?

S = (1 + g) ^ n ==> n = ln(S) / ln ( 1 + g)

I omitted the size of the initial drive on both sides of initial equation., but you can put a 1 or 100.

We do not know what the cost is but we need to optimize by finding s so we have minimal cost per time period

Cost / Time Unit = ( S * c ) / n Substitute n from above and find the local minimum

Minimize { (S * c) * ln( 1+g)} / ln (S) ==> K * S/ln(S)

K= Constant and disappears.

Therefore we need to minimize S/ln(S) ==>  S= e ~ 2.72

Summary

The optimal buying decision is to buy a disk where the size S of the new drive is e= 2.72 times the size of the one you have now.
If you old disk is 100GB buy a new one at approx. 300GB, don’t spring for the fancy 500GB that just came out

* See Paper on CPU procurement.