Comment of the Day - October 25, 2005 Calculate the Optimal Size of a New Disk Drive

You asked about how I figured out the optimal size of a new disk drive?

You don't need to worry about cache size and the like. At any given time the disk drives performance is within a very narrow band. Without going into details it 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 drivers accordingly leaving this to various niches. The disk market is performance wise very homogeneous.

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

Here comes the complex part. 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,. namely S



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

This is an excellent assumption if you think it through, and you can look at historical trends etc. What I mean by thinking it through is that most of the increase in storage needs is outside driven as we discussed. (This goes for companies and institutions as well)
You start to encode mp3's at higher bitrate as storage prices falls, Video etc. Most importantly for this assumption to hold is that there is a feedback loop. If storage cost gets ahead of the curve ie being cheaper than expected we change our behavior. Same if it falls behind.

With 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.

An obvious thing to understand and articulate is the two forces that pulls in opposite the directions thereby creating a cost minimum. For the disk drives it is the cost of unused space and the decline of cost of storage plus scrapping of drive. Very large drive and the cost of unused space is very 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:
**
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, presto
Minimize s/ln(s) ==> s = e



Presto Disk drive should be 2.7 times the one you have at the point where you want to replace it.

It is a little hard to read as I can't use math notation here, but hope you got the gist of it.