Cache size: T p o s i t i o n T_{position} TpositionRepresents the rotation and seek overhead time before write operation, disk transfer rate R p e a k R_{peak} RpeakMB/s, therefore transfer D D DTime required for MB T w r i t e T_{write} Twrite
T w r i t e = T p o s i t i o n + D R p e a k T_{write} = T_{position} + \frac{D}{R_{peak}} Twrite=Tposition+RpeakD
Writing efficiency
R e f f e c t i v e = D T w r i t e = D T p o s i t i o n + D R p e a k R_{effective} = \frac{D}{T_{write}} = \frac{D}{T_{position} + \frac{D}{R_{peak}}} Reffective=TwriteD=Tposition+RpeakDD
Write efficiency is based on the transmission rate and is expressed by F(0<F<1)
R e f f e c t i v e = D T p o s i t i o n + D R p e a k = F × R p e a k R_{effective} = \frac{D}{T_{position} + \frac{D}{R_{peak}}} = F \times R_{peak} Reffective=Tposition+RpeakDD=F×Rpeak
D = F × R p e a k × ( T p o s i t i o n + D R p e a k ) D = F \times R_{peak} \times (T_{position} + \frac{D}{R_{peak}}) D=F×Rpeak×(Tposition+RpeakD)
Transform to obtain
D = F 1 − F × R p e a k × T p o s i t i o n D = \frac{F}{1-F} \times R_{peak} \times T_{position} D=1−FF×Rpeak×Tposition
Assume that you want to get 90% efficiency, assuming T p o s i t i o n T_{position} TpositionIt is 0.01 seconds, the disk transfer rate is 100MB/s, so the calculated D is 9MB, so the cache size is 9MB