Continuous Media Storage and Retrieval - Scalable Continuous Media Streaming Systems - page 55

T avg = Q / R

. . .

. . .

Transmission

. . .

. . .

. . . . . .

. ..

Disk 0

. . .

. . . . ..

. . .

Disk 1

. ..

. . . . . .

. . .

Disk 2

One service round retrieving up to dk media blocks

Figure 3.11 Retrieval and transmission under the concurrent scheduler (blocks of the same pattern
belongs to the same media stream)

own seek optimization using CSCAN and the data blocks retrieved from all disks are then
sequentially transmitted in the next service round as shown in Figure 3.11.
As the disks effectively operate in parallel serving the same set of media streams, it seems
that the throughput will then be increased by d times for a d -disk array. Interestingly, the
actual achievable throughput under concurrent schedule can in fact be larger than d times the
throughput of a single disk.
Recall that in capacity dimensioning the constraint is that the worst-case time spent in a
service round must not exceed the transmission time for the data retrieved:

Q
R =

t max

round ( k )

≤ T avg (3.10)

Extending this to a d -disk striped disk array we have

dQ
R =

t max

round ( k )

≤ dT avg (3.11)

Now since the number of tracks in the disk is the same no matter a disk is used independently
or in a striped disk array, the seek distance on a per-data-block basismust decrease if we increase
the number of blocks retrieved in a service round. In other words, we have in general

t max dt max

≤ ≤

round ( dk ) round ( k )

dT avg (3.12)

Thus, by retrieving more data blocks in a service round, the striped disk array can achieve
lower seek overhead than operating the disks independently. This performance gain, however,
is offset by two trade-offs.
Assume that each disk has a capacity to retrieve k data blocks in a service round of T avg
seconds. Ignoring the performance gains due to reduced seek overhead, then in a d -disk array
each disk will retrieve dk data blocks in a service round of dT avg seconds, or d 2 k data blocks
for the entire disk array. Therefore, the buffer requirement is equal to 2 d 2 kQ bytes, half of