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Retrieval Multimedia Data from Disks

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RAID 5: Independent Data Disks with Distributed Parity Blocks ... 1 parity. D2. D3. D4. 0 parity. E1. E2. E3. E4. Router. Server1. d1. d2. d3. dm. Server n. d1 ... – PowerPoint PPT presentation

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Title: Retrieval Multimedia Data from Disks


1
Retrieval Multimedia Data from Disks Presented by
Yuni Xia
2
  • Fundamental characteristics
  • Real-time storage and retrieval
  • Large data transfer rate and storage space
    requirement
  • Why choosing magnetic disk
  • Storage capacity
  • Speed
  • Moderate cost / Random access / Writing

3
Side View
platters
Read/write head
Spindle
Top View
Tracks
Sector
4
Symbol Meaning
tnum snum itd ss rv dtr
total of tracks total of sectors intertrack
distance spin speed radial velocity data
transfer rate
Suppose Wish to read data sector i on track ti,
read head is currently over sector j in track tj
Readtime seek(ti, tj) rotation(si,sj)data/dt
r seek(ti, tj) abs(ti-tj) / rv
rotation(si,sj) (abs(si-sj) / snum ) / ss
5
  • Raid arrays and Placement methods
  • By spreading data across several hard disks
  • faster performance
  • greater storage capacity
  • higher data security
  • Six standards 0-5
  • (cross-type variations, such as 0/1, 3/5)
  • Implemented by software and hardware

6
RAID 0 Striped Disk Array without Fault
Tolerance RAID Level 0 requires a minimum
of 2 drives to implement
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
etc..
7
RAID 1 Mirroring and Duplexing RAID
Level 1 requires a minimum of 2 drives to
implement
A
E
I
M
A
E
M
I
B
F
J
N
B
F
N
J




C
G
K
O
C
G
O
K
D
H
L
P
D
H
P
L
8
RAID 5 Independent Data Disks with Distributed
Parity Blocks RAID Level 5 requires a
minimum of 3 drives to implement
A0
B0
C0
D0
0 parity
A1
B1
C1
1 parity
E1
A2
B2
2 parity
D2
E2
A3
3 parity
C3
D3
E3
4 parity
B4
C4
D4
E4
9
A model of heterogeneous disk servers
Router
Server1
Server n
. . .
...
...
d1
d2
d3
dm
d1
d2
d3
dn
10
  • What needs to be modeled?
  • The intrinsic characteristics of each disk
    server
  • The intrinsic characteristics/capabilities of
    each client
  • The relationship between the disk servers and
    clients
  • The distribution of data across the disk server

11
Disk Server Characteristics 1. Dtr(i)
Total disk bandwidth of disk server i 2. Buf(i)
Total buffer space associated with server
i 3. Switchtime(i, t) Time required for si
to switch between clients at time t 4. Cyctime(i,
t) One cycle of read operation to be
executed by si at time t
12
Client Characteristics 1. Cons(i,t) The
consumption rate of client Ci at time t 2.
Data(i, t) (M, b) Play data(i, t)
(m,b), (m, b1), FF data(i, t)
(m,b), (m, bffs), (m, b2ffs), RW
data(i, t) (m,b), (m, b-rws), (m, b-2rws),
Pause data(i, t) (m,b)
13
Client Characteristics Data(i, t) (M, b, len,
step) b, (bstep), (b2step), . ,
(b(len-1)step) 1. Play step 1
2. FF step ffs 3. RW step -rws
4. Pause step 0
14
  • Client-Server Characteristics
  • 1. Timealloc(i,j,t)
  • In any given cycle of disk server i, each
    client
  • cj has a time-slice, timealloc(i, j, t)
  • cyctime(i, t) gt sum( timealloc(i,j,t))
  • (ni,t switchtime(i,t))
  • 2. active(t)
  • The set of all clients that are active at time
    t.
  • 3. d_active(i, t)
  • active(t) Union(d_active(i, t))

15
Client-Server Characteristics 4. Ut (i) The
set of servers which are handling the requests of
client Ci. Ut (i) S Ci d_active(s,
t) 5. Bufreq(j, i, t) The amount of buffer
that is required at server Si so that data that
client Cj needs to read doesnt get
overwritten. Buf(i) gt sum(bufreq(j, i, t)
16
  • Distribution of Data
  • M (mi , b) placement mapping
  • The set of all servers that contain block b of
    mi
  • M ( Sound of Music, 20 ) 2, 4, 5
  • Placement constraint
  • data (C, t) (m, b, len, step) i (0lt
    i ltlen)
  • ( j Ut (i) ) ( j M (mi , b ( i
    step ) )

17
Suppose Data (C, t) (M, 5, 5, 3),
Ut (C) 1, 3, 4 5, 8, 11, 14, 17
must be in S1, S3, S4
Definition State of an MOD System S(t) 1.
Active ( t ) 2. Cyctime (i, t) 3. Cons ( i,
t ) 4. Timealloc ( i, j, t) 5. Data ( i,
t) 6. Ut
18
Disk availability constraint 1. Consumption Rate
Constraint Sum(cons(j,t)) switchtime(i,t)
dtr(i)/ cyctime(i,t) ltdtr(i) 2. Buffer
requirement constraint sum(buf(j, i, t)) lt
buf (i) timealloc (i, j, t) cyctime (i, t)
cons (j, t) / dtr(i) bufreq(j,i,t)
(dtr(i)-cons(j,t)) timealloc(i, j, t)
19
Router
Server1 B 1-150
Server 2 B 151-250
Server 3 B 200-300
...
...
...
(mi, 140, 2, 5)
(140, 145)
(150, 155)
(mi, 199, 2, 1)
(199, 200)
(201, 202)
20
Trans Transaction type
Priority
tr1 tr2 tr3 tr4 tr5 tr6
exiting client continuing client-normal continuing
client-needs switching continuing client- needs
splitting new client new client -needs splitting
5 4 3 3 2 1
21
  • An event-based algorithm QuickSOL
  • FindSOL
  • OptimizeSOL

FindSOL Phase 1. Split EV(t) into 6 sets
new(t), exit(t), cont(t), pause(t), ff(t),
rew(t) 2. (handle exiting clients)
For each clients Ci in exit(t) do 1) free the
resources 2) delete Ci from state table
22
3. (Handle Continuing Clients) For each
clients Ci in cont(t) or ff(t) or rew(t) do
If servers currently assigned to C satisfy
.. then modify the state table else 1)
re-set Cs priority to 3 2) Move it
into new(t) 3) update the resource
table 4. (Handle New Clients) For each
clients Ci in new(t) do 1) Identify the
servers that have the data required by C
23
  • 2) Determine which server have enough bandwidth
    ..
  • IF no such server is available,
  • split the event into 2 sub-events
  • data(C, t) (m, s, l/2, 2step) and
  • data(C, t) (m, sstep, l/2,
    2step)
  • Keep splitting till for both sub-events ...
  • Update state table
  • 3) Do the same as 2) in terms of buffer
    requirement

OptimizeSOL phase 1. Switching 2. Splitting
Balancing the load, Maximizing the of
clients ...
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