1.Rotational Latency
a. r = ½ x (( 60 x 1000 ) / RPM) = ½ x (( 60 x 1000 ) / 2500) = ½ x 24 = 12
b. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 2000) = ½ x 30 = 15
c. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 7000) = ½ x 8,57 = 4,285
d. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 8000) = ½ x 7,5 = 3,75
e. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 10000) = ½ x 6 = 3
2.Metode Fixed Blocking
Diketahui :
- Seek Time = 10 ms
- Kecepatan Putar Disk = 3000 Rpm
- Transfer Rate = 1024 byte/s
- Ukuran Blok = 2048 byte
- Ukuran Record = 128 byte
- Ukuran Gap = 64 byte
Ditanya :
a) Bfr
b) r
c) TR
d) btt
e) W
f) t'
Jawaban
a.Bfr = B/R = 2048 / 128 =16
b.r = ½ x (( 60 x 1000 ) / RPM)
= ½ x (( 60 x 1000 ) / 3000)
= ½ x 20
= 10
c.TR = R/t = 128 / 1024 = 0,125
d.Btt = B/t = 2048 / 1024 = 2
e.W = G/Bfr = 64 / 16 = 4
f.t’ = (t/2)*(R / (R + W))
= (1024 / 2) * (128 / (128 + 4))
= 512 * (128 / 132)
= 512 * 0,96
= 496,48
3. Tugas
Diketahui :
- Seek Time = 10 ms
- Kecepatan Putar Disk = 6000 Rpm
- Transfer Rate = 2048 byte/s
- Ukuran Blok = 2048 byte
- Ukuran Record = 250 byte
- Ukuran Gap = 256 byte
- Ukuran Pointer = 8
Ditanya :
a) Blocking Faktor
b) Record Transfer Rate
c) Block Transfer Time
d) Pemborosan Ruang
e) Bulk Transfer Rate
Jika Metodenya :
a) Fixed Bloking
b) Spanned Bloking
c) Unspanned Bloking
a. r = ½ x (( 60 x 1000 ) / RPM) = ½ x (( 60 x 1000 ) / 2500) = ½ x 24 = 12
b. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 2000) = ½ x 30 = 15
c. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 7000) = ½ x 8,57 = 4,285
d. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 8000) = ½ x 7,5 = 3,75
e. r = ½ x (( 60 x 1000 ) / RPM) =½ x (( 60 x 1000 ) / 10000) = ½ x 6 = 3
2.Metode Fixed Blocking
Diketahui :
- Seek Time = 10 ms
- Kecepatan Putar Disk = 3000 Rpm
- Transfer Rate = 1024 byte/s
- Ukuran Blok = 2048 byte
- Ukuran Record = 128 byte
- Ukuran Gap = 64 byte
Ditanya :
a) Bfr
b) r
c) TR
d) btt
e) W
f) t'
Jawaban
a.Bfr = B/R = 2048 / 128 =16
b.r = ½ x (( 60 x 1000 ) / RPM)
= ½ x (( 60 x 1000 ) / 3000)
= ½ x 20
= 10
c.TR = R/t = 128 / 1024 = 0,125
d.Btt = B/t = 2048 / 1024 = 2
e.W = G/Bfr = 64 / 16 = 4
f.t’ = (t/2)*(R / (R + W))
= (1024 / 2) * (128 / (128 + 4))
= 512 * (128 / 132)
= 512 * 0,96
= 496,48
3. Tugas
Diketahui :
- Seek Time = 10 ms
- Kecepatan Putar Disk = 6000 Rpm
- Transfer Rate = 2048 byte/s
- Ukuran Blok = 2048 byte
- Ukuran Record = 250 byte
- Ukuran Gap = 256 byte
- Ukuran Pointer = 8
Ditanya :
a) Blocking Faktor
b) Record Transfer Rate
c) Block Transfer Time
d) Pemborosan Ruang
e) Bulk Transfer Rate
Jika Metodenya :
a) Fixed Bloking
b) Spanned Bloking
c) Unspanned Bloking
Jawab:
1. Fixed
Blocking
a. Bfr
= B/R = 2048 / 250 = 8,192
b. TR
= R/t = 250 / 20148 = 0,122
c. Btt
= B/t = 2048 / 2048 = 1
d. WG
= G/ Bfr = 256 / 8,192 = 31,25
WR = B / Bfr = 2048 /8,192 = 250
W = WG
+ WR
=250
+ 31,25
=
281,25
e. t’ = (t / 2) * (R / (R + W))
= (2048 / 2) * (250 / (250 + 281,25))
= 1024 * (250 / 531,25)
= 1024 * 0,47
= 481,28
a. Bfr = (B-P)/(R+P)
=
(2048 – 8) / (250 + 8)
=
2040 / 258
=
7,906
b. TR
= R/t = 250 / 20148 = 0,122
c. Btt
= B/t = 2048 / 2048 = 1
d. W = P + (P + G) / Bfr
= 8 + (8 + 256) / 7,906
= 8 + 264/7,906
= 8 + 33,39
= 41,39
e. t’ = (t / 2) * (R / (R + W))
= (2048 / 2) * (250 / (250 + 41,39))
= 1024 * (250 / 291,39)
= 1024 * 0,85
= 870,4
3. Variabel
Unspanned
a. Bfr = (B–½R)/(R+P)
= (2048 – ½ . 250) / (250 / 8)
= 1923 / 258
= 7,453
b. TR
= R/t = 250 / 20148 = 0,122
c. Btt
= B/t = 2048 / 2048 = 1
d. W = P + ( ½ R + G) / Bfr
= 8 + ( ½ . 250 + 256) / 7,453
= 8 + 381 / 7,453
= 8 + 51,12
= 59,12
e. t’ = (t / 2) * (R / (R + W))
= (2048 / 2) * (250 / (250 + 59,12))
= 1024 * (250 / 309,12)
= 1024 * 0,808
= 897,392