quiz
PERTEMUAN 7
QUIZ
Soal C
1. Diketahui
sebuah hardisk memiliki karakteristik :
a.) Seek
time (s) = 10 ms
b.) Kecepatan
putaran disk = 6000 rpm
c.) Transfer
rate (t) = 1024
byte/s
d.) Ukuran
block (B) = 4096 byte
e.) Ukuran
record (R) = 400 byte
f. d.) Ukuran
gab (G) = 256 byte
g.) Ukuran
pointer (P) = 8 byte
Hitung :
a.) Bloking
factor
b.) Pemborosan
ruang
c.) Bulk
transfer rate
Jika
metode blokingnya
a.) Fixed
blocking
b.) Spanned
blocking
c.) Unspanned
blocking
2.
hitung rotation latency bila kecepatan putar disk (rpm) adalah sebagai berikut
:
a.)
3000 rpm
b.)
10000 rpm
c.)
7500 rpm
jawab
fixed blocking
Bfr
= B/R
=
4096/400 = 10,24
WG
= G/Bfr
=
256/10,24 = 25
WR
= B/Bfr
=
4096/10,24 = 400
W
= WG + WR
= 25 + 400 = 425
Btt
= B/t
=
4096/1024 = 4
TR
= R/t
=
400/1024 = 0,39
t’ = (t/2) * {R/(R+W)}
=(1024/2) * {400/(400+425)}
=512 * 400/825
= 512 * 0,48
= 245,76
Spanned blocking
Bfr
= (B-P) / (R+P)
=
(4096-8) / (400+8)
=
4088 / 408
=
10,01
W
= P+(P+G) / Bfr
= 8+(8+256) / 10,01
= 8+264 / 10,01
= 8+26,37
= 34,37
Btt
= B/t
=
4096/1024 = 4
TR
= R/t
=
400/1024 = 0,39
t’ = (t/2) * {R/(R+W)}
= (1024/2) * {400/(400+34,37)}
=512 * 400/434,37
= 512 * 0,92
= 471,04
Unspanned blocking
Bfr
= (B – ½ R) / (R + P)
=
(4096 – ½ 400) / (400 + 8)
=
4096 – 200 / 408
=
4096 - 0,49
= 4,09
W
= P + (1/2 R + G) / Bfr
= 8
+ (1/2 400 + 256) / 4,09
= 8
+ 456 / 4,09
= 8
+ 111,49
=
119,49
Btt
= B/t
=
4096/1024 = 4
TR
= R/t
=
400/1024 = 0,39
t’ = (t/2) * {R/(R+W)}
= (1024/2) * {400/(400+119,49)}
=512 * 400/519,49
= 512 *
= 0,76
Rotation latency
a.) r
= ½ x ((60x100) / Rpm)
= ½ x 60000 / 3000
= 30.000 / 3000
= 100
b.) r
= ½ x ((60x100) / Rpm)
= ½ x 60000 / 10000
= 30.000 / 10000
= 3
c.) r
= ½ x ((60x100) / Rpm)
= ½ x 60000 / 7500
= 30.000 / 7500
= 4
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