Pertemuan
7 – 25 April 2017
Dosen
: Andriani Kala’Lembang, S.Kom, M.M
SOAL C !
No. 1
- Diketahui sebuah hardisk memiliki
karakteristik:
- Seek
time (s) =
10 ms
Kecepatan putaran disk =
6000 rpm
Transfer rate (t) =
1024 byte/ms
Ukuran blok (B) =
4096 byte
Ukuran Record (R) =
400 byte
Ukuran gap (G) =
256 byte
Ukuran pointer (P) =
8 byte
·
Hitunglah :
a. Bloking factor
b. Pemborosan ruang
c. Bulk transfer rate
·
Jika metode blockingnya
:
a. Fixed blocking
b. Variabel spanned
c. Variabel unspanned
JAWABAN :
· Fixed Blocking
a. Bfr =
B/R
= 4096/400
= 10,24
b. W = WG
+ WR
WG = G/Bfr
= 256/10,24
= 25
WR = B/Bfr
= 4096/10,24
= 400
= 425
c. t’ = (t/2)×(R/(R+W))
= (1024/2)×(400/(400+425))
= (512)×(400/825)
= 512 ×0,48
= 245,76
·
Variable
Length Spanned Bloking
a.
Bfr = (B-P)/(R+P)
= (4096-8)/(400+8)
= (4088)/(408)
= 10,01
b.
W = P+(P+G)/Bfr
=
8+(8+256)/10,01
=
8+264/10,01
=
8+26,37
=
34,37
c.
t’ = (t/2)×(R/(R+W))
=
(1024/2)×(400/(400+34,37))
=
(512)×(400/434,37)
=
512 ×0,92
=
471,04
·
Variable
Length Unspanned Bloking
a.
Bfr = (B-1/2 R)/(R+P)
= (4096-1/2×400)/(400+8)
= (4096-200)/(408)
= (3896)/(408)
= 9,54
b.
W = P+(1/2 R+G)/Bfr
=
8+(200+256)/9,54
=
8+456/9,54
=
8+47,79
=
55,79
c.
t’ = (t/2)×(R/(R+W))
=
(1024/2)×(400/(400+55,79))
=
(512)×(400/455,79)
=
512 ×0,87
=
445,44
No. 2
·
Hitunglah Rotation Latency bila kecepatan putar
disk (RPM) adalah :
a. 3000 RPM
b. 10000 RPM
c. 7500 RPM
JAWABAN :
a. r = 1/2×((60×1000)/RPM)
= 1/2×((60×1000)/3000)
= 1/2×(60000/3000)
= 1/2×20
= 10 detik
b. r = 1/2×((60×1000)/RPM)
= 1/2×((60×1000)/10000)
= 1/2×(60000/10000)
= 1/2×6
= 3 detik
c. r = 1/2×((60×1000)/RPM)
= 1/2×((60×1000)/7500)
= 1/2×(60000/7500)
= 1/2×8
= 4 detik
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