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 Question No 1 10 Marks A hard drive has 4 surfaces, with 512 tracks per surface and constant 64 sectors per track. Sector size is 2K byte. The average seek time is 3 ms, the drive runs at 3,600 rpm.   What is the drive capacity? (5) What is average access time for the drive? (5) Question No 2 10 Marks Use the hamming code to encode the following 4-bit values (8 marks) Also write down the encoding scheme (2 mark)    Total (8+2 = 10 marks)   A.                 0110 B.                 1001 C.                 0000 D.                1111   Note : Use odd bit parity to encode above values

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Cs501_5th Assignment idea solution

Q: 1 Convert the following base12 numbers to decimal numbers.

Assume that

they are unsigned.

1. 9A6B12

x=0

x=x+9=9

x = 12x9 + A = 118

x = 12x118 + 6 = 1422

x = 12x1422 + B = 17075

So, 9A6B12 = 1707510

2. AABB12

x=0

x = x + A (=10) = 10

x = 12x10 + A (=10) = 130

x = 12x130 + B (=11) = 1571

x = 12x1571 + B (=11) = 18863

So, AABB12 = 1886310

Q: 2 Convert the following base10 number to base2 number.

1. 0.7510

0.75x2 = 1.5, f-1 = 1

0.5x2 = 1.0, f-2 = 1

So 0.7510 = (0.11)2

Q: 3 Convert the following base16 number to decimal number.

1. 0.ABC16

F=0

F = (0 + 12) / 16 = 0.75

F = (0.75 + 11) / 16 = 0.734375

F = (0.734375 + 10) / 16 = (0.6708984)10

See the attached file please CS501 Assignment#05 Solution
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wrong solution

Wrong Solution plz share right solution.
it is not rele
what is progress only 7 hours left please

Hey friends,

Just follow this example and you will get through Question no: 01 !!!

Just don't forget to change your format otherwise, your assignment will be caught as a copied one !!!

A hard drive has 8 surface, with 521 tracks per surface and a constant 64 sectors per track.

Sector size is 1KB. The average seek time is 8 m sec, the track to track access time

is 1.5 m sec and the drive runs at 3600 rpm. Successive tracks in a cylinder can

a) What is the drive capacity?

b) What is the average access time for the drive?

c) Estimate the time required to transfer a 5MB file?

d) What is the burst transfer rate?

Solution:-

a) Capacity  = Surfaces x tracks x sectors x sector size

= 8 x 521 x 64 x 1k = 266.752MB

b) Rotational latency : Rotation time/2 = 60/3600x2 = 8.3 m sec

Average Access time = Seek time + Rotational latency

= 8 m sec + 8.3 m sec = 16.3 m sec

c) Assume the file is stored in successive sectors and tracks starting at sector #0,

track#0 of cylinder #i. A 5MB file will need 1000 blocks and occupy from cylinder #1,

track #0, sector #0 to cylinder #(i+9), track #6, sector #7, we also assume the size of disk buffer is unlimited.

The disk needs 8ms, which is the seek time, to find the cylinder  #i, 8.3 ms

to find sector #0 and 8 x (60/3600) seconds to read all 8 tracks data of this cylinder.

Then the time needed for the read to move to next adjoining track will be only 1.5 m sec.

Which is the track to track access time. Assume a rotational latency before each new track.

Access time = 8+9(8.3¸8x16.6+1.5)+8.3+6x16.6+8/64x16.6

= 1406.6 m sec

d) Burst rate = rows/sec x sectors/row x bytes/sector

= 3600/60 x 64 x 1k = 3.84 MB/sec

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