CS501 - Assignment 5 Solution Spring 2011 - June 2011

[paki daramas]

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Assignment No. 05 Semester Spring 2011 Advance Computer Architecture - CS501
Question No 1



10 Marks

e 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)



Sponsored Links

Total (8+2 = 10 marks)



A. 0110

B. 1001

C. 0000

D. 1111



Note : Use odd bit parity to encode above values

[Vuhelper]

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

[moon13]

please explain what is this ?

[Xpert]

i dont know also.......

[faisal mahmood]

xprt plz upload the solution of cs501

[Xpert]

Dear i am busy now a days. I am not giving full time to assignments I am really sorry.

[moon13]

very good this is good answer yes brothers visit other websites for assignment solutions and do not disturb Xpert bhi hhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[Vuhelper]

ust 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
be read without head movement.
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

[moon13]

this is wrong solution please check it again

[Vuhelper]

it is only idea