PEC is announcing the result of Class - 5 and Class 8th on Saturday 31, March 2012.
Result Class 5 and 8 : Toba Tek Singh
Download From the Links Given BelowResult: Grade: 5
Result : Grade: 8
Thank you for visiting.
Farmaya NABI_E_KAREEM (SAW)ne.qabr ka azaab barhaq hai.insaan ke paida hone se pehle uski maut muqarrer kardi jati hai.
Din me 4 martaba hamari qabr hame pukarti hai lakin ham maut ko kabhi yaad nahi karte.har momin ko chahia ke woh maut ko yaad kare.gunah se perhazz kare,maafi mange aur her din yeah dua 25 martaba padhe.
Allahumma baarikli filmauti wafeema baadal maut.InshaAllah maut ki sakhti aasan hogi ....
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Assignment No. 05 SOLUTION
Semester Spring 2011 Advance Computer Architecture - CS501 |
Total Marks: 20
Due Date: 01/07/2011
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Objective:
To learn and understand basic concepts of disk drives, and
error control
Instructions:
Please read
the following instructions carefully before solving & submitting
assignment:
Assignment should be in
your own wordings not copied from net, handouts or books.
It should be clear that
your assignment will not get any credit (zero marks) if:
Note: You have to provide solution with all the calculations and
formulations involved, else no marks will be awarded.
For any query about the assignment, contact at cs501@vu.edu.pk
GOOD LUCK
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Question No 1
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10 Marks
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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.
Solution:
Average access time= Seek time +
Rotational latency = 3 + 8.3 = 11.3 ms
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Question No 2
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10 Marks
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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
Solution:
The encoding scheme is P1P2D3P4D5D6D7
, using odd parity
0110 → 0001110
1001 → 1110001
0000 → 1101000
1111 → 0010111
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