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Question number One:

 

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.

 

  1. What is the drive capacity?
  2. What is average access time for the drive?

 

Solution:

 

The drive capacity will be as follows:

 

Hard drive has surfaces = 4

Tracks per surface = 512

Constant sector per track = 64

Sector size = 2K byte

What is the drive capacity=?

 

Capacity = 4 x 512 x 64 x 2k

               = 262.144MB

 

The second part of the question is as follows:

 

The Average seek time =3ms

Drives runs at = 3,600 rpm

What is average access time for the drive=?

Rotational latency = Rotational time/2

 

=      60      = 8.3ms

  3,600x2

Average access time= Seek time + Rotational latency

                                 = 3ms + 8.3

                                 =11.3ms

 

 

 

Question number Two:

 

Use the hamming code to encode the following 4-bit values.

Also write down the encoding scheme.

 

  1. A.                 0110
  2. B.                 1001
  3. C.                 0000
  4. D.                1111

 

 

Solution:

 

Encoding:

Traditional Hamming code are (7, 4) codes, encoding four bits of data into seven bit blocks (a Hamming code word). The extra three bits are the parity bits. Each of the three parity bits is parity for three of the four data bits, and no parity bits are for the same three data bits. All of the parity bits are even parity bits.

 

Step 1:

Represent each data bit with a column vertex as follows:

 

             0

d1 =     1

            1

            0

 

 


          1

            0

d2 =     0

            1

 

 

0

d3 =     0

0

0

 

            1

            1

d4 =     1

            1

 

 

Step 2:

Represent each parity bit with a column vector containing a 1 in the row corresponding to each data bit included in the computation and a zero in all other rows. Using the parity bit definition:

 

         1

P1=   0

         0

         1

 

 

            0

P2 =    1

            1

            0

 


            1

P3 = `   1

 1

 1

 

 

 

Step 3:

 

Now we will encode the data value 0110 using the Hamming code :

 

 


1   0    1    0    1    0    1

0   1    1    1    0    0    1

                              [0110]             0   1    1    1    0    0    1

1   0    1    0    1    0    1

 

 

= (0*1) + (1*0) + (1*1) + (0*0)

(0*1) + (1*0) + (1*0) + (0*1)

(0*1) + (1*0) + (1*0) + (0*1)

(0*1) + (1*0) + (1*1) + (0*0)

(0*1) + (1*1) + (1*0) + (0*1)

(0*1) + (1*1) + (1*0) + (0*1)

(0*1) + (1*1) + (1*0) + (0*1)

 

0 + 0 + 1 + 0 = 1

0 + 0 + 0 +0 = 0

0 + 0 + 0 + 0= 0

0 + 0 + 1 + 0 = 1

0 + 1 + 0 + 0 = 1

0 + 1 + 0 + 0 = 1

0 + 1 + 0 + 0 = 1

 

Hence the encoded value of the data value of 0110 is = 100111

 

 


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CS501 Advance Computer Architecture Assignment#05 Solution

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