Computer Organization

Posted: November 26th, 2013

Computer Organization

Assignment #2        Name: _________________________________________                         Spring, 2011

 

Circle True or False for each statement (1 point each):

 

1.  True False           Pipelining improves performance by duplicating portions of a processor so that multiple parts can be doing the same function at the same time. Pipelining improves performance by increasing the number of instructions that can be executed at the same time. The processor processes two or more instructions concurrently by dividing the executions among several sub units.

2.  True False           “Big endian” and “little endian” are methods of encoding or encrypting data for secure transmission over a network. Endians are used to order bytes in a certain way. Big endian arranges words in such a way that the last byte is the most significant. Little endian arranges bytes in such a way that the last byte is the least significant.

3.  True False           The use of one parity bit can detect an error in a 16-bit data value assuming that only one bit could be in error at the most. Parity detects but cannot correct errors. One parity bit is not enough to detect an error since it cannot tell which bit is bad.

4.  True False           A Hamming code using three parity bits is enough to detect and correct a one-bit error in a data value of 4 bits. The three parity bits ensure that there is an even number of bits. One can find out where an error is if the total number is not even.

5.  True False           Cache memory is added to a CPU to provide reliability in case of data transmission errors. The cache is located between the central processing unit and the main memory. It is small fast and expensive. It is used to store data and instructions that are used frequently. In case the data is missing from the cache it is read from the main memory.

6.  True False           RAID is used to improve efficiency and/or add error-correction features. RAID stands for redundancy array independent disk and is used in storage. Redundant disk capacity stores parity information which is used for error detection. This ensures that data can be recovered in case the disk fails.

7.  True False           A typical single-sided, single-layer DVD can store roughly 4.7 GB of data. Compared to a CD, a DVD has smaller pits, a more closely spaced track and a shorter wavelength red laser. Its capacity enables it to have more resolutions making it more adaptable for video adaptation

8.  True False           Cable-based Internet service will always provide consistent speeds, regardless of how many people in your neighborhood are using the Internet at the same time as you are. If all the people serviced by the same signal were to access the internet at the same time, they would get an equal share of the bandwidth. They will therefore use the same consistent but slow speed.

9.  True False           Most color printers use an RGB system to produce colors made up of varying amounts of red, green, and blue inks. Different colors are made by combining different shades of red blue and green.

10.  True False         LCD display panels are made using special liquid crystals that produce glowing light when electricity is applied to them. The liquid form twists the helix which in turn bends the light entering the display and forms a thin low power display

 

 

 

 

 

 

 

 

 

 

Problems: Show all steps of your work for each question ( 5 points each ):

 

1.     A 4-bit data value followed by 3 bits of Hamming Code (using even parity) is received as the value 1111000.  Assume that at most one bit is incorrect.  Use the Venn Diagramming method explained in the text to diagram this code with error-correction bits, being sure to follow the text’s ordering for entering data bits and parity bits in alphabetical order for consistency.  Is the value received correct or incorrect?  If it is not correct, show the corrected 4-bit value.

 

Before correction, the regions will have the values, AB -1, ABC – 1, AC – 1, BC – 1. With this in mind, it is noted that all circles A, B and C will have wrong (odd) parity. To correct, this, the region ABC undergoes a single-bit change restoring it to 0. Therefore, all regions A, B and C will have numbers 0,1,0,1 with the total being 2, an even number. The corrected 4-bit value is shown below:

 

 

 

 

 

 

 

 

Corrected 4 – Bit Value

 

AB              – 1

ABC           – 0

AC              – 1

BC              – 1

 

Value: 1011000

 

 

 

 

 

 

2.     The value shown below is transmitted using a Hamming Code for 16 data bits with 5 parity bits (positions 1, 2, 4, 8, and 16 – shown shaded below). Using even parity and the sample on pages 76 and 77, indicate if the value is correct or incorrect assuming that at most one bit is incorrect.  If the value is incorrect, indicate which bit is incorrect and show how you determined it.

 

1

0

1

1

1

1

1

0

0

0

0

0

1

0

0

0

0

1

1

1

1

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

 

 

Parity bit 1 incorrect (1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, contain seven 1s)

Parity bit 2 incorrect (2, 3, 6, 7, 10, 11, 14, 15, 18, 19 contain five 1s)

Parity bit 4 incorrect (4, 5, 6, 7, 12, 13, 14, 15, 20, 21 contain seven 1s)

Parity bit 8 incorrect (8, 9, 10, 11, 12, 13, 14, 15 contain one 1s)

Parity bit 16 correct (16, 17, 18, 19, 20, 21 contains four 1s)

 

Since even parity is being used, then the total number of 1s covered by the parity bits should be an even number. Therefore, the incorrect bit must be one of the bits checked by parity bits 1, 2, 4 and 8. Since the bits covered by parity bit 16 are correct, then we can eliminate 16, 17, 18, 19, 20 and 21).

The incorrect bit can be computed by adding the incorrect bits: namely, 1, 2, 4, 8.

Therefore, the incorrect bit is 1 + 2 + 4 + 8 = 15.

It is noted that this bit appears in all the incorrect lists.

Since it is a 0, it is changed to a 1 to correct the anomaly.

 

The correct codeword is therefore changed from 1111000010001111 to 1111000010101111.

 

 

 

 

 

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