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Advance Computer Architecture (CS501) Assignment#01(GRADED) Total marks = 20 Deadline Date = 24-November-2015

Advance Computer Architecture (CS501)

Assignment#01(GRADED)

Total marks = 20

                                                                                       Deadline Date = 24-November-2015

 

Please carefully read the following instructions before attempting the assignment.

Assignment Guidelines:

It should be clear that your assignment would not get any credit if:

  • The assignment is submitted after due date.
  • The submitted assignment does not open or file is corrupt.
  • The assignment is copied. Note that strict action would be taken if the submitted assignment is copied from any other student. Both students will be punished severely.

 

1)     You should consult recommended books to clarify your concepts as handouts are not sufficient.

2)     You are supposed to submit your assignment in .doc or docx format. Any other formats like scan images, PDF, Zip, rar, bmp etc will not be accepted.

3)     You are advised to upload your assignment at least two days before due date.

Important Note:

Assignment comprises of 20 Marks. Note that no assignment will be accepted after due date via email in any case (whether it is the case of load shedding or emergency electric failure or internet malfunctioning etc.). Hence, refrain from uploading assignment in the last hour of the deadline, and try to upload Solutions at least 02 days before the deadline to avoid inconvenience later on.

For any query please contact: CS501@vu.edu.pk

 

The objective of this assignment is:

  • To understand the implementation of 4address, 3address, 2address, 1address , 0address  instruction format.
  • To performcomparison among different instruction formats with help of code size and number of bytes memory accessed.

 

 

 

Question:

 

You need to perform the comparison among 4address, 3address, 2address, 1address and 0address instruction formats with the help of code size and number of bytes accessed from memory. After that, find out the minimum code size and number of bytes memory accessed.

 

You are required to perform the above task with the help of following two steps:

 

a)     First of all, calculate the code size and number of memory access for each 4address, 3address, 2address, 1address and 0address instruction formats separately from the following information:

 

  • Op-code takes 2 bytes it means that 2 bytes is used for the op code.
  • Size of each source operand is 5 bytes.
  • Size of destination operand is 5 bytes.
  • 2 bytes are used to store next instruction address.

 

b)     At the last, perform the comparison among 4address, 3address, 2address, 1address and 0address instruction formats.

Note: Two parameters i.e. code size and numbers of bytes accessed from memory are used for comparison.

 

 

 

Good Luck

 


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yar ye complete nai ha plz.

complete send karen.

kis ne kr li he assignment???

100% complete solution ....

4-Address Instruction Format:
Op code destination source1 source2 next address
2 bytes 5 bytes 5 bytes 5 bytes 2 bytes
Memory access=19 bytes for instruction fetch + 10 bytes for source operand fetch + 5 bytes for storing destination operand = 34 bytes
Code size=2+5+5+5+2=19 bytes,Memory access=34 bytes

3-Address Instruction Format:
Op code destination source1 source2
2 bytes 5 bytes 5 bytes 5 bytes
Memory access=17 bytes for instruction fetch + 10 bytes for source operand fetch + 5 bytes for storing destination operand =32 bytes
Code size=2+5+5+5+=17 bytes,Memory access=32 bytes

2-Address Instruction Format:
Op code destination source1 source2
2 bytes 5 bytes 5 bytes
Memory access=12 bytes for instruction fetch + 10 bytes for source operand fetch + 5 bytes for storing destination operand=27 bytes
Code size=2+5+5=12 bytes,Memory access=27 bytes

1-Address Instruction Format:
Op code source2
2 bytes 5 bytes
Memory access=7 bytes for instruction fetch + 5 bytes for source operand fetch + 0 bytes for storing destination operand=12 bytes
Code size=2+5=7 bytes,Memory access=12 bytes

0-Address Instruction Format:
Op code
2 bytes
Memory access=2 bytes for instruction fetch + 10 bytes for source operand fetch + 5 bytes for storing destination operand=17 bytes
Code size=2 bytes,Memory access=17 bytes

Instruction Format Code Size Number of Memory Bytes
4-Address instruction 19 34
3-Address instruction 17 32
2-Address instruction 12 27
1-Address instruction 7 12
0-Address instruction 2 17

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