CS501 Advance Computer Architecture Assignment 1 Solution Spring 2012

Question:

Dear Student, you are given typical view of selected Memory and Processor registers in Figure 1. In memory, you can see three instruction codes in upper area, while two data values are given in lower part. These instruction codes represent an addition operation.

Addition Operation:  You are required to perform an addition operation in which you have to add the contents of memory word at address 781 to the contents of memory word at 782. After performing this addition, result should be stored at 782. Your task is to perform step-by-step execution of these three instructions and show registers configuration at each step.

For your understanding, Step-1 of this sequence of execution is shown in Figure 1.  In this Figure, you can see PC=200 which indicate that instruction stored at address 200 is in execution and the same instruction (1781) is loaded in IR register.

Memory

  CPU register

200

1 7 8 1

200

PC

201

5 7 8 2

AC

202

2 7 8 2

1 7 8 1

IR

203

:

7 8 1

0 0 0 4

7 8 2

0 0 0 2

(STEP 1)

Figure-1

Where,

AC is Accumulator Register.

PC is Program Counter.

IR is Instruction Register.

 

Submission:

You need to complete execution of the three given instructions and provide step wise registers and memory configuration. Please submit your solution in MS Word format by completing/filling the provided steps on page-4.

 

Hint Regarding Three Instructions:

  • First instruction 1781 means Load value stored at address 781 into accumulator register.
  • Second instruction 5782 means perform addition of value stored at address 781 with value in accumulator register and store the result in accumulator register.
  • Third instruction 2782 means store value of accumulator register at address 782.

Solution:


 

Memory

  CPU register  

Memory

  CPU register

200

1 7 8 1

 

200

PC  

200

1 7 8 1

 

201

PC

201

5 7 8 2

 

 0 0 0 4

AC  

201

5 7 8 2

 

 0 0 0 4

AC

202

2 7 8 2

 

1 7 8 1

IR  

202

2 7 8 2

 

5 7 8 2

IR

203

 

   

203

 

 

:

 

   

:

 

 

7 8 1

0 0 0 4

 

   

7 8 1

0 0 0 4

 

 

7 8 2

0 0 0 2

 

(STEP 2)

   

7 8 2

0 0 0 2

 

(STEP 3)

 

 

   

 

 
       

             
     

             

Memory

  CPU register  

Memory

  CPU register

200

1 7 8 1

 

 201

PC  

200

1 7 8 1

 

PC

201

5 7 8 2

 

0006

AC  

201

5 7 8 2

 

AC

202

2 7 8 2

 

 5 7 8 2

IR  

202

2 7 8 2

 

IR

203

 

   

203

 

 

:

 

   

:

 

 

7 8 1

0 0 0 4

 

   

7 8 1

0 0 0 4

 

 

7 8 2

0 0 0 2

 

(STEP 4)

   

7 8 2

0 0 0 2

 

(STEP 5)

 

 

   

 

 

 

 

 

Memory

  CPU register

200

1 7 8 1

 

PC

201

5 7 8 2

 

AC

202

2 7 8 2

    IR

203

 

 

:

 

 

7 8 1

0 0 0 4

 

 

7 8 2

0 0 0 6

 

(STEP 6)

 

 

 

 

 

 

 

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