Friday, 19 April 2013
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.
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Memory
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CPU register
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200
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1 7 8 1
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200
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PC
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201
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5 7 8 2
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AC
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202
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2 7 8 2
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1 7 8 1
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IR
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 203 |
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:
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7 8 1
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0 0 0 4
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7 8 2
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0 0 0 2
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(STEP 1)
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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.
Answer:
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Memory
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CPU register
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Memory
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CPU register
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200
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1 7 8 1
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PC
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200
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1 7 8 1
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PC
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201
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5 7 8 2
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AC
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201
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5 7 8 2
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AC
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202
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2 7 8 2
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IR
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202
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2 7 8 2
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IR
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203
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203
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:
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:
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7 8 1
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0 0 0 4
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7 8 1
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0 0 0 4
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7 8 2
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0 0 0 2
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(STEP 2)
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7 8 2
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0 0 0 2
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(STEP 3)
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Memory
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CPU register
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Memory
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CPU register
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200
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1 7 8 1
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PC
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200
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1 7 8 1
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PC
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201
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5 7 8 2
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AC
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201
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5 7 8 2
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AC
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202
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2 7 8 2
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IR
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202
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2 7 8 2
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IR
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203
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203
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:
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:
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7 8 1
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0 0 0 4
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7 8 1
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0 0 0 4
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7 8 2
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0 0 0 2
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(STEP 4)
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7 8 2
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0 0 0 2
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(STEP 5)
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Memory
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CPU register
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200
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1 7 8 1
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PC
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201
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5 7 8 2
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AC
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202
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2 7 8 2
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IR
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203
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:
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7 8 1
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0 0 0 4
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7 8 2
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0 0 0 2
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(STEP 6)
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