Suppose the location X in an ADD X instruction is 105 which contains the value o
ID: 664427 • Letter: S
Question
Suppose the location X in an ADD X instruction is 105 which contains the value of -23dand the instruction stored at 101 in the memory, complete the following table:
Please explain clearly how to do this exercise
Step
RTN
PC
IR
MAR
MBR
AC
(initial values)
101
1104
104
0023
0023
Fetch
MAR ¬ PC
101
101
101
0023
0023
IR ¬ M[MAR]
101
3105
101
0023
0023
PC ¬ PC + 1
102
3105
101
0023
0023
Decode
MAR ¬ IR[11-0]
102
3105
105
0023
0023
(decode IR[15-12])
102
3105
105
0023
0023
Get operand
MBR ¬ M[MAR]
102
3105
105
FFE9
0023
Execute
AC ¬ AC + MBR
102
3105
105
FFE9
000C
Step
RTN
PC
IR
MAR
MBR
AC
(initial values)
101
1104
104
0023
0023
Fetch
MAR ¬ PC
101
101
101
0023
0023
IR ¬ M[MAR]
101
3105
101
0023
0023
PC ¬ PC + 1
102
3105
101
0023
0023
Decode
MAR ¬ IR[11-0]
102
3105
105
0023
0023
(decode IR[15-12])
102
3105
105
0023
0023
Get operand
MBR ¬ M[MAR]
102
3105
105
FFE9
0023
Execute
AC ¬ AC + MBR
102
3105
105
FFE9
000C
Explanation / Answer
Step
RTN
PC
IR
MAR
MBR
AC
(initial values)
101
1104
104
0023
0023
Fetch
MAR ßPC
101
101
101
0023
0023
IR ßM[MAR]
101
3105
101
0023
0023
PC ß PC + 1
102
3105
101
0023
0023
Decode
MAR ß IR[11-0]
102
3105
105
0023
0023
(decode IR[15-12])
102
3105
105
0023
0023
Get operand
MBR ß M[MAR]
102
3105
105
FFE9
0023
Execute
AC ß AC + MBR
102
3105
105
FFE9
000C
Here is the explanation.
1)So initially PC IR MAR MBR AC are loaded. For initial values.
Once ACC is set then it will store after execution statement gets executed.
So next Fetching the initialized values.. Program counter is 101 and now Instruction pointer and MAR register will Point to 101. MBR (Master boot record) and Acc are same no change.
2) In next Instruction we are loading MAR register effective address to Instruction pointer. So
IR is loaded with effective address 3105
3) Incrementing program pointer to execute next instruction. So PC changed 102 and add operation updates MAR to 105 (already given in question itself).
4) Next is decoding micro operation instruction…NO changes in registers. We are already changed in above
point 3.
5) Next instruction is operation fetching. In this instruction storing effective address of MAR register to MBR…so Now MBR contains FFE9 since output is 16 bit external bus (Hexa decimal Number).
6) So finally execute statement….already I mentioned in first point…Acc will change after exec statement. So now Acc will contain 000C.
Step
RTN
PC
IR
MAR
MBR
AC
(initial values)
101
1104
104
0023
0023
Fetch
MAR ßPC
101
101
101
0023
0023
IR ßM[MAR]
101
3105
101
0023
0023
PC ß PC + 1
102
3105
101
0023
0023
Decode
MAR ß IR[11-0]
102
3105
105
0023
0023
(decode IR[15-12])
102
3105
105
0023
0023
Get operand
MBR ß M[MAR]
102
3105
105
FFE9
0023
Execute
AC ß AC + MBR
102
3105
105
FFE9
000C
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