0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h. Add the contents of memory location 451h to AC. Store AC to memory location 8h. Store AC to I/O device number 8h.
0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h. Add the contents of memory location 451h to AC. Store AC to memory location 8h. Store AC to I/O device number 8h.
Computer Networking: A Top-Down Approach (7th Edition)
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Question
0001 = Load AC from memory
0010 = Store AC to memory
0101 = Add to AC from memory
0011 = Load AC (the accumulator register) from an I/O device
0111 = Store AC to an I/O device
With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number.
What is the hexadecimal string that expresses the following instructions?
- Load AC from memory location 62h.
- Add the contents of memory location 451h to AC.
- Store AC to memory location 8h.
- Store AC to I/O device number 8h.
![пуроше
IImachiIne in
Memory
300 1 9 4 0
301 5 9 4 1
302 29 41
CPU registers
30 0 PC
Memory
300 1 9 4 0
AC 301 5 9 4 1
1940 IR 302 29 4 1
CPU registers
30 1 PC
0 00 3 AC
194 0 IR
940 0 0 0 3
941 0 0 0 2
940 0 0 0 3
941 0 00 2
Step 1
Step 2
Memory
300 1 9 4 0
301 5 9 41
302 29 4 1
CPU registers
30 1 PC
0 00 3 AC 301 5 9 4 1
5 94 1 IR 302 29 4 1
CPU registers
30 2 PC
0 0 0 5 AC
5 9 4 1IR
Memory
300 19 4 0
940 0 0 0 3
941 0 0 0 2
940 0 00 3
941 0 0 0 2
Step 3
Step 4
Memory
300 1 9 4 0
301 5 9 4 1
302 2 9 4 I
CPU registers
30 2 PC
0 0 0 5 AC 301 5 9 4 1
294 1 IR 302 2 9 4 1
Memory
CPU registers
30 3 PC
0 00 5 AC
294 1 IR
300 19 4 0
940 0 0 0 3
941 0 0 0 2
940 0 0 03
941 0 0 0 5
Step 5
Step 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7751392e-ac03-4538-847a-24a343f2f0bd%2F9cff4aa0-8ca1-4e59-8813-1dd70e706a05%2F9d3e1d_processed.png&w=3840&q=75)
Transcribed Image Text:пуроше
IImachiIne in
Memory
300 1 9 4 0
301 5 9 4 1
302 29 41
CPU registers
30 0 PC
Memory
300 1 9 4 0
AC 301 5 9 4 1
1940 IR 302 29 4 1
CPU registers
30 1 PC
0 00 3 AC
194 0 IR
940 0 0 0 3
941 0 0 0 2
940 0 0 0 3
941 0 00 2
Step 1
Step 2
Memory
300 1 9 4 0
301 5 9 41
302 29 4 1
CPU registers
30 1 PC
0 00 3 AC 301 5 9 4 1
5 94 1 IR 302 29 4 1
CPU registers
30 2 PC
0 0 0 5 AC
5 9 4 1IR
Memory
300 19 4 0
940 0 0 0 3
941 0 0 0 2
940 0 00 3
941 0 0 0 2
Step 3
Step 4
Memory
300 1 9 4 0
301 5 9 4 1
302 2 9 4 I
CPU registers
30 2 PC
0 0 0 5 AC 301 5 9 4 1
294 1 IR 302 2 9 4 1
Memory
CPU registers
30 3 PC
0 00 5 AC
294 1 IR
300 19 4 0
940 0 0 0 3
941 0 0 0 2
940 0 0 03
941 0 0 0 5
Step 5
Step 6
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