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gisters 0001 Load AC from memory Multiply AC by memo Store AC to memory Add AC from memory 33 1506 PC 0010 34 4503 AC 0100 35 1505 1000 MQ 36 2503 IR 37 8503 1506 MBR 38 2504 MAR 39 4505 40 8506 503 04 504| 12 505 05 506 16

gisters 0001 Load AC from memory Multiply AC by memo Store AC to memory Add AC from memory 33 1506 PC 0010 34 4503 AC 0100 35 1505 1000 MQ 36 2503 IR 37 8503 1506 MBR 38 2504 MAR 39 4505 40 8506 503 04 504| 12 505 05 506 16

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Memory CPU Registers 0001 Load AC from memory 0010 Multiply AC by memory PC 33 1506 34 4503 0100 Store AC to memory AC 1000 Add AC from memory 35 1505 MQ 36 2503 IR 37 8503 1506 MBR 38 2504 MAR 39 4505 40 8506 ... 503 04 504 12 505 05 506 16 Assume the memory fragment and CPU registers of hypothetical machine shown above. What will be the content of MAR while executing the instruction loaded into MBR. Notes: 1- The stored program starts at address 33 2- All numbers shown are unsigned. 1- MAR =
Transcribed Image Text:Memory CPU Registers 0001 Load AC from memory 0010 Multiply AC by memory PC 33 1506 34 4503 0100 Store AC to memory AC 1000 Add AC from memory 35 1505 MQ 36 2503 IR 37 8503 1506 MBR 38 2504 MAR 39 4505 40 8506 ... 503 04 504 12 505 05 506 16 Assume the memory fragment and CPU registers of hypothetical machine shown above. What will be the content of MAR while executing the instruction loaded into MBR. Notes: 1- The stored program starts at address 33 2- All numbers shown are unsigned. 1- MAR =
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