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JESTION 1 Given that the PC contains 511, the address of the first instruction, and the partial list of opcodes as shown in Figure 1: | Address Content Instruction Meaning Load AC from memory Store AC to memory Add to AC from memory Subtract memory from AC 511 1824 0001 512 5822 0010 513 2823 0101 0110 822 21AB Figure 1. Partial List Opcodes 823 FFDE 824 AB12 Original memory contents i. Illustrate the execution of the instructions from memory content 512 and 513 in the diagram (511, 512 and 513). ii. Show the contents of the registers (Program Counter – PC, Accumulator – AC, and Instruction Register – IR).

JESTION 1 Given that the PC contains 511, the address of the first instruction, and the partial list of opcodes as shown in Figure 1: | Address Content Instruction Meaning Load AC from memory Store AC to memory Add to AC from memory Subtract memory from AC 511 1824 0001 512 5822 0010 513 2823 0101 0110 822 21AB Figure 1. Partial List Opcodes 823 FFDE 824 AB12 Original memory contents i. Illustrate the execution of the instructions from memory content 512 and 513 in the diagram (511, 512 and 513). ii. Show the contents of the registers (Program Counter – PC, Accumulator – AC, and Instruction Register – IR).

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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QUESTION 1 a) Given that the PC contains 511, the address of the first instruction, and the partial list of opcodes as shown in Figure 1: Address Content Instruction Meaning 511 1824 0001 Load AC from memory 512 5822 0010 Store AC to memory Add to AC from memory Subtract memory from AC 513 2823 0101 0110 ... 822 21AB Figure 1. Partial List Opcodes 823 FFDE 824 AB12 Original memory contents i. Illustrate the execution of the instructions from memory content 512 and 513 in the diagram (511, 512 and 513). ii. Show the contents of the registers (Program Counter – PC, Accumulator – AC, - and Instruction Register – IR).
Transcribed Image Text:QUESTION 1 a) Given that the PC contains 511, the address of the first instruction, and the partial list of opcodes as shown in Figure 1: Address Content Instruction Meaning 511 1824 0001 Load AC from memory 512 5822 0010 Store AC to memory Add to AC from memory Subtract memory from AC 513 2823 0101 0110 ... 822 21AB Figure 1. Partial List Opcodes 823 FFDE 824 AB12 Original memory contents i. Illustrate the execution of the instructions from memory content 512 and 513 in the diagram (511, 512 and 513). ii. Show the contents of the registers (Program Counter – PC, Accumulator – AC, - and Instruction Register – IR).
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