Consider the following instruction format and the list of opcodese Opcode 4 bits (15 – 12) Addresse 12 bits (11 – 0)<- Opcodee 0101e Instruction Load AC from memory Ae Store AC from memory Ae Add AC from memory Ae Subtract AC from memory A Multiply AC from memory Ae Divide AC from memory Ae 0110e 0001e 0010e 0011e 0100e 1. Run the following program where PC starts from 500- Memorye addresse contentse 500e 5800e 501e 1801e 502e 6800e 800- FFFEe 801e 000A- 802e 000Ce 2. Run the following program where PC starts from 500 Memorye addresse contentse 500e 5801e 501e 2802e 502e 6801e 800e FFFE 801e 000A- 802e FFFC-
Consider the following instruction format and the list of opcodese Opcode 4 bits (15 – 12) Addresse 12 bits (11 – 0)<- Opcodee 0101e Instruction Load AC from memory Ae Store AC from memory Ae Add AC from memory Ae Subtract AC from memory A Multiply AC from memory Ae Divide AC from memory Ae 0110e 0001e 0010e 0011e 0100e 1. Run the following program where PC starts from 500- Memorye addresse contentse 500e 5800e 501e 1801e 502e 6800e 800- FFFEe 801e 000A- 802e 000Ce 2. Run the following program where PC starts from 500 Memorye addresse contentse 500e 5801e 501e 2802e 502e 6801e 800e FFFE 801e 000A- 802e FFFC-
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Question
![Consider the following instruction format and the list of opcodes
Opcode
4 bits (15 – 12) e
Addresse
12 bits (11 – 0)
Opcode
Instructione
Load AC from memory Ae
Store AC from memory A
Add AC from memory Ae
Subtract AC from memory A
Multiply AC from memory A
Divide AC from memory Ae
0101e
0110e
0001e
0010e
0011e
0100e
1. Run the following program where PC starts from 500-
Memory
contents
5800e
addresse
500e
501e
1801e
502e
6800-
800e
FFFE
801e
000A
802
000C
2. Run the following program where PC starts from 500
Memorye
addresse
contentse
500e
5801e
501e
2802e
502e
6801e
800e
FFFEe
801e
000A-
802e
FFFC-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b64befb-94a8-42a1-adf4-44168b6b8246%2F5cb29661-eec6-4aa4-8b72-44cd0fd16f32%2Fy1su6e_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the following instruction format and the list of opcodes
Opcode
4 bits (15 – 12) e
Addresse
12 bits (11 – 0)
Opcode
Instructione
Load AC from memory Ae
Store AC from memory A
Add AC from memory Ae
Subtract AC from memory A
Multiply AC from memory A
Divide AC from memory Ae
0101e
0110e
0001e
0010e
0011e
0100e
1. Run the following program where PC starts from 500-
Memory
contents
5800e
addresse
500e
501e
1801e
502e
6800-
800e
FFFE
801e
000A
802
000C
2. Run the following program where PC starts from 500
Memorye
addresse
contentse
500e
5801e
501e
2802e
502e
6801e
800e
FFFEe
801e
000A-
802e
FFFC-
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