### Program Execution and Memory Analysis#### 1. Program Execution Starting at Memory Address 100**Memory Table:**| Address | Contents ||---------|----------|| 100 | 5800 || 101 | 1801 || 102 | 6800 || ... | ... || 800 | 001F || 801 | 000A || 802 | 000C |**Question 1:** Upon the completion of the program, what will be the contents of the following memory address?- Memory address 800 contents: [____] ---#### 2. Program Execution Starting at Memory Address 210**Memory Table:**| Address | Contents ||---------|----------|| 210 | 5801 || 211 | 2802 || 212 | 6801 || ... | ... || 800 | FFFF || 801 | 000A || 802 | 0005 |**Question 2:** Upon the completion of the program, what will be the contents of the following memory address?- Memory address 801 contents: [____] - Answer all questions with hexadecimal. ---These exercises involve understanding how a program interacts with memory and how operations alter memory addresses. Remember to analyze how each instruction affects the given memory contents. ### Instruction Format and OpcodesIn this section, we illustrate the instruction format and provide a list of opcodes used in a hypothetical computing system. This foundational understanding is essential for decoding and executing machine-level instructions.#### Instruction FormatThe instruction format is divided into two parts:- **Opcode (15..12)**: This segment of the instruction specifies the operation to be performed. It consists of the four most significant bits.- **Address (11..0)**: The remaining bits define the memory address, which is used in conjunction with the opcode to execute the instruction.> *Note that all memory contents and addresses are hexadecimal.*#### A Portion of the Instruction SetThe following table presents a portion of the instruction set, detailing various opcodes and their associated operations:| Opcode | Instruction ||--------|------------------------------------------|| 0101 | Load AC from memory || 0110 | Store AC to memory || 0001 | Add AC from memory - AC + memory || 0010 | Subtract memory from AC - AC - memory || 0011 | Multiply AC by memory - AC * memory || 0100 | Divide AC by memory - AC / memory |These opcodes demonstrate basic arithmetic and data transfer operations using an accumulator (AC) and memory interaction. This set is crucial for programmers working on low-level system programming to understand the manipulation and processing of data at the hardware level.

Answered: 00 01 02 00 01 02 Memory contents 5800…

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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### Program Execution and Memory Analysis

#### 1. Program Execution Starting at Memory Address 100

**Memory Table:**

| Address | Contents |
|---------|----------|
| 100 | 5800 |
| 101 | 1801 |
| 102 | 6800 |
| ... | ... |
| 800 | 001F |
| 801 | 000A |
| 802 | 000C |

**Question 1:**
Upon the completion of the program, what will be the contents of the following memory address?

- Memory address 800 contents: [____]

---

#### 2. Program Execution Starting at Memory Address 210

**Memory Table:**

| Address | Contents |
|---------|----------|
| 210 | 5801 |
| 211 | 2802 |
| 212 | 6801 |
| ... | ... |
| 800 | FFFF |
| 801 | 000A |
| 802 | 0005 |

**Question 2:**
Upon the completion of the program, what will be the contents of the following memory address?

- Memory address 801 contents: [____]

- Answer all questions with hexadecimal.

---

These exercises involve understanding how a program interacts with memory and how operations alter memory addresses. Remember to analyze how each instruction affects the given memory contents.

### Instruction Format and Opcodes

In this section, we illustrate the instruction format and provide a list of opcodes used in a hypothetical computing system. This foundational understanding is essential for decoding and executing machine-level instructions.

#### Instruction Format

The instruction format is divided into two parts:
- **Opcode (15..12)**: This segment of the instruction specifies the operation to be performed. It consists of the four most significant bits.
- **Address (11..0)**: The remaining bits define the memory address, which is used in conjunction with the opcode to execute the instruction.

> *Note that all memory contents and addresses are hexadecimal.*

#### A Portion of the Instruction Set

The following table presents a portion of the instruction set, detailing various opcodes and their associated operations:

| Opcode | Instruction |
|--------|------------------------------------------|
| 0101 | Load AC from memory |
| 0110 | Store AC to memory |
| 0001 | Add AC from memory - AC + memory |
| 0010 | Subtract memory from AC - AC - memory |
| 0011 | Multiply AC by memory - AC * memory |
| 0100 | Divide AC by memory - AC / memory |

These opcodes demonstrate basic arithmetic and data transfer operations using an accumulator (AC) and memory interaction. This set is crucial for programmers working on low-level system programming to understand the manipulation and processing of data at the hardware level.

Expert Solution

Step 1

Solution 1-

Upon completion of program the value of 800 content inside the memory location is 0028H.

For First content 5800 the opcode is 5 and adress is 800 .Now converting opcode 5 to binary we get 0101, which is given in table as load accumulator in memory .The initial 101 instructions load the memory address 800 to AC.
For second content 1801 the opcode is 1 so instruction by converting 1 to binary we get 0001 which is the second instructions that adds the contents at memory location 801 to AC.
For 6800 , the third instruction Stores current value of AC to memory location.
So value inside 800 is result of addition of contents of 800 and 801 which can be seen by proper solution of given program.

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