MCS–012, Question & Answer December 2023

May 17, 2024

MCS–012, Question & Answer December 2023

1. (a) Convert the following decimal numbers to binary and hexadecimal : 6

(i) (28)10

(ii) (256)10

(iii) (42.25)10

### (i) \( (28)_{10} \)

\[
\begin{align*}
28 \div 2 &= 14 \quad \text{remainder: } 0 \\
14 \div 2 &= 7 \quad \text{remainder: } 0 \\
7 \div 2 &= 3 \quad \text{remainder: } 1 \\
3 \div 2 &= 1 \quad \text{remainder: } 1 \\
1 \div 2 &= 0 \quad \text{remainder: } 1 \\
\end{align*}
\]

Binary: \( (28)_{10} = (11100)_2 \)

#### Conversion to Hexadecimal:

\[
\begin{align*}
28 \div 16 &= 1 \quad \text{remainder: } 12 \quad (12 = C) \\
1 \div 16 &= 0 \quad \text{remainder: } 1 \\
\end{align*}
\]

Hexadecimal: \( (28)_{10} = (1C)_{16} \)

### (ii) \( (256)_{10} \)

#### Conversion to Binary:
\[
\begin{align*}
256 \div 2 &= 128 \quad \text{remainder: } 0 \\
128 \div 2 &= 64 \quad \text{remainder: } 0 \\
64 \div 2 &= 32 \quad \text{remainder: } 0 \\
32 \div 2 &= 16 \quad \text{remainder: } 0 \\
16 \div 2 &= 8 \quad \text{remainder: } 0 \\
8 \div 2 &= 4 \quad \text{remainder: } 0 \\
4 \div 2 &= 2 \quad \text{remainder: } 0 \\
2 \div 2 &= 1 \quad \text{remainder: } 0 \\
1 \div 2 &= 0 \quad \text{remainder: } 1 \\
\end{align*}
\]

Binary: \( (256)_{10} = (100000000)_2 \)

#### Conversion to Hexadecimal:
\[
\begin{align*}
256 \div 16 &= 16 \quad \text{remainder: } 0 \\
16 \div 16 &= 1 \quad \text{remainder: } 0 \\
1 \div 16 &= 0 \quad \text{remainder: } 1 \\
\end{align*}
\]

Hexadecimal: \( (256)_{10} = (100)_{16} \)

### (iii) \( (42.25)_{10} \)

#### Conversion to Binary:
**Integer part:**
\[
\begin{align*}
42 \div 2 &= 21 \quad \text{remainder: } 0 \\
21 \div 2 &= 10 \quad \text{remainder: } 1 \\
10 \div 2 &= 5 \quad \text{remainder: } 0 \\
5 \div 2 &= 2 \quad \text{remainder: } 1 \\
2 \div 2 &= 1 \quad \text{remainder: } 0 \\
1 \div 2 &= 0 \quad \text{remainder: } 1 \\
\end{align*}
\]

Integer part in binary: \( 101010_2 \)

**Fractional part:**
1. Multiply the fractional part by 2 and record the integer part.
2. Repeat for the fractional part until it becomes 0.

\[
\begin{align*}
0.25 \times 2 &= 0.5 \quad \text{integer part: } 0 \\
0.5 \times 2 &= 1.0 \quad \text{integer part: } 1 \\
\end{align*}
\]

Fractional part in binary: \( .01_2 \)

Binary: \( (42.25)_{10} = (101010.01)_2 \)

#### Conversion to Hexadecimal:
**Integer part:**
\[
\begin{align*}
42 \div 16 &= 2 \quad \text{remainder: } 10 \quad (10 = A) \\
2 \div 16 &= 0 \quad \text{remainder: } 2 \\
\end{align*}
\]

Integer part in hexadecimal: \( 2A_{16} \)

**Fractional part:**
1. Multiply the fractional part by 16 and record the integer part.
2. Repeat for the fractional part until it becomes 0.

\[
\begin{align*}
0.25 \times 16 &= 4.0 \quad \text{integer part: } 4 \\
\end{align*}
\]

Fractional part in hexadecimal: \( .4_{16} \)

Hexadecimal: \( (42.25)_{10} = (2A.4)_{16} \)

### Summary
- \( (28)_{10} = (11100)_2 = (1C)_{16} \)
- \( (256)_{10} = (100000000)_2 = (100)_{16} \)
- \( (42.25)_{10} = (101010.01)_2 = (2A.4)_{16} \)

(b) What is half-adder ? Make the truth table and logic diagram of a half-adder.

A half-adder is a basic digital circuit used for adding two single-bit binary numbers. It produces two outputs: the sum (S) and the carry (C). The sum represents the addition result, and the carry represents the overflow that is carried to the next higher bit in multi-bit addition.

Truth Table

Logic Diagram

The half-adder uses two logic gates:
1. **XOR Gate** for the sum (S).
2. **AND Gate** for the carry (C).

The logic diagram is as follows:

- **Inputs:** A and B
- **Sum (S):** \( A \oplus B \) (XOR gate)
- **Carry (C):** \( A \cdot B \) (AND gate)

In this diagram:

The XOR gate outputs the sum (S) of A and B.

The AND gate outputs the carry (C) of A and B.

(c) Explain the purpose of RAID. Explain the features of any two levels of RAID.

### Purpose of RAID

RAID (Redundant Array of Independent Disks) is a data storage technology that combines multiple physical disk drives into a single unit to improve performance, reliability, and data redundancy. The primary purposes of RAID are to:

1. **Enhance Performance**: By distributing data across multiple disks, RAID can improve read and write speeds.

2. **Increase Fault Tolerance**: RAID provides data redundancy, allowing data recovery in case of disk failure.

3. **Improve Storage Capacity**: RAID can combine the storage capacity of multiple disks into one large volume.

### RAID 0 (Striping)

**Features**:

- **Performance**: RAID 0 stripes data across all disks, significantly improving read and write speeds as data is accessed in parallel.

- **No Redundancy**: There is no data redundancy; if one disk fails, all data in the array is lost.

**Calculation Example**:

- **Number of Disks**: 4

- **Disk Capacity**: 1 TB each

- **Total Capacity**: 4 disks * 1 TB = 4 TB

- **Performance Improvement**: Data is split into stripes, enhancing read/write speeds by a factor equal to the number of disks.

### RAID 1 (Mirroring)

**Features**:

- **Data Redundancy**: RAID 1 mirrors data across two disks. Each disk contains a complete copy of the data, providing high fault tolerance.

- **Read Performance Improvement**: Read operations can be performed on both disks simultaneously, improving read speed. Write speed remains similar to a single disk since data must be written to both disks.

**Calculation Example**:

- **Number of Disks**: 2

- **Disk Capacity**: 1 TB each

- **Total Capacity**: 2 disks * 1 TB = 2 TB / 2 (due to mirroring) = 1 TB

- **Fault Tolerance**: One disk can fail without data loss.

In summary, RAID 0 offers enhanced performance without redundancy, while RAID 1 provides data redundancy and improved read performance at the cost of storage efficiency.

(d) Explain the interrupt driven I/O technique with the help of a diagram. 5

Interrupt-Driven I/O Technique

Interrupt-driven I/O improves efficiency by allowing the CPU to execute other tasks while waiting for I/O operations. When an I/O device is ready, it sends an interrupt signal to the CPU. The CPU then pauses its current task, saves its state, and executes an Interrupt Service Routine (ISR) to handle the I/O operation. Once the ISR completes, the CPU restores its state and resumes the interrupted task.

(e) Explain the subroutine call and return instructions of a computer system with the help of an example. 5

Subroutine call and return instructions manage the flow of control between the main program and a subroutine. A subroutine call instruction, such as CALL in assembly language, saves the address of the next instruction on the stack and jumps to the subroutine's address. The return instruction, like RET, retrieves the saved address from the stack and jumps back to it, resuming the main program.

**Example:**

assembly

main:

; Main program code

CALL subroutine

; Code to execute after subroutine returns

HLT

subroutine:

; Subroutine code

RET

When CALL subroutine is executed, the address of the next instruction (`HLT`) is pushed onto the stack, and control transfers to subroutine. Upon executing `RET`, the address (`HLT`) is popped from the stack, and control returns to the main program, continuing after the call. This mechanism ensures structured program flow and reusability of code.

(f) Explain the concept of instruction pipelining with the help of a diagram. 5

Instruction Pipelining

Instruction pipelining is a technique used in the design of modern microprocessors to improve performance by overlapping the execution of multiple instructions. The main idea is to break down the instruction execution process into several stages, with each stage being handled by a different part of the processor simultaneously. This allows for multiple instructions to be processed at once, significantly increasing the throughput of the CPU.

Concept

1. **Stages of Pipelining**:

1. Fetch (F) , 2. Decode (D) , 3. Execute (E), 4. Memory Access (M), 5. Write Back (WB)

Diagram

Below is a simplified illustration of instruction pipelining with five stages:

Cycle: 1 2 3 4 5 6 7 8 9

----------------------------------------------------------------------------------

Instr 1: [Fetch][Decode][Execute][Memory][Write B]

Instr 2: [Fetch][Decode][Execute][Memory][Write B]

Instr 3: [Fetch][Decode][Execute][Memory][Write B]

Instr 4: [Fetch][Decode][Execute][Memory][Write B]

Instr 5: [Fetch][Decode][Execute][Memory][Write B]

The 8086 microprocessor has four segment registers:

Code Segment (CS):

Purpose: Holds the base address of the code segment, where the executable instructions are stored.
Example: If CS=0x1000 and the instruction pointer (IP) is 0x0005, the CPU fetches the instruction from address 0x10005 (CS:IP).

Data Segment (DS):

Purpose: Contains the base address of the data segment, where variables and data structures are stored.
Example: If DS=0x2000 and an instruction like MOV AX, [1234H] is executed, it fetches data from address 0x21234.

Stack Segment (SS):

Purpose: Holds the base address of the stack segment, used for stack operations like push and pop.
Example: If SS=0x3000 and the stack pointer (SP) is 0xFFFE, a PUSH AX operation stores the value of AX at address 0x3FFFE, and SP is decremented.

Extra Segment (ES):

Purpose: Used as an additional data segment, often for string operations.
Example: If ES=0x4000, and MOVS instruction is executed to move a string from DS:SI to ES:DI, it transfers data starting from address 0x40000.

These segment registers enable organized and efficient access to different memory areas, facilitating code execution, data manipulation, and stack management.

(h) What is interrupt INT 21 h in 8086 microprocessor ? Explain its use with the help of an example.

INT 21h is a DOS interrupt in the 8086 microprocessor, providing various system services like file operations, input/output handling, and program termination. It's a software interrupt used for interfacing with the operating system. Each service is accessed by setting the AH register to a specific function number before invoking INT 21h.

For example, to display a character on the screen:

Load the character into the DL register.
Set AH to 02h (function number for displaying a character).
Execute INT 21h.

MOV AH, 02h ; Function to display a character

MOV DL, 'A' ; Character to display

INT 21h ; Call DOS interrupt

2. (a) Simplify the following Boolean functions in SOP and POS forms using K-map. Also draw the logic diagrams : 10 F (A, B, C, D) =  (0, 2, 8, 9, 10, 11, 14, 15)

To simplify the given Boolean function \( F(A, B, C, D) \) using a Karnaugh Map (K-map) and then represent it in Sum of Products (SOP) and Product of Sums (POS) forms, we need to follow these steps:

1. **Construct the K-map for the given minterms.**

2. **Simplify the expression using K-map grouping for both SOP and POS forms.**

3. **Draw the logic diagrams for the simplified expressions.**

1. Construct the K-map

The given function is \( F(A, B, C, D) = \sum(0, 2, 8, 9, 10, 11, 14, 15) \).

Let's create a 4-variable K-map. We'll label the variables as \( A \), \( B \), \( C \), and \( D \), with \( A \) and \( B \) representing the rows, and \( C \) and \( D \) representing the columns.

Here is the K-map layout:

Now, populate the K-map with 1's in the cells corresponding to the given minterms (0, 2, 8, 9, 10, 11, 14, 15).

2. Simplify the Expression

Sum of Products (SOP):

We need to find the largest possible groups of 1's.

- Group 1: Minterms 8, 9 (AB'CD') and (AB'CD)

- Group 2: Minterms 10, 11, 14, 15 (ABCD', ABCD, AB'CD', AB'CD)

- Group 3: Minterms 0, 2 (A'B'C'D' and A'BC'D')

These can be grouped as:

1. \( \overline{A}\overline{B}\overline{C}\overline{D} + \overline{A}\overline{B}C\overline{D} \) (Group of 2)

2. \( \overline{A}\overline{B}D + \overline{A}\overline{B}C \) (Group of 4)

3. \( \overline{A}B + AB \) (Group of 4)

So, the SOP form is:

\[ F(A, B, C, D) = \overline{A}\overline{B}\overline{C}\overline{D} + \overline{A}\overline{B}C\overline{D} + \overline{A}\overline{B}D + \overline{A}B + AB \]

#### Product of Sums (POS):

We find the groups of 0's. Since it's already in minimized form, we can directly find:

\[ F(A, B, C, D) = (A + B)(A + \overline{B} + \overline{D})(\overline{A} + \overline{C} + D)(\overline{A} + \overline{C} + \overline{D}) \]

### 3. Draw the Logic Diagrams

**SOP Logic Diagram:**

- \( F(A, B, C, D) = \overline{A}\overline{B}\overline{C}\overline{D} + \overline{A}\overline{B}C\overline{D} + \overline{A}\overline{B}D + \overline{A}B + AB \)

**POS Logic Diagram:**

- \( F(A, B, C, D) = (A + B)(A + \overline{B} + \overline{D})(\overline{A} + \overline{C} + D)(\overline{A} + \overline{C} + \overline{D}) \)

Let's summarize with the K-map and simplified expressions:

K-map

SOP Form:

\[ F(A, B, C, D) = \overline{A}\overline{B}\overline{C}\overline{D} + \overline{A}\overline{B}C\overline{D} + \overline{A}\overline{B}D + \overline{A}B + AB \]

POS Form:

\[ F(A, B, C, D) = (A + B)(A + \overline{B} + \overline{D})(\overline{A} + \overline{C} + D)(\overline{A} + \overline{C} + \overline{D}) \]

Logic Diagrams

**SOP Logic Diagram:**

1. AND gates for each product term.

2. OR gate to combine the outputs of the AND gates.

**POS Logic Diagram:**

1. OR gates for each sum term.

2. AND gate to combine the outputs of the OR gates.

This completes the simplification and logic diagram representation.

(b) What is the need of Cache memory ? Explain the direct mapping cache organisation and associative mapping cache organisation with the help of a diagram.

The Need for Cache Memory

Cache memory is a small, high-speed storage located close to the CPU. It temporarily holds frequently accessed data and instructions to reduce the average time to access data from the main memory. Here are the primary reasons why cache memory is needed:

1. **Speed**: Cache memory is faster than main memory (RAM), thus it speeds up data access time for the CPU.

2. **Efficiency**: By storing frequently used data and instructions, it reduces the need to fetch data from slower main memory.

3. **Performance**: It significantly improves the overall performance of the system by reducing latency and increasing the rate of instruction execution.

4. **Reduced Bottlenecks**: Cache helps in minimizing the bottleneck caused by the slower speeds of main memory compared to the CPU.

Direct Mapping Cache Organization

**Direct Mapping** is a simple cache mapping technique where each block of main memory maps to only one possible cache line.

#### Characteristics:

- Each memory block is assigned to a specific line in the cache.

- The mapping is determined by the formula: \( \text{Cache Line} = (\text{Block Address}) \mod (\text{Number of Cache Lines}) \).

#### Diagram:

In direct mapping:

- Memory blocks 0, 4, 8, etc., map to cache line 0.

- Memory blocks 1, 5, 9, etc., map to cache line 1.

- Memory blocks 2, 6, 10, etc., map to cache line 2.

#### Advantages:

- Simple and easy to implement.

- Less hardware required compared to other mapping techniques.

#### Disadvantages:

- High chance of collisions, as multiple blocks map to the same line.

- May lead to inefficient use of cache if the program accesses data blocks that map to the same cache line frequently.

### Associative Mapping Cache Organization

**Associative Mapping** allows any block of main memory to be loaded into any line of the cache.

#### Characteristics:

- A memory block can be placed in any line of the cache, providing greater flexibility.

- Requires searching the entire cache to find a block, which can be time-consuming and requires additional hardware.

Diagram:

In associative mapping:

- Any memory block can be placed into any cache line, avoiding fixed positions.

#### Advantages:

- Reduces the possibility of collisions because blocks can occupy any line.

- Utilizes the cache more efficiently.

#### Disadvantages:

- More complex and expensive to implement due to the need for hardware to search the entire cache.

- Longer access times compared to direct mapping due to the search process.

### Summary

- **Direct Mapping** is simple and less hardware-intensive but can lead to high collision rates.

- **Associative Mapping** is more flexible and efficient in cache usage but requires more complex hardware and longer access times.

Understanding these cache mapping techniques is crucial for optimizing computer system performance and designing efficient memory systems.

3. (a) Write a program using 8086 assembly language that exchanges the byte values given in two memory locations.

Sure, I can help you with that. Here is an 8086 assembly language program that exchanges the byte values in two memory locations. This example assumes the memory locations are at addresses `DS:1000h` and `DS:1001h`.

; Assembly program to exchange byte values in two memory locations

.model small

.stack 100h

.data

; Memory locations

mem_loc1 db 20h ; Example value at memory location 1000h

mem_loc2 db 30h ; Example value at memory location 1001h

.code

main:

; Initialize data segment

mov ax, @data

mov ds, ax

mov es, ax

; Load the value from the first memory location into AL

mov si, offset mem_loc1

mov al, [si]

; Load the value from the second memory location into BL

mov di, offset mem_loc2

mov bl, [di]

; Exchange the values

mov [si], bl

mov [di], al

; Exit the program

mov ah, 4Ch

int 21h

end main

Explanation

Initialization:

The .model small directive sets up a small memory model where both code and data are in the same segment.
The .stack 100h directive allocates 256 bytes for the stack.
The .data section is where the variables mem_loc1 and mem_loc2 are declared with initial values.

Loading Segment Registers: