Model Questions and Solutions BIM 3rd Semester

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Model Question And Solution -2012

Ans: Here are the steps to detect the overflow while adding 2’s complement data:

  • Addition of Two Numbers.
  • Check Carry from MSB
  • Determine Overflow.

Ans: An operating system (OS) is a software program that serves as an intermediary between computer hardware and the applications running on it. It manages the hardware resources of a computer system and provides a set of services to software applications

Ans: A crossbar switch is a type of switching network used in digital systems, particularly in computer hardware and telecommunications, to enable efficient communication between multiple input and output ports.

Ans: Here are some of the uses of daisy chaining and they are :

  • Data Transfer.
  • Signal Propagation.
  • Power Distribution
  • Control and Configuration

Ans:

Ans: The uses of Control Address Register(CAR) are :

  • Microinstruction Execution.
  • Microinstruction Addressing.
  • Control Flow.
  • Branching and Jumping.
  • Program Control.

Ans: The completeness of a set of instructions depends on the context in which the instructions are being executed. Here are few scenarios where you can determine that a set of instructions is complete:

  1. Sequential Program Execution.
  2. Batch Processing.
  3. Parallel Processing.
  4. Pipeline Execution.

Ans: A bootstrap loader, is defined as a small program that is responsible for loading the operating system (OS) into a computer’s memory and initiating its execution

Ans: The difference between RISC and CISC are:

CISC.

  1. CISC architectures typically have a complex instruction set with a large variety of instructions, some of which may perform complex operations or access memory directly.
  2. In CISC architectures, instructions may vary in length and complexity, and a single instruction may perform multiple operations or address multiple memory locations

RISC:

  1. RISC architectures have a reduced and simplified instruction set with fewer instructions, each performing a simple and well-defined operation.
  2. RISC architectures emphasize simple instructions that execute quickly. Each instruction typically performs only one operation, making instruction decoding simpler and more efficient.

Ans: In Relative Addressing Mode, content of the program counter is added to the address part of the instruction in order to obtain the effective address.

Ans: A microprogram control unit, also known as microprogrammed control unit, which is defined as a component within a computer’s central processing unit (CPU) responsible for generating control signals that coordinate the operations of the CPU’s functional units based on a sequence of microinstructions stored in a control memory.

The advantages of using floating point representation are:

  • Expressiveness: Floating-point representation allows for the representation of a wide range of real numbers, including very large and very small numbers, with a high level of precision. This makes it suitable for applications that require a broad range of numerical values, such as scientific computing, engineering, and graphics.
  • Accuracy: Floating-point representation can provide high precision for numerical calculations, allowing for accurate results in computations involving real numbers. This precision is essential for applications that demand high levels of accuracy, such as simulations, numerical analysis, and financial modeling.
  • Efficiency: Modern CPUs and hardware architectures include dedicated floating-point units (FPUs) optimized for performing floating-point arithmetic operations efficiently. This hardware support accelerates numerical computations and improves overall performance in applications that heavily rely on floating-point calculations.
  • Error Handling: Floating-point representation includes mechanisms for handling special values such as infinity, NAN (Not a Number), and denormalized numbers, allowing for robust error handling and graceful degradation in numerical computations.
  • Standardization: Floating-point representation follows standard formats, such as IEEE 754, which ensures interoperability and portability across different computing platforms and programming languages. This standardization simplifies data exchange and computation across diverse systems.

Here are different types of conflicts that can occur in pipelining:

a. Resource Conflict: Occurs when multiple pipeline stages require the same hardware resource simultaneously. For example, if two instructions in different stages of the pipeline attempt to access the same functional unit (e.g., ALU, multiplier), a resource conflict arises, and one instruction must wait until the resource becomes available.

b. Data Hazards:

  • Read-After-Write (RAW) Hazard: Occurs when a pipeline stage attempts to read data from a register or memory location that is being written to by a previous instruction in an earlier pipeline stage. This results in a data dependency and may require stalling or forwarding mechanisms to resolve.
  • Write-After-Read (WAR) Hazard: Occurs when a pipeline stage attempts to write data to a register or memory location that is being read by a subsequent instruction in a later pipeline stage. This can lead to incorrect results if not properly handled.
  • Write-After-Write (WAW) Hazard: Occurs when multiple pipeline stages attempt to write data to the same register or memory location in overlapping clock cycles. This can result in data corruption if not properly managed.

c. Bubble: A pipeline bubble occurs when a pipeline stage is idle due to a conflict or hazard, causing a delay in instruction execution. Bubbles reduce pipeline efficiency and throughput, leading to performance degradation.

d. Pipeline Flushes:

  • Instruction Abort: Occurs when an instruction in the pipeline encounters an exception, such as a divide-by-zero error or a memory access violation. In such cases, the pipeline must be flushed, and the partially completed instruction must be discarded to maintain program correctness.
  • Mispredictions: Occurs when the pipeline makes incorrect predictions about the outcome of a branch instruction. In such cases, the pipeline must be flushed, and the correct instruction path must be re-fetched and executed.
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  • Fetch cycle: The processor retrieves the next instruction from memory. The program counter (PC) holds the address of the next instruction to be fetched. The instruction is typically stored in the instruction cache or main memory.
  • Decode cycle: The fetched instruction is decoded to determine its opcode and operands. The CPU identifies the operation to be performed and prepares the necessary control signals to execute the instruction.
    1. Store (Write Back)/ Road cycle: If the instruction produces a result that needs to be stored in memory or a register, the CPU writes the result back to the appropriate destination. This phase completes the execution of the instruction.

  • Execute cycle: The decoded instruction is executed, and the required operation is performed by the CPU’s functional units. This phase may involve arithmetic or logical calculations, data transfers, or control flow operations.

Ans: The difference between strobe and handshaking are as followed:

AspectsStrobeHandshaking
DefinitionStrobe is defined as a pulse or a signal that occurs at regular intervals, marking the beginning or end of a data transfer cycle.Handshaking is defined as a sequence of signals or actions exchanged between the sender and receiver to ensure that both are ready to send or receive data.
PurposeStrobe is a signal used to indicate the timing or synchronization of data transfer.Handshaking is a method used to establish and confirm communication between two devices or components before data transfer.
Timing and SynchronizationStrobe signals are typically associated with synchronous systems where data transfer is synchronized to a clock signal. Handshaking protocols are often used in both synchronous and asynchronous systems.
UsageStrobe signals are commonly used in scenarios where data is transferred in bursts or packets, and precise timing is important.Handshaking protocols are used in scenarios where data transfer may be more sporadic or asynchronous, and where confirmation of communication readiness is crucial.
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Ans: Cache coherence is defined as a consistency and synchronization of data stored in different caches within a multiprocessor or multicore system. In such systems, each processor or core typically has its own cache memory to improve performance.

There are various cache mapping techniques and some of them are as:

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  1. Direct Mapping:
    • Each memory block is mapped to exactly one cache line.
    • The mapping is determined by a modulo function, which maps memory addresses to specific cache lines.
    • Simple and easy to implement but may lead to cache conflicts.
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2. Fully Associative Mapping:

  • Each memory block can be placed in any cache line.
  • No restrictions on the placement of memory blocks.
  • Offers maximum flexibility and minimizes cache conflicts but requires complex hardware for searching cache contents.
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3. K-way Set-Associative Mapping:

  • Combines aspects of both direct mapping and fully associative mapping.
  • Memory is divided into a number of sets, and each set contains multiple cache lines.
  • Each memory block can be mapped to any cache line within its corresponding set.
  • Allows for more flexibility and reduces the chance of cache conflicts compared to direct mapping.
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These mapping techniques help optimize cache performance by balancing factors such as cache access time, cache size, and complexity of cache hardware. The choice of mapping technique depends on the specific requirements and constraints of the system design.

Ans:

Given:

In the conventional machine:

  • Time taken to complete one task = 25 ns
  • Number of tasks to be completed = 100

In the pipelined machine:

  • One task is divided into 5 segments, each taking 4 ns.
  • Since the tasks are pipelined, there’s an overlap in the execution of tasks, so the effective time per task is reduced.

Let’s first calculate the time taken to complete one task in the pipelined machine:

Total time taken in pipelined machine = Time per segment = 4 ns

Now, since there are 5 segments for each task, the total time taken to complete one task in the pipelined machine is:

Total time per task in pipelined machine = 5 segments * 4 ns/segment = 20 ns

Now, let’s calculate the speedup:

Speedup = Time taken in conventional machine / Time taken in pipelined machine

Speedup = 25 ns / 20 ns = 1.25

So, the speedup achieved by the pipelined machine compared to the conventional machine is 1.25.

This means the pipelined machine completes tasks 1.25 times faster than the conventional machine.

Ans: Addressing modes are used in computer architecture and assembly language programming to specify how the operand of an instruction is determined. Different addressing modes provide flexibility in accessing data or operands from memory or registers. They allow programmers to write more efficient and concise code by providing various ways to specify operands.

Here are some common addressing modes:

  1. Immediate Addressing:
    • The operand is a constant value or immediate data.
    • Example: MOV AX, 5 (Move the value 5 into register AX).
  2. Direct Addressing:
    • The operand specifies the address of the data directly.
    • Example: MOV AX, [1234H] (Move the value stored at memory address 1234H into register AX).
  3. Register Addressing:
    • The operand specifies a register.
    • Example: ADD AX, BX (Add the value of register BX to register AX).
  4. Indirect Addressing:
    • The operand contains the address of the memory location which holds the actual operand.
    • Example: MOV AX, [BX] (Move the value stored at the memory address pointed to by register BX into register AX).
  5. Indexed Addressing:
    • The effective address of the operand is obtained by adding a constant value or value in a register to a base address.
    • Example: MOV AX, [SI + 10] (Move the value stored at the memory address pointed to by the sum of the value in register SI and 10 into register AX).
  6. Base-Displacement Addressing:
    • Similar to indexed addressing, but the displacement value is added to a base address register.
    • Example: MOV AX, [BX + 20] (Move the value stored at the memory address pointed to by the sum of the value in register BX and 20 into register AX).
  7. Relative Addressing:
    • The operand is specified relative to the program counter (PC) or instruction pointer (IP).
    • Example: JMP LABEL (Jump to the memory address specified by the label).

These addressing modes provide flexibility and efficiency in accessing data and operands, allowing programmers to write code that is both readable and optimized for performance. The choice of addressing mode depends on the specific requirements of the instruction and the programming task at hand.

Ans: The hardware implementation of addition and subtraction for signed 2’s complement data typically involves a combination of basic arithmetic and logic operations performed by digital logic circuits .The hardware implementation of addition/subtraction operation for signed 2’s complement data are:

Addition:

  1. Binary Addition Circuit:
    • The addition of two binary numbers follows the same principles as adding decimal numbers but is carried out using binary addition circuits.
    • The basic building block for binary addition is a full adder circuit.
    • Full adder circuits take three input bits (two bits to be added and a carry-in) and produce two output bits (the sum and a carry-out).
  2. Cascade of Full Adders:
    • To perform addition on multi-bit binary numbers, multiple full adder circuits are cascaded together.
    • Each full adder takes care of adding a pair of corresponding bits from the two operands along with the carry from the previous stage.
  3. Overflow Detection:
    • In signed 2’s complement arithmetic, overflow occurs when the result of addition exceeds the range that can be represented by the number of bits used.
    • Overflow detection logic is added to the addition circuit to detect when an overflow occurs, which indicates that the result is not valid due to the limited number of bits available.

Subtraction:

  1. Binary Subtraction Circuit:
    • Subtraction can be performed using binary addition by utilizing 2’s complement representation.
    • To subtract one number from another, the second number (subtrahend) is negated (flipped) and then added to the first number (minuend).
    • The negation can be performed using an inverter circuit (NOT gate) along with the addition circuit.
  2. Overflow Detection:
    • Similar to addition, overflow detection logic is necessary for subtraction to detect when the result goes beyond the representable range.

Hardware Implementation Considerations:

  1. Operand Width: The number of bits allocated to represent operands affects the precision and range of numbers that can be handled.
  2. Carry Propagation: Carry signals need to propagate through the cascade of full adders efficiently to ensure correct addition or subtraction.
  3. Overflow Detection: Circuitry is needed to detect when the result of an addition or subtraction operation exceeds the representable range of the number of bits used.
  4. 2’s Complement Conversion: For subtraction, converting the subtrahend to its 2’s complement form is necessary.

performing the multiplication of -23 * -9 using sign-magnitude representation involves a series of microoperations. Here’s how you can do it step by step:

  1. Convert to Sign-Magnitude Representation:
  • Convert -23 to sign-magnitude: -23 = 100101 (-23 in binary, with the leftmost bit denoting the sign, and the rest representing magnitude).
  • Convert -9 to sign-magnitude: -9 = 1001 (-9 in binary, with the leftmost bit denoting the sign, and the rest representing magnitude).
  1. Perform Bitwise Multiplication:
  • Multiply the magnitudes:
    • 100101 (-23) * 1001 (-9) = 000000000000000100101 (Partial product).
  1. Determine the Sign:
  • Since both numbers are negative, the result will be positive.
  1. Convert Back to Decimal:
  • Convert the binary result back to decimal: 000000000000000100101 = 145 (in decimal).

So, -23 * -9 = 145.

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