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Using different instruction formats write pseudo code to evaluate the following expression Z = 4(A + B ) - 16(C +58)

  1. Infix notation:
Z = 4
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Using different instruction formats (0 - address , 1 - address ,2 - address, 3 - address) write pseudo code to evaluate the following expression Z = 4(A + B ) - 16(C +58)

  1. 0-address format:
LOAD 4 ADD A ADD B STORE TEMP1 LOAD 16 LOAD C ADD 58 STORE TEMP2 MUL TEMP1 MUL TEMP2 SUB STORE Z
  1. 1-address format:
LOAD 4 A ADD B STORE TEMP1 LOAD 16 C ADD 58 STORE TEMP2 MUL TEMP1 MUL TEMP2 SUB STORE Z
  1. 2-address format:
LOAD 4 A B ADD STORE TEMP1 LOAD 16 C 58 ADD STORE TEMP2 MUL TEMP1 MUL TEMP2 SUB STORE Z
  1. 3-address format:
MUL A B 4 TEMP1 MUL C 58 16 TEMP2 SUB TEMP1 TEMP2 Z

Note that the above pseudo codes are examples of how the expression can be represented using different instruction formats and the format of the instruction may vary based on the architecture or programming language you are using.

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Please explain the pseudo code answers for each instruction sets above

  1. 0-address format: In this format, all operations are represented using opcodes and operands are implicitly referenced using registers or memory locations.
  • The instruction "LOAD 4" loads the value 4 into a register.
  • The instruction "ADD A" adds the value stored in memory location A to the register.
  • The instruction "ADD B" adds the value stored in memory location B to the register.
  • The instruction "STORE TEMP1" stores the result of the previous addition operation in memory location TEMP1.
  • The instruction "LOAD 16" loads the value 16 into a register.
  • The instruction "LOAD C" loads the value stored in memory location C into a register.
  • The instruction "ADD 58" adds the value 58 to the register.
  • The instruction "STORE TEMP2" stores the result of the previous addition operation in memory location TEMP2.
  • The instruction "MUL TEMP1" multiplies the value stored in memory location TEMP1 with the value in the register.
  • The instruction "MUL TEMP2" multiplies the value stored in memory location TEMP2 with the value in the register.
  • The instruction "SUB" subtracts the value in the register from the previous multiplication result.
  • The instruction "STORE Z" stores the result of the previous subtraction operation in memory location Z.
  1. 1-address format: In this format, all operations are represented using opcodes and one explicit operand, the other operand is implicitly referenced using a register or memory location.
  • The instruction "LOAD 4 A" loads the value 4 into a register and adds the value stored in memory location A to it.
  • The instruction "ADD B" adds the value stored in memory location B to the register.
  • The instruction "STORE TEMP1" stores the result of the previous addition operation in memory location TEMP1.
  • The instruction "LOAD 16 C" loads the value 16 into a register and adds the value stored in memory location C to it.
  • The instruction "ADD 58" adds the value 58 to the register.
  • The instruction "STORE TEMP2" stores the result of the previous addition operation in memory location TEMP2.
  • The instruction "MUL TEMP1" multiplies the value stored in memory location TEMP1 with the value in the register
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Please explain the pseudo code answers for each instruction sets above

  1. 1-address format (continued):
  • The instruction "SUB" subtracts the value in the register from the previous multiplication result.
  • The instruction "STORE Z" stores the result of the previous subtraction operation in memory location Z.
  1. 2-address format: In this format, all operations are represented using opcodes and two explicit operands.
  • The instruction "LOAD 4 A B" loads the value 4 into a register, adds the value stored in memory location A to it and add the value stored in memory location B to it.
  • The instruction "ADD" stores the result of the previous addition operation in register.
  • The instruction "STORE TEMP1" stores the result in memory location TEMP1.
  • The instruction "LOAD 16 C 58" loads the value 16 into a register, adds the value stored in memory location C to it and add the value 58 to it.
  • The instruction "ADD" stores the result of the previous addition operation in register.
  • The instruction "STORE TEMP2" stores the result in memory location TEMP2.
  • The instruction "MUL TEMP1" multiplies the value stored in memory location TEMP1 with the value in the register.
  • The instruction "MUL TEMP2" multiplies the value stored in memory location TEMP2 with the value in the register.
  • The instruction "SUB" subtracts the value in the register from the previous multiplication result.
  • The instruction "STORE Z" stores the result of the previous subtraction operation in memory location Z.
  1. 3-address format: In this format, all operations are represented using opcodes and three explicit operands.
  • The instruction "MUL A B 4 TEMP1" multiplies the value stored in memory location A with the value stored in memory location B and multiplies the result with 4, and stores the result in memory location TEMP1
  • The instruction "MUL C 58 16 TEMP2" multiplies the value stored in memory location C with the value 58 and multiplies the result with 16, and stores the result in memory location TEMP2
  • The instruction "SUB TEMP1 TEMP2 Z" subtracts the value in memory location TEMP1 with the value in memory location TEMP2, and stores the result in memory location Z

In all the above instruction formats, the instruction set is used to evaluate the given expression: Z = 4(A + B ) - 16(C +58)

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write detailed thesis on Bounding Worst-Case Performance for Real-Time Applications

In the field of real-time systems, ensuring predictable and reliable performance is crucial. One important aspect of real-time performance is the ability to bound the worst-case execution time (WCET) of a task, which represents the maximum amount of time that a task can take to complete. Bounding the WCET of a task allows for the design of real-time systems that can meet their deadlines and avoid errors caused by exceeding the specified time limits.

There are several approaches to bounding the WCET of a task, including static analysis, dynamic analysis, and measurement-based analysis. Static analysis involves analyzing the code of a task to determine the worst-case execution time without actually running the code. This approach is useful for small and simple programs, but it becomes increasingly difficult to perform as the size and complexity of the program increases. Dynamic analysis involves executing the code of a task multiple times and measuring the execution time to determine the WCET. This approach is useful for large and complex programs, but it can be time-consuming and may not provide accurate results if the program includes non-deterministic elements. Measurement-based analysis involves measuring the execution time of a task during normal operation and using this information to estimate the WCET. This approach is useful for systems that are already in operation and provides accurate results, but it may not be feasible for systems that are being designed or developed.

Another important aspect of real-time performance is the ability to bound the worst-case memory usage of a task, which represents the maximum amount of memory that a task can use to complete. Bounding the worst-case memory usage of a task allows for the design of real-time systems that can meet their memory constraints and avoid errors caused by exceeding the specified memory limits.

One of the most popular methodologies to bound the worst-case execution time is the "path-based analysis" which consists in analyzing the control flow graph of the program, identifying the longest path in it and then use the information of the instructions on this path to compute the WCET. This method is considered effective because it is based on the actual execution of the program, however, it has a high computational cost and can only provide an upper bound of the WCET.

Another popular methodology is "timing-analysis" which is based on the use of measurement-based analysis and the use of statistical methods to estimate the WCET. This method is considered effective because it is based on the actual execution of the program and it is less computationally expensive than path-based analysis. However, it can only provide a lower bound of the WCET.

In conclusion, bounding the worst-case execution time and memory usage of a task is an important aspect of real-time systems design. There are several approaches to bounding these values, including static analysis, dynamic analysis, and measurement-based analysis. Each approach has its own advantages and disadvantages and the choice of approach depends on the specific requirements of the system being designed. Additionally, the use of multiple methodologies can provide more accurate and realistic results.

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write detailed project on Bounding Worst-Case Performance for Real-Time Applications

Project Title: Bounding Worst-Case Performance for Real-Time Applications

Introduction: The ability to predict and guarantee the performance of real-time systems is crucial for their correct functioning. One of the most important aspects of real-time performance is the ability to bound the worst-case execution time (WCET) of a task, which represents the maximum amount of time that a task can take to complete. The goal of this project is to develop a methodology for bounding the WCET of real-time tasks and evaluate its effectiveness.

Background: There are several approaches to bounding the WCET of a task, including static analysis, dynamic analysis, and measurement-based analysis. Static analysis involves analyzing the code of a task to determine the worst-case execution time without actually running the code. Dynamic analysis involves executing the code of a task multiple times and measuring the execution time to determine the WCET. Measurement-based analysis involves measuring the execution time of a task during normal operation and using this information to estimate the WCET.

Methodology: The methodology proposed in this project is based on a combination of static analysis and measurement-based analysis. The first step is to perform static analysis on the code of the task to identify the longest execution path and the instructions on this path. Next, the task is executed multiple times under different inputs and the execution time is measured. The execution time of the instructions on the longest path is then used to estimate the WCET.

Experiments: The methodology proposed in this project will be evaluated using a set of real-time tasks taken from different domains such as control systems, embedded systems, and avionics. The effectiveness of the methodology will be measured by comparing the estimated WCET with the actual WCET obtained by executing the task.

Expected Results: The results of this project are expected to show that the proposed methodology is effective in bounding the WCET of real-time tasks. The estimated WCET should be close to the actual WCET for most tasks, with a small margin of error. Additionally, the results will also demonstrate how the methodology can be useful for real-time systems design and can help to avoid errors caused by exceeding the specified time limits.

Conclusion: Bounding the worst-case execution time of real-time tasks is crucial for ensuring predictable and reliable performance. The methodology proposed in this project is based on a combination of static analysis and measurement-based analysis and is expected to be effective in bounding the WCET of real-time tasks. The results of this project will be useful for real-time systems design and can help to avoid errors caused by exceeding the specified time limits.

Note: This project proposal is a high-level overview and would require further details, feasibility, and scope analysis. Also, the expected results are theoretical and the actual results may vary depending on the specific tasks and the system being evaluated.

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1. Briefly explain the following: (i) latency (ii) throughput (iii). Instruction set

(i) Latency: Latency is the time it takes for a system or a process to respond to a request. In a computer system, latency can refer to the time it takes for a storage device to access data, the time it takes for a network packet to travel from one endpoint to another, or the time it takes for a process to complete a specific task. Latency is an important metric in real-time systems, as it can affect the ability of the system to meet its deadlines.

(ii) Throughput: Throughput is the rate at which a system or a process can handle and process data. In a computer system, throughput can refer to the number of requests that can be handled by a storage device, the number of packets that can be transmitted by a network, or the number of tasks that can be completed by a process. Throughput is an important metric in systems that handle large amounts of data, such as data centers and cloud computing systems.

(iii) Instruction Set: An instruction set is a set of instructions that a computer's central processing unit (CPU) can execute. The instruction set defines the operations that the CPU can perform, such as arithmetic and logical operations, memory access, and control flow operations. The instruction set architecture (ISA) defines the interface between the software and the hardware, and different ISAs have different instruction sets, which affects the programming of the systems. Different CPU's have different instruction sets architectures, and it's important to choose the right instruction set for the application to optimize performance.

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In response to feedback from customers of M&B cloud Nigeria on response time and throughput of their server, the system engineers of the organization proposed two solutions: (i) a GPU server with a faster clock speed, (ii) separate servers to handle for different tasks. Discuss the effect of these enhancement on response time, throughput or both.

(i) GPU server with faster clock speed:

  • Effect on response time: A GPU server with a faster clock speed is likely to have a positive impact on response time as the faster clock speed will enable the server to process requests more quickly. The GPU server will also be able to perform parallel processing which can also improve the response time.

  • Effect on throughput: A GPU server with a faster clock speed may also have a positive impact on throughput as the server will be able to handle more requests in a given period of time. The GPU server will also be able to perform parallel processing which can also improve the throughput.

(ii) Separate servers to handle different tasks:

  • Effect on response time: Having separate servers to handle different tasks can have a positive impact on response time. By separating the tasks, the servers can be optimized to perform specific tasks, which can improve the response time.

  • Effect on throughput: Separating the servers to handle different tasks may also have a positive impact on throughput. By separating the tasks, the servers can be optimized to perform specific tasks, which can improve the throughput. The separated servers can handle more requests in a given period of time and also have a better performance.

In summary, both solutions proposed by the system engineers have the potential to improve both response time and throughput. However, the actual impact will depend on the specific implementation and how well the servers are optimized for the specific tasks.

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Briefly discuss the effects of advancements in computer hardware technology on computer software industry.

Advancements in computer hardware technology have a significant impact on the computer software industry. The following are some of the effects of advancements in computer hardware technology on computer software industry:

  1. Performance improvements: Advancements in computer hardware technology, such as faster processors, more memory, and faster storage devices, result in improved performance of computer systems. This allows software developers to create more complex and sophisticated software that can perform tasks more quickly and efficiently.

  2. Increased functionality: Advancements in computer hardware technology also result in new capabilities for computer systems. For example, the development of graphics processing units (GPUs) has led to the creation of software that can perform complex visual and graphical computations.

  3. New application domains: Advancements in computer hardware technology open up new application domains for software development. For example, the development of mobile devices with powerful processors and sophisticated sensors has led to the creation of new software applications for these devices, such as augmented reality and machine learning.

  4. Changes in software development: Advancements in computer hardware technology also result in changes in the way software is developed. For example, the development of parallel processing technologies has led to the development of parallel programming languages, libraries, and frameworks that allow software developers to take advantage of these technologies.

  5. Hardware-software integration: Advancements in computer hardware technology also make it possible for software developers to create more powerful and sophisticated software by integrating hardware and software in new ways. For example, the development of the Internet of Things (IoT) devices has led to the creation of software that can interact with and control these devices.

Overall, advancements in computer hardware technology drive innovation in software development and open up new opportunities for software developers to create more powerful, sophisticated and efficient software applications.

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4. Describe In-Order Issue with Out-of-Order completion

In-order issue with out-of-order completion refers to a technique used in computer processors to improve performance by allowing instructions to be executed out of order, but still maintaining the original order of the instructions as they were received by the processor.

In-order issue refers to the process of fetching and decoding instructions in the order they were received by the processor. The processor will fetch and decode the first instruction, then the second instruction and so on. This is the traditional way in which processors have executed instructions, and it is known as in-order issue.

Out-of-order completion, on the other hand, refers to the process of completing instructions in a different order than they were received by the processor. For example, the processor may execute the second instruction before the first instruction, or the third instruction before the second instruction. This is done to take advantage of the dependencies and latencies of the instructions, allowing the processor to execute instructions more efficiently.

By using both in-order issue and out-of-order completion, the processor can fetch and decode instructions in the original order, but execute them in a more efficient order, resulting in improved performance.

In-order issue with out-of-order completion can help to reduce pipeline stalls and improve the overall performance of the processor. It can also improve the performance of the processor by reducing the amount of time spent waiting for dependencies and latencies to be resolved. However, the implementation of this technique can be complex and may require additional hardware and software support.

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. Briefly explain how the following has contributed to system execution or response time and throughput: i. Cache Memory ii. Main Memory iii. Bus Address System.

i. Cache Memory: Cache memory is a small, high-speed memory that stores frequently-accessed data for quick access by the CPU. This improves the system execution and response time by reducing the time needed to access data from main memory. The CPU can access the data stored in cache memory much faster than it can access data from main memory, resulting in faster system execution and response time.

ii. Main Memory: Main memory, also known as random-access memory (RAM), stores data that is currently being used by the CPU. The larger the main memory, the more data can be stored and accessed quickly by the CPU, which can improve system execution and response time.

iii. Bus Address System: The bus address system is responsible for managing the communication between the CPU, main memory, and other components of the system. A faster bus address system can improve the system's execution and response time by allowing the CPU to access data from main memory more quickly. A bus address system with a higher bandwidth can also improve the throughput of the system by allowing more data to be transferred between the CPU, main memory, and other components at a faster rate.

In summary, the implementation of cache memory, main memory and bus address system can contribute to improve system execution and response time by providing faster data access and transfer and also improve throughput by allowing more data to be processed in a given time.

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In a microcomputer, the addresses of memory locations are binary numbers that identify each memory circuit where a byte is stored. The number of bits that make up an address depends on how many memory locations there are. Since the number of bits can be very large, the addresses are often specified in hex instead of binary. i. If a microcomputer uses a 20-bit address, how many different memory locations are there? ii. How many hex digits are needed to represent the address of a memory location there? iii. What is the hex address of the 255th memory location? (Note: The first address is always 0) iv. The computer program is stored in the lowest 2kbyte block of memory. Give the start and end address of this block.

i. If a microcomputer uses a 20-bit address, there would be 2^20= 1,048,576 different memory locations.

ii. 20-bits is equal to 20/4 = 5 hex digits. Therefore, 5 hex digits are needed to represent the address of a memory location there.

iii. The hex address of the 255th memory location would be 0x00FF. (Note: 255 in decimal is equal to FF in hexadecimal)

iv. The lowest 2Kbyte block of memory is 2KByte= 2048 bytes = 2048 * 8 = 16384 bits. Since the computer uses 20-bit addresses, so the start address of the block would be 0x0000 and the end address would be 0x03FF. This block will contain the address from 0x0000 to 0x03FF

Note: The hex address calculation above is based on the assumption that the memory addresses start from 0, if it starts from 1 then the address would be different.

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7. What is the major concept of RISC architecture? Discuss any five characteristics of RISC and CISC processors.

The major concept of RISC (Reduced Instruction Set Computing) architecture is to simplify the instruction set of a computer's central processing unit (CPU) in order to improve performance. RISC processors use a smaller number of simple instructions, which are executed quickly, to perform the same tasks that a larger number of complex instructions would be required to perform on a CISC (Complex Instruction Set Computing) processor.

Five characteristics of RISC processors:

  1. Simple instruction set: RISC processors use a small number of simple instructions that can be executed quickly, which results in improved performance.

  2. Large number of registers: RISC processors have a large number of registers to hold data, which reduces the need to access memory and improves performance.

  3. Fixed-length instructions: RISC processors use fixed-length instructions, which simplifies instruction decoding and improves performance.

  4. Reduced instruction-level parallelism: RISC processors have reduced instruction-level parallelism, which means that the instructions are executed sequentially and this improves the predictability of the performance.

  5. High clock speed: RISC processors have a high clock speed, which allows them to execute instructions more quickly and improve performance.

Five characteristics of CISC processors:

  1. Complex instruction set: CISC processors use a large number of complex instructions that can perform a wide variety of tasks, but can take longer to execute.

  2. Smaller number of registers: CISC processors have a smaller number of registers, which means that more data needs to be accessed from memory, which can slow down performance.

  3. Variable-length instructions: CISC processors use variable-length instructions, which can make instruction decoding more complex and slow down performance.

  4. Increased instruction-level parallelism: CISC processors have increased instruction-level parallelism, which means that instructions can be executed in parallel, improving performance but also making it harder to predict the performance.

  5. Lower clock speed

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continue

: CISC processors generally have a lower clock speed compared to RISC processors. This means that they may take longer to execute instructions, which can slow down performance.

In summary, RISC processors are designed to have a simple instruction set, a large number of registers, fixed-length instructions, reduced instruction-level parallelism, and high clock speed in order to improve performance. CISC processors, on the other hand, have a more complex instruction set, a smaller number of registers, variable-length instructions, increased instruction-level parallelism, and generally lower clock speed, which can make performance less predictable.

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8. Consider a 32-bit machine with 64KB cache, 32B block and two-way set-associative cache, obtain the offset bits, index bits and tag bits for mapping addresses to frames on this machine.

In a 32-bit machine with 64KB cache, 32B block and two-way set-associative cache:

  • The size of the cache is 64KB = 64 * 2^10 = 65536 bytes = 2^16 bytes.

  • The size of a block is 32B = 32 bytes = 2^5 bytes.

  • The number of blocks in the cache is the cache size divided by the block size: 2^16 / 2^5 = 2^11 blocks.

  • The number of sets in the cache is the number of blocks divided by the associativity: 2^11 / 2 = 2^10 sets.

  • The number of offset bits is the number of bits needed to represent the size of a block: log2(32) = 5 bits.

  • The number of index bits is the number of bits needed to represent the number of sets: log2(2^10) = 10 bits.

  • The number of tag bits is the number of bits in the memory address minus the number of offset and index bits: 32 - 5 - 10 = 17 bits.

So, the offset bits are 5, index bits are 10 and tag bits are 17 for mapping addresses to frames on this machine.

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9. Write short notes on memory hierarchy considering capacity, access time and cost.

Memory hierarchy is the arrangement of different types of memory in a computer system, organized based on their capacity, access time, and cost. The goal of the memory hierarchy is to provide the CPU with fast access to a large amount of data while minimizing the cost.

  1. Registers: Registers are the smallest and fastest type of memory, located inside the CPU. They have the smallest capacity, usually only a few bytes, but have the fastest access time, usually less than a nanosecond. Registers are also the most expensive type of memory.

  2. Cache: Cache is a small, fast memory located on or near the CPU. It has a larger capacity than registers, usually a few kilobytes, and a slightly slower access time, usually a few nanoseconds. Cache is more expensive than main memory but less expensive than registers.

  3. Main memory: Main memory, also known as random-access memory (RAM), is located on the motherboard and holds the data that is currently being used by the CPU. It has a larger capacity than cache, usually a few gigabytes, and a slower access time, usually a few tens of nanoseconds. Main memory is less expensive than cache but more expensive than storage.

  4. Storage: Storage, such as hard drives and solid-state drives, holds data that is not currently being used by the CPU but needs to be retained for later use. It has the largest capacity, usually several terabytes, but also the slowest access time, usually a few milliseconds. Storage is the most economical type of memory.

In summary, memory hierarchy is designed to provide the CPU with fast access to a large amount of data while minimizing the cost. Registers are the smallest and fastest but most expensive type of memory, while storage is the largest and cheapest but slowest type of memory. Cache and main memory are located in between, with varying degrees of capacity, access time and cost.

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. For a car ignition to be turned on: (i) the driver must be in the car having all the doors locked having the gear in parking or neutral state, or (ii) the car can be put on using a remote starter from outside without the driver in the car, but all the car doors must be locked with hand brake on and gear in parking state or neutral. Any deviations from these two conditions will set the car security alarm on. i. What are the variables to be used in representing the states? ii. Design a logic circuit for the scenario iii. Modify the design to set the car alarm on for a second after the ignition must have been on for five minutes provided the doors, gear and handbrake are at the initial state and then off the ignition if the states remain the same after the last second.

i. The variables to be used in representing the states are:

  • Driver in car (D)
  • Doors locked (L)
  • Gear in parking or neutral state (G)
  • Hand brake on (B)
  • Remote starter used (R)

ii. A logic circuit for the scenario can be designed using Boolean logic gates. The circuit could use AND gates to check that the driver is in the car and the doors are locked, or that the remote starter is used and the doors, gear, and hand brake are in the correct state. An OR gate can then be used to combine these two conditions. If either of these conditions is met, the circuit can output a signal to turn the ignition on.

iii. To set the car alarm on for a second after the ignition has been on for five minutes and then off the ignition if the states remain the same after the last second, a timer circuit can be added to the design. An additional AND gate can be used to check that the doors, gear, and hand brake are in their initial state. A timer can then be used to keep track of how long the ignition has been on. If the ignition has been on for five minutes and the doors, gear, and hand brake are in their initial state, the timer can output a signal to turn on the alarm for one second. If the states remain the same after the alarm has sounded, the timer can output a signal to turn off the ignition.

It's a complex system and the design can be different based on the different microcontroller or microprocessor used, the above explanation is just a high-level overview of how the system could be designed.

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11. A common transformation required in graphics processors is square root. Implementations of floatingpoint (FP) square root vary significantly in performance, especially among processors designed for graphics. Suppose FP square root (FPSQR) is responsible for 20% of the execution time of a critical graphics benchmark. One proposal is to enhance the FPSQR hardware and speed up this operation by a factor of 10. The other alternative is just to try to make all FP instructions in the graphics processor run faster by a factor of 1.6; FP instructions are responsible for half of the execution time for the application. The design team believes that they can make all FP instructions run 1.6 times faster with the same effort as required for the fast square root. Compare these two design alternatives.

The first design alternative is to enhance the FPSQR hardware and speed it up by a factor of 10. This would result in a significant reduction in the execution time for the FPSQR operation, which is currently responsible for 20% of the execution time for the critical graphics benchmark. This would result in a reduction of 20% * 10 = 2% in the total execution time for the application.

The second design alternative is to try to make all FP instructions in the graphics processor run faster by a factor of 1.6. FP instructions are currently responsible for half of the execution time for the application. This would result in a reduction of 0.5 * 1.6 = 0.8 in the total execution time for the application.

Comparing the two design alternatives, the first design alternative would result in a larger reduction in the total execution time for the application (2% vs 0.8%). However, the design team believes that they can make all FP instructions run 1.6 times faster with the same effort as required for the fast square root. So the second alternative would be the best option if the design team can achieve the performance improvement in FP instructions as expected.

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12. A computer uses a memory unit with 256k words of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts, an indirect bit, an operation code, a register code part to specify one of 64 registers, and an address part i. How many bits are there in the operation code, the register code part, and the address ii. Draw the instruction word format and indicate the number of bits in each part. iii. How many bits are there in the data and address inputs of the memory?

i. The instruction has a total of 32 bits, with an indirect bit, an operation code, a register code part, and an address part. Since there are 64 registers, the register code part requires log2(64) = 6 bits. The number of bits in the operation code and the address part will depend on the specific instruction set architecture and the number of instructions and memory addresses that need to be represented. Let's say the operation code is 8 bits and the address part is 18 bits.

ii. The instruction word format can be represented as follows:

Indirect BitOperation CodeRegister CodeAddress
18618

iii. The memory unit has a total of 256k words of 32 bits each, so the data input has 32 bits and the address input has log2(256k) = 18 bits.

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3. The memory unit of a computer has 256k words of 32 bits each. The computer has an instruction format with for fields: an operation code field, a mode field to specify one of seven addressing modes, a register address field to specify one of 60 processor registers and a memory address. Specify the instruction format and the number of bits in each field if the instruction is in one memory word.

The instruction format for the computer can be represented as follows:

| Operation Code | Mode | Register Address | Memory Address |

The instruction has a total of 32 bits, with an operation code field, a mode field, a register address field and a memory address. Since there are 60 registers, the register address field requires log2(60) = 6 bits. The number of bits in the operation code, the mode field and memory address will depend on the specific instruction set architecture and the number of instructions, addressing modes and memory addresses that need to be represented. Let's say the operation code is 8 bits, the mode field is 4 bits and the memory address is 14 bits.

Operation CodeModeRegister AddressMemory Address
84614

So, The instruction format is 8 bits for operation code, 4 bits for mode field, 6 bits for register address and 14 bits for memory address.

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14. Suppose you have a machine used in an I/O intensive environment; the CPU is working 75% of the and the rest is waiting for I/O operations to complete. You may consider an improvement of the CPU by a factor of 2 (it will run twice as fast as it runs now) for a fivefold increase in cost. The present cost of the CPU is 20% of the machine's cost. Is the suggested improvement cost effective?

To determine if the suggested improvement of the CPU by a factor of 2 is cost-effective, we need to compare the cost of the improvement to the potential increase in performance.

Currently, the CPU is working 75% of the time and waiting for I/O operations to complete for 25% of the time. If the CPU is improved by a factor of 2, it will run twice as fast, which means the CPU will be working 75% * 2 = 150% of the time.

The cost of the improvement is a fivefold increase in the cost of the CPU, which is currently 20% of the machine's cost. Therefore, the cost of the improvement is 20% * 5 = 100% of the machine's cost.

Since the cost of the improvement is equal to the entire cost of the machine, it is not cost-effective. It is not worth spending the entire cost of the machine to improve the CPU's performance by a factor of 2, when it will only increase the CPU's utilization by 50%. Additionally, you need to consider the other factors like benefits, risk and other factors that might influence the decision.

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15. Suppose you have enhanced your machine with a floating-point coprocessor; all floating-point operations are faster by a factor of 10 when the coprocessor is in use. What percent of the time should be spent in floating point operations such that the overall speedup is 2?

Let's call the time spent in floating-point operations as "T" and the overall speedup as "S".

When the coprocessor is in use, the speedup for floating-point operations is a factor of 10. Therefore, the time spent in floating-point operations with the coprocessor is T/10.

The overall speedup is defined as the ratio of the execution time of the machine with the coprocessor to the execution time of the machine without the coprocessor. S = (T + (T/10)) / T

We are given that the overall speedup is 2, so we can set up the equation: 2 = (T + (T/10)) / T

Solving for T, we get: T = T/5

The percent of time spent in floating-point operations is T/5, or 20%. Therefore, if 20% of the time is spent in floating-point operations, the overall speedup will be 2.

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16. Briefly describe Pipelining and its limitation.

Pipelining is a technique used to increase the performance of computer processors by breaking down the execution of instructions into multiple stages. Each stage performs a specific task, such as fetching an instruction, decoding it, executing it, and writing the result. While one instruction is being executed in one stage, another instruction can be fetched and decoded in the next stage. This allows for multiple instructions to be processed simultaneously, increasing the overall performance of the processor.

The limitation of pipelining is that it can introduce stalls or pipeline hazards. These occur when an instruction depends on the result of a previous instruction and cannot be executed until that instruction completes. This causes the pipeline to stall, and can result in a significant reduction in performance. To mitigate these stalls, processors use techniques such as branch prediction and instruction reordering to minimize the number of stalls and maximize the performance of the pipeline. Another limitation is that it increases the complexity of the processor and the design process.

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17. A 8 Megapixel digital camera stores an eight-bit number for the brightness of each of the primary colours (red, green, blue) found in each picture element (pixel). If every bit is stored (no data compression), how many pictures can be stored on a 256 Megabyte memory card.

An 8 Megapixel digital camera stores an eight-bit number for the brightness of each of the primary colours (red, green, blue) found in each picture element (pixel), which means that each pixel requires 24 bits (8 bits for red, 8 bits for green, 8 bits for blue).

A 256 Megabyte memory card can store 256 * 2^20 bytes = 268,435,456 bytes.

To store one picture with 8 Megapixel, we need 810^6 pixels. To store this amount of pixels we need 810^6*24 bits = 192,000,000 bits.

So, the number of pictures that can be stored on a 256 Megabyte memory card is: 268,435,456 bytes * 8 bits/byte / 192,000,000 bits/picture = 14 pictures

So, a 256 Megabyte memory card can store 14 pictures if every bit is stored (no data compression).

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18. Discuss computer memory hierarchy in terms of size, speed and cost.

Computer memory hierarchy refers to the different types of memory used in a computer system, arranged in order of increasing size and decreasing speed. Each level of the memory hierarchy serves a different purpose, and the cost of each level also varies.

  1. Register: At the top of the memory hierarchy are the registers. They are the smallest and fastest memory in a computer, and are located within the CPU. They are used to hold temporary data and instructions for the CPU to execute. They are the most expensive memory per byte.

  2. Cache: Cache memory is the next level of the memory hierarchy. It is smaller and faster than main memory, but larger and slower than registers. It is used to store frequently accessed data and instructions, so that the CPU can quickly access them without having to go to the main memory. It is more expensive than main memory, but less expensive than registers.

  3. Main Memory (RAM): Main memory is the largest and fastest memory that is directly accessible to the CPU. It is used to store data and instructions for the CPU to access while the computer is running. It is slower than cache, but faster than storage. It is less expensive than cache, but more expensive than storage.

  4. Secondary storage (Hard Disk, SSD): At the bottom of the memory hierarchy, we have secondary storage. It is the largest and slowest memory, and is used to store data and instructions for long-term storage. It is slower than main memory, but it is less expensive per byte.

In summary, the computer memory hierarchy is arranged in order of decreasing speed and increasing size as we move down the hierarchy. The cost per byte also increases as we move up the hierarchy. Register is the fastest and most expensive memory, while secondary storage is the slowest and least expensive memory.

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19. Difference between the two types of control units used to execute an instruction

There are two main types of control units used to execute an instruction in a computer: the hardwired control unit and the microprogrammed control unit.

  1. Hardwired control unit: A hardwired control unit uses combinational logic to implement the control logic required to execute instructions. The control signals are generated by the control unit based on the instruction opcode. The control unit consists of a set of combinational logic circuits that are designed to generate the control signals required to execute the instruction. The hardwired control unit is simple, fast and efficient but it is difficult to modify the instruction set and it is less flexible.

  2. Microprogrammed control unit: A microprogrammed control unit uses a microcode stored in a control memory to implement the control logic required to execute instructions. The control signals are generated by the control unit based on the microcode. The control unit consists of a control memory that stores the microcode, a microsequencer that reads and interprets the microcode, and a set of control signals that are generated based on the microcode. The microprogrammed control unit is more flexible and can be modified easily but it is less efficient and slower than a hardwired control unit.

In summary, the main difference between the two types of control units is that a hardwired control unit uses combinational logic to execute instructions, while a microprogrammed control unit uses microcode stored in a control memory to execute instructions. The hardwired control unit is simple, fast and efficient, but less flexible, while the microprogrammed control unit is more flexible, but less efficient and slower.

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20. A processor has a five-stage pipeline. If a branch is taken, then four cycles are needed to flush the pipeline. The branch penalty b is thus 4. The probability Pb that a particular instruction is a branch is 0.25. The probability Pt that the branch is taken is 0.5. Compute the average number of cycles needed to execute an instruction, and the execution efficiency.

The average number of cycles needed to execute an instruction can be calculated using the following formula:

Cycles per instruction (CPI) = 1 + (Pb * Pt * b)

Where b is the branch penalty, Pb is the probability that a particular instruction is a branch, and Pt is the probability that the branch is taken.

Given the values, the CPI is: CPI = 1 + (0.25 * 0.5 * 4) = 1 + (0.25 * 2) = 1.5

The execution efficiency is the ratio of the number of instructions executed per cycle to the number of instructions that could be executed per cycle in the absence of any pipeline stalls.

The execution efficiency can be calculated using the following formula:

Execution efficiency = 1 / CPI

Given the CPI of 1.5, the execution efficiency is: Execution efficiency = 1 / 1.5 = 0.66 or 66%

So, the average number of cycles needed to execute an instruction is 1.5 and the execution efficiency is 66%.

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21. What is meant by sub-word parallelism?

Sub-word parallelism is a technique used to increase the performance of computer processors by processing multiple smaller pieces of data, called sub-words, in parallel. This can be done by using specialized instructions, such as SIMD (Single Instruction, Multiple Data) instructions, that can operate on multiple sub-words at the same time.

For example, a 64-bit processor can process 8 bytes in parallel using SIMD instructions. This allows for a significant increase in performance for certain types of computations, such as image and video processing, where the same operation is applied to multiple pixels or data points simultaneously.

Another example is in VLIW (Very Long Instruction Word) processors, which can execute multiple instructions in parallel, but they are not as flexible as in SIMD instructions, since they are designed to execute specific instructions in parallel.

In summary, sub-word parallelism is a technique that allows a computer processor to process multiple smaller pieces of data simultaneously in order to increase the performance. This can be done using specialized instructions, such as SIMD and VLIW instructions.

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22. What is Computer Architecture? Briefly explain the following terms: Interrupts (ii) Instruction formats (iii) Multiplexers (iv) Decoders

Computer Architecture refers to the fundamental design and organization of a computer system, including the data paths, control unit, memory, and I/O systems. It defines the way in which a computer system handles and processes data and instructions.

  1. Interrupts: An interrupt is a signal sent to the computer processor by a device or program, indicating that it needs the processor's attention. When an interrupt occurs, the processor stops its current task and begins executing the code associated with the interrupt. This allows the processor to handle multiple tasks simultaneously, and is an important feature for real-time systems and for handling asynchronous events.

  2. Instruction Formats: An instruction format is the layout of an instruction in memory, including the operation code, operands, and addressing modes. The instruction format defines the way in which the processor interprets and executes instructions. Different instruction formats may have different sizes, numbers of operands, and addressing modes.

  3. Multiplexers: A multiplexer (MUX) is a digital circuit that selects one of several input lines and directs it to a single output line. It is used in computer architecture to select between different sources of data or instructions, and to route them to the appropriate destination.

  4. Decoders: A decoder is a digital circuit that converts a binary input into a set of outputs. In computer architecture, decoders are used to convert instruction opcodes into control signals that are used by the control unit to execute instructions. Decoders are used to interpret instructions and generate control signals for the processor.

In summary, Computer Architecture is the fundamental design and organization of a computer system, which defines the way in which a computer system handles and processes data and instructions. Interrupts, Instruction Formats, Multiplexers, and Decoders are all important components of computer architecture that play a role in the way a computer system processes and executes instructions.

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23. Consider the following quadratic equation: 2x 2 – 40x + 150 = 0. A trusted mathematician tells us that the roots for this equation are 15 and 75. However, when you try to solve it, the roots turns out to be 15 and 5. Explain why 75 could be a solution while 5 is not.

The equation 2x^2 - 40x + 150 = 0 can be factored into (2x - 15)(x - 75) = 0. From this factored form, we can see that the roots of the equation are x = 15 and x = 75.

In solving the equation using the quadratic formula, we can get the roots by using the following formula: x = (-b ± √(b^2 - 4ac)) / 2a.

For the given equation, a = 2, b = -40 and c = 150, so the roots are: x = (-(-40) ± √((-40)^2 - 42150)) / 2*2 x = (40 ± √(1600 - 1200)) / 4 x = (40 ± √(400)) / 4 x = (40 ± 20) / 4 x = 15 and 5

The root x = 75 is not a solution because it is not a root of the equation. The reason that the quadratic formula gives a different result than the factoring is that the equation may have been rounded or approximated to get the roots. Also, it is possible that the equation has been manipulated or simplified in some way, which can affect the final result. And also the difference may be due to the way the quadratic formula is applied.

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24. Four memory hierarchy questions are as follows: Where can a block be placed (block placement)? How is a block found (block identification)?; Which block is to be replaced on a miss (block replacement)?; and What happens on a write (write strategy)? Provide answers to each of the four common memory hierarchy questions for cache memory.

  1. Block Placement: In cache memory, blocks can be placed in different ways, depending on the cache design. The most common block placement policies are:
  • Direct mapping: A block is placed in a specific cache location based on its memory address.
  • Associative mapping: A block can be placed in any location in the cache.
  • Set-associative mapping: A block can be placed in any location within a set of cache locations.
  1. Block Identification: In cache memory, block identification refers to the process of determining whether a block is present in the cache. This is typically done by comparing the memory address of the block with the memory addresses of the blocks in the cache.

  2. Block Replacement: In cache memory, block replacement refers to the process of replacing a block in the cache when a new block needs to be stored and the cache is full. Common replacement policies include:

  • Least Recently Used (LRU): The block that has been least recently accessed is replaced.
  • Least Frequently Used (LFU): The block that has been least frequently accessed is replaced.
  • Random Replacement: A random block is replaced
  1. Write Strategy: In cache memory, the write strategy refers to the method used to update the data in the cache when a write operation occurs. The two common write strategies are:
  • Write-through: The data is written to both the cache and the main memory at the same time.
  • Write-back: The data is written to the cache only and the main memory is updated later when the block is replaced.

In summary, for cache memory the four common memory hierarchy questions are: Where a block can be placed, How is a block found, Which block is to be replaced on a miss and What happens on a write. Different policies can be used for each question to optimize cache performance.

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25. What are the disadvantages of increasing the number of stages in pipelined processing?

There are several disadvantages of increasing the number of stages in pipelined processing:

  1. Increased complexity: As the number of stages in a pipeline increases, the complexity of the pipeline also increases. This makes the pipeline more difficult to design, implement and debug.

  2. Increased latency: The more stages in a pipeline, the longer it takes for data to pass through the pipeline. This means that there is an increase in the latency between the time data is loaded into the pipeline and the time it is processed.

  3. Increased power consumption: As the number of stages in a pipeline increases, the power consumption also increases. This can be a significant issue for systems that need to operate on battery power or have limited power supply.

  4. Limited throughput: When a pipeline has more stages, it becomes more difficult to keep the pipeline full. This can lead to a decrease in throughput, as the pipeline is not able to process as many instructions per clock cycle.

  5. Increased branch penalty: With a larger number of pipeline stages, it takes longer to flush the pipeline when a branch instruction is encountered. This leads to increased branch penalty, which degrades performance.

  6. More difficult to predict and schedule: With more stages, it becomes more difficult to predict how long it will take for an instruction to be executed and for the pipeline to be full again. This makes it more difficult to schedule instructions and achieve optimal performance.

In summary, the main disadvantage of increasing the number of stages in pipelined processing is that it can lead to increased complexity, latency, power consumption, limited throughput, increased branch penalty, and more difficult to predict and schedule.

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26. Consider a nonpipelined machine with 6 execution stages of lengths 50ns, 50ns, 60ns, 60ns, 50ns, and 50ns. i. Find the instruction latency on this machine. ii. How much time does it take to execute 100 instructions?

  1. Instruction latency on this machine is the time it takes for an instruction to pass through all the stages of the pipeline. In this case, the instruction latency is the sum of the length of all the stages.

i. The instruction latency = (50ns + 50ns + 60ns + 60ns + 50ns + 50ns) = 320ns

  1. To find the time it takes to execute 100 instructions, you need to multiply the instruction latency by the number of instructions.

ii. Time to execute 100 instructions = instruction latency x number of instructions = 320ns x 100 = 32,000ns = 32 microseconds.

It takes 32 microseconds to execute 100 instructions on this non-pipelined machine with 6 execution stages.

It's important to note that non-pipelined machines are less efficient than pipelined machines because the execution of one instruction must be completed before the next instruction can be started. This makes the overall performance less efficient.

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27. Describe, with the aid of diagram, Flynn’s Classification of Computer Architecture.

Flynn’s classification of computer architecture is a framework for classifying computers based on the number of concurrent instruction streams and the number of concurrent data streams. It is represented using a diagram known as Flynn’s Taxonomy.

The diagram has two axes: the horizontal axis represents the number of concurrent instruction streams, and the vertical axis represents the number of concurrent data streams.

There are four main categories in Flynn’s Classification:

  1. Single Instruction Single Data (SISD): This category represents computers that execute one instruction at a time and operate on one data stream at a time. These computers are typically single-core, non-parallel computers such as early personal computers.

  2. Single Instruction Multiple Data (SIMD): This category represents computers that execute one instruction at a time but operate on multiple data streams simultaneously. These computers are typically used for vector processing and multimedia applications.

  3. Multiple Instruction Single Data (MISD): This category represents computers that execute multiple instructions at a time but operate on a single data stream. These computers are typically not found in practical use.

  4. Multiple Instruction Multiple Data (MIMD): This category represents computers that execute multiple instructions at a time and operate on multiple data streams simultaneously. These computers are typically used for parallel processing and high-performance computing.

In Flynn's Taxonomy, the SISD is at the bottom left, SIMD is at the bottom right, MISD is at the top left, and MIMD is at the top right.

It's important to note that the Flynn's Classification doesn't include the concept of superscalar architecture and pipelining, which can be implemented in SISD and SIMD architectures.

In summary, Flynn’s Classification of computer architecture is a framework for classifying computers based on the number of concurrent instruction streams and the number of concurrent data streams. It is represented using a diagram known as Flynn’s Taxonomy which has four main categories: SISD, SIMD, MISD and MIMD.

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28. What are the basic differences between a branch instruction, a call subroutine and program interrupt?

A branch instruction, a call subroutine and a program interrupt are all control flow instructions that change the flow of execution in a program, but they are different in their purpose and functionality.

  1. Branch instruction: A branch instruction is a type of instruction that allows the program to jump to a different location in memory to execute a different set of instructions. A branch instruction can be either conditional, where the branch is taken only if a certain condition is met, or unconditional, where the branch is always taken. The branch instruction is used to implement control structures like if-else, loops etc.

  2. Call Subroutine: A call subroutine instruction is used to call a specific subroutine or function in a program. The subroutine is a separate block of code that can perform a specific task. When the subroutine is called, the program execution is temporarily transferred to the subroutine and then returns to the main program after the subroutine is finished executing. This allows for code reuse and modularity.

  3. Program Interrupt: A program interrupt is an event that occurs when the computer's hardware or software generates a signal that interrupts the normal flow of execution. This can happen for a variety of reasons, such as a hardware error, a timer event, or a user input. Interrupts are used to handle events that require immediate attention, such as a key press, or a device request. The program execution is temporarily transferred to a specific interrupt handler routine, where the interrupt is serviced, and then returns to the main program.

In summary, a branch instruction allows the program to jump to a different location in memory to execute a different set of instructions. A call subroutine instruction is used to call a specific subroutine or function in a program, while a program interrupt is an event that occurs when the computer's hardware or software generates a signal that interrupts the normal flow of execution, allowing the system to handle external events.

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29. Write a program to evaluate the arithmetic statement  = − + ( ∗  − 4 ∗ ∗ ) +  i. Using a general register computer with 3 address constructions ii. Using a stack organized computer with zero address operation instructions.

i. Using a general register computer with 3 address constructions:

load A, R1 load B, R2 add R1, R2, R3 multiply R3, C, R4 load D, R5 multiply R4, R5, R6 subtract R6, E, R7 add R7, F, R8 store R8, Z

Explanation:

  • The program first loads the values of A, B, C, D, E, and F into separate registers (R1, R2, R3, R4, R5, and R6 respectively).
  • Then it performs the arithmetic operations as specified in the statement using 3-address instructions, storing the intermediate results in different registers (R3, R4, R5, R6, R7)
  • Finally, it stores the final result in the register R8 and store it in the memory location Z.

ii. Using a stack organized computer with zero address operation instructions:

load A load B add load C multiply load D multiply load E subtract load F add store Z

Explanation:

  • The program first loads the values of A, B, C, D, E, and F onto the stack.
  • Then it performs the arithmetic operations as specified in the statement using zero-address instructions, which operate on the top elements of the stack.
  • Finally, it stores the final result on the top of the stack into the memory location Z.

It's important to note that the program is written in a hypothetical assembly language, and the specific instructions and registers may vary depending on the specific architecture of the computer.

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29. Write a program to evaluate the arithmetic statement x = (-b + (b*b - 4*a*c))/ C + H i. Using a general register computer with 3 address constructions ii. Using a stack organized computer with zero address operation instructions.

i. Using a general register computer with 3 address constructions:

load b, R1 multiply R1, R1, R2 load a, R3 multiply R3, c, R4 multiply 4, R4, R5 subtract R2, R5, R6 add R6, -b, R7 load c, R8 divide R7, R8, R9 load h, R10 add R9, R10, R11 store R11, x

Explanation:

  • The program first loads the values of a, b, c, and h into separate registers (R1, R2, R3, R4).
  • Then it performs the arithmetic operations as specified in the statement using 3-address instructions, storing the intermediate results in different registers (R2, R5, R6, R7, R9, R11)
  • Finally, it stores the final result in the register R11 and store it in the memory location x.

ii. Using a stack organized computer with zero address operation instructions:

load b dup multiply load a load c multiply 4 multiply subtract -b add load c divide load h add store x

Explanation:

  • The program first loads the values of a, b, c, and h onto the stack.
  • Then it performs the arithmetic operations as specified in the statement using zero-address instructions, which operate on the top elements of the stack.
  • Finally, it stores the final result on the top of the stack into the memory location x.

It's important to note that the program is written in a hypothetical assembly language, and the specific instructions and registers may vary depending on the specific architecture of the computer.

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30. Differentiate in-order execution from out-of-order execution.

In-order execution and out-of-order execution are two different ways of processing instructions in a computer.

In-order execution is a method of processing instructions in the order in which they are received. The computer retrieves instructions from memory, processes them in sequence, and then stores the results back to memory. This method of execution is simple and easy to implement, but it can be slow because the computer has to wait for each instruction to be completed before it can move on to the next one.

Out-of-order execution, on the other hand, is a method of processing instructions in a different order than they were received. The computer retrieves instructions from memory, processes them as soon as they become available, and then stores the results back to memory. This method of execution is more complex and difficult to implement, but it can be faster because the computer can take advantage of the parallelism inherent in the instruction set.

In out-of-order execution, the CPU can reorder the instructions to optimize the pipeline and reduce pipeline stalls. The CPU uses a reorder buffer and a reservation station to hold the instructions and track their progress. The CPU also uses a scheduler to determine the order of instructions execution and a register renaming technique to avoid register contention.

In summary, In-order execution is a simple and easy method of processing instructions in the order in which they are received, while Out-of-order execution is a more complex method of processing instructions in a different order than they were received, which can take advantage of parallelism in the instruction set, to optimize the pipeline and reduce pipeline stalls.

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31. Write the three ways to improve MTTF.

Mean Time To Failure (MTTF) is a measure of the reliability of a system or component, and it can be improved in the following ways:

  1. Design improvements: By improving the design of a system or component, the MTTF can be increased. This can include using higher-quality materials, implementing better manufacturing processes, and incorporating more robust testing procedures.

  2. Maintenance and upkeep: Regular maintenance and upkeep of a system or component can help to extend its lifespan and improve its MTTF. This can include regular cleaning, calibration, and replacement of worn or damaged parts.

  3. Environmental controls: MTTF can be improved by controlling the environmental conditions that a system or component is exposed to. This can include controlling temperature, humidity, vibration, and other factors that can contribute to wear and tear on a system or component.

It's important to note that MTTF is a statistical measure, and it's not a guarantee that a device or system will not fail, but it's a way to measure the reliability of a system or component.