It depends on the processor's register size, core frequence and some other stuff. Every instruction goes through a pipleline, if we are talking about an array of instructions. And for every instruction execution, these threads are passed and processed ALU (Arithmetic Logic Unit) etc. So there is no direct answer to this.
One cycle, in a computer, takes a time that is the reciprocal of the computer's clock rate - that which is expressed in MHz, or - more commonly nowadays - in GHz. Example: if your computer has a clock rate of 3 GHz, the time for a single cycle is (1 / 3 billion) seconds, i.e., 1/3 of a nanosecond, or 333 picoseconds.
It depends on your architecture but could be as little as 1 instruction, which on a 1 GHz processor would take roughly 1 nanosecond
That is an instruction and not a question that can be answered.
The bit time - the time it takes to transmit a single bit - is 1/10,000,000,000 second (1/10 nanosecond). Convert the 64 bytes into bits (just multiply by 8), and multiply this by the bit time. This gives you the time it takes to transmit the frame, from beginning to end. Then multiply this by the speed of the signal, which is typically about 68% of the speed of light in a vacuum, or roughly 200,000,000 m/s.
"Once" means a single time. He visited the zoo, but he only went there once.
The execution cycle.
One cycle, in a computer, takes a time that is the reciprocal of the computer's clock rate - that which is expressed in MHz, or - more commonly nowadays - in GHz. Example: if your computer has a clock rate of 3 GHz, the time for a single cycle is (1 / 3 billion) seconds, i.e., 1/3 of a nanosecond, or 333 picoseconds.
The time to execute a 3 clock cycle instruction in a 25MHz processor is 120ns. One clock cycle is 40ns, 1/25Mhz, so three of them are 120ns.
"SIMD, which stands for 'single instruction, multiple data,' is a process that allows the CPU to receive a single instruction and then execute it on multiple pieces of data rather than receiving the same instruction each time each piece of data is received."(Pg. 434, A+ Guide to Managing and Maintaining Your PC)
Timing Diagram is a graphical representation. It represents the execution time taken by each instruction in a graphical format. The execution time is represented in T-states.Instruction Cycle:The time required to execute an instruction is called instruction cycle.Machine Cycle:The time required to access the memory or input/output devices is called machine cycle.T-State:The machine cycle and instruction cycle takes multiple clock periods.A portion of an operation carried out in one system clock period is called as T-state.MACHINE CYCLES OF 8085:The 8085 microprocessor has 5 (seven) basic machine cycles. They areOpcode fetch cycle (4T)Memory read cycle (3 T)Memory write cycle (3 T)I/O read cycle (3 T)I/O write cycle (3 T)Each instruction of the 8085 processor consists of one to five machine cycles, i.e., when the 8085 processor executes an instruction, it will execute some of the machine cycles in a specific order.The processor takes a definite time to execute the machine cycles. The time taken by the processor to execute a machine cycle is expressed in T-states.One T-state is equal to the time period of the internal clock signal of the processor.The T-state starts at the falling edge of a clock.
Each instruction requires specific time for the execution of instruction and this time is called instruction cycle. Each instruction cycle consists 1 to 5 machine cycle -- opcode fetch, memory read, memory write, IO read, IO write and each machine cycle consist 3 to 6 T - states. Time required to execute 1 T-state = 1/ operating frequency of 8085 Microprocessor for example operating frequency = 2MHz then time required to execute 1 T-state = 0.5 uSec example: Calculate time required to execute instruction MOV C, A sol: This instruction has one machine cycle i.e. opcode fetch (In any instruction 1st cycle is always opcode fetch and opcode fetch consists 4 to 6 T state depend on the operation of particular instruction) so to execute MOV C, A required 4T states so time required to execute this instruction is 4*0.5usec = 2usec any other queries pls contect: nileshbahadure2000@yahoo.co.in example:Calculate the time required to execute LXI H,2000H sol:Here we have to draw opcode fetch and two memory reads as two bytes 00H and 20H have to be read from memory. i.e, opcode fetch+Memory reads *2(bytes address) =4+3+3 so to execute LXI H,2000H,the required T-states is 10T and time is 10*0.5usec=5usec
The time between a program starting and finishing. ie, the time it takes to execute the program.
The time it takes to retrieve an instruction and complete the respective command is known as instruction latency. This includes factors like the fetching of the instruction, decoding it, executing it, and storing the result. The overall latency can impact the performance of a computer system.
It is a 3 byte instruction, with one byte for opcode and the other two for the 16bit address. It takes four machine cycles (one to fetch opcode, one to fetch lower order address, one to fetch higher order address and another one to fetch the data from the memory)... i.e. it takes 13 time states to perform the LDA instruction
The system that execute the one process at the time and the next job when process is completed it is called the single process system
The time it takes to execute an action after identifying a hazard depends on the individual's reaction time, the complexity of the action required, and the nature of the hazard. In general, the quicker you can recognize a hazard and respond with the appropriate action, the more effective you will be in mitigating the risk.
In a 'single tasking' environment, only one task is given to the processor at a time & the processor takes millisecond to execute that task. For example, a student is writing a program which takes say, 10 minutes. During this time, the processor is sitting idle. When the program is given to the processor, it finishes its execution in a millisecond. This way, processor is wasting a lot of time which is a dawback in 'single tasking'. In a 'Multi tasking' environment, processor time is divided equally. For example, there are 4 tasks to execute. This is divided into 1/4 millisecond time for each job. This small part is called 'time slice'. Within this time slice, it will try to execute each job. If it can't, intermediate result will be stored in temporary memory. Once executing the fourth task, it will come back to the first task, in a circular manner. This is called 'round robin' method. The main difference is that, in single tasking, the processor time is wasted, but in multitasking, we can utilize the processor time in an optimum way.