Pipelining improves CPU performance due to
In pipelining, a number of functional units are employed in sequence to perform a single computation.
An instruction cycle refers to
An instruction cycle is the basic operational process of a computer. It is the process by which a computer retrieves a program instruction from its memory, determines what actions the instruction dictates and carries out those actions. Also called as fetch, decode-execute cycle.
Given a 5 stage pipeline with stages taking 1,2, 3, 1, 1 units of time, the clock period of the pipeline is
Clock period of pipeline = Maximum delay of stages
= Max ( 1, 2, 3, 1, 1) = 3
Which of following registers processor used for fetch and execute operations?
1. Program counter
2. Instruction register
3. Address register
μ-instruction for fetch cycle:
So, PC, IR and address register are used.
A ________ is required to translate such microprogram into executable programs that can be stored in the control memory in microprogramming.
It is the definition of microassembler.
Which of the following statements is false about CISC architectures?
In RISC instruction set all arithmetic/logic instructions must be register-based but not in CISC.
The register which holds the address of the location to or from which data are to be transferred is known as
MAR is a register that holds the address of location to or from which data are to be transferred.
Consider a case where 4-segment pipeline with a clock cycle time 20 ns in each sub operation to execute 100 tasks. Assume that a non pipeline unit that can perform the same operation. Pipelined system will take how much time to complete the task?
For pipeline with K segment and n instruction
Time (pipeline ) = ( K + n - i ) tP
= ( 4 + 100 - 1 ) x 20 nsec
= (4 + 99) x 20 nsec
= (103) x 20 nsec = 2060 nsec
Find out the speed-up ratio between pipelined and non-pipelined system?
The needed for non pipelined processor
Speedup = 8000/2060 = 3.88
Assume that the time required for the eight functional units, which operate in each of the eight cycles, are as follows 5 ns, 8 ns, 6 ns, 10 ns, 15 ns, 12 ns, 6 ns, 8 ns. Assume that pipe lining adds 1 ns of overhead. Find the speedup versus the single cycle data path.
Since the unpipelined machine executes all instructions in a single clock cycle, its average time per instruction is simply clock time. The dock is equal to the sum of the times for each step in the execution.
∴ Average instructions execution time
= 5 + 8 + 6 + 1 0 + 15 + 12 + 6 + 8
The dock cycle time on the pipelined machine must be the largest time for any stage in the pipeline (15 ns) + the overhead of 1 ns for a total of 16 ns.
∴ Speed from pipelining