CASE II AziTech is considering the design of a new CPU for its new model of computer systems for 2021. It is considering choosing between two (2) CPU (CPUA and CPUB) implementations based on their performance. Both CPU are expected to have the same instruction set architecture. CPUA has a clock cycle time of 60 ns and CPUB has a clock cycle time of 75 ns. The same number of a particular instruction type is expected to be executed on both CPUs in order to determine which CPU would executes more instructions. CPUA is able to execute 2MB of instructions in 5*106 clock cycles. CPUB executes the same number of instructions in 3*106 clock cycles. a) Using the MIPS performance metric, which of the two (2) CPU should be selected to be implemented in the new computer system. Justify your choice. b) Compute the execution time for both CPUs. Which CPU is faster?

(a)The CPU B should be  selected for the new computer as it has a low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A.

(b) The CPU B is faster because it executes the same number of instruction in a lesser time than the CPU A .

Explanation:

Solution

(a)With regards to  the MIPS performance metric the CPU B should be chosen for the new computer as it low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A and when we look at the amount of process done by the system , the CPU B is faster when compared to other CPU and carries out same number of instruction in time.

The metric of response time for CPU B is lower than the CPU A and it has advantage over the other CPU and it has better amount as compared to CPU A, as CPU B is carrying out more execution is particular amount of time.

(b) The execution can be computed as follows:

Clock cycles taken for a program to finish * increased by the clock cycle time = the Clock cycles for a program * Clock cycle time

Thus

CPU A= 5*10^6 * 60*10^-9 →300*10^-3 →0.3 second (1 nano seconds =10^-9 second)

CPU B= 3 *10^6 * 75*10^-9 → 225*10^-3 → 0.225 second

Therefore,The CPU B is faster as it is executing the same number of instruction in a lesser time than the CPU A

answer: internal fragmentation is the difference between the (fixed) size of a memory hole and the data within in that memory. for example, if a 1,024 byte block of memory holds 800 bytes of data, then the 224 bytes difference is considered lost to internal fragmentation.

external fragmentation occurs when the sum of memory fragments is large enough to satisfy a request, but the holes are non-contiguous and the request cannot be satisfied. this can occur when memory is dynamically allocated and freed.

explanation:

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