Please show the specific calculation !!!!!!! 11. (21) Two CPU manufacturers (XYZ
ID: 3741945 • Letter: P
Question
Please show the specific calculation !!!!!!!
11. (21) Two CPU manufacturers (XYZ and ABC) implemented the same ISA (instruction set architecture). They each have their own datapath and control unit (multi-cycle) designs. Let us assume that there are only three kinds of instructions (A, B, and C) here XYZ 100 MHz 4 ABC 50 MHz Clock Freque CPI A CPI B CPI C 4 11.1 (3) Consider each instruction type and find out which CPU executes that instruction faster (i)Instr. A a.) Both are sameb.) XYZ is fasterc.) ABC is faster (ii)Instr. Ba.) Both are sameb.) XYZ is fasterc.) ABC is faster iii)Instr. C a.) Both are same b.) XYZ is fasterc.) ABC is faster 11.2 There is a standard third party compiler, and also there are special compilers produced by XYZ and ABC. The compilers use the A, B, and C instructions in the percentages shown below. Assume that the instruction counts of binaries produced by the three compilers are all the same. Notice that the special compilers are designed to make the competitor look relatively bad Standard Compiler 50% 30% 20% Special for XYZ 30% 50% 20% Special for ABC 30% 20% 50%Explanation / Answer
11.1:
Performance of CPU is measured as
Performance = Tasks/second = (Tasks/instruction) x (Instructions/cycle) x (Cycles/Second)
Since it is given that they have implemented the same ISA , so (Tasks/Instruction) will be same for both CPU for a given Instruction.
It is also given that for ABC Cyles/second = 50 MHz and for XYZ = 100 MHz
Now For instruction A
IN CPU ABC = (Instructions/cycle) x (Cycles/Second)
= (1/2) x 50 Mhz= 25 MHz
IN CPU XYZ = (Instructions/cycle) x (Cycles/Second)
=(1/4) x 100 Mhz= 25 MHz
Hence the performance of Instruction A will be same in both CPU
So the answer of 11.1 .(i) (a) Both are same
Now For instruction B
IN CPU ABC = (Instructions/cycle) x (Cycles/Second)
= (1/4) x 50 Mhz= 12.5 MHz
IN CPU XYZ = (Instructions/cycle) x (Cycles/Second)
=(1/6) x 100 Mhz= 16.67 MHz
Hence the performance of Instruction A will be better in CPU XYZ
So the answer of 11.1 .(ii) (b) XYZ is faster
Now For instruction C
IN CPU ABC = (Instructions/cycle) x (Cycles/Second)
= (1/3) x 50 Mhz= 16.33 MHz
IN CPU XYZ = (Instructions/cycle) x (Cycles/Second)
=(1/8) x 100 Mhz= 12.5 MHz
Hence the performance of Instruction A will be better in CPU ABC
So the answer of 11.1 .(iii) (C) Better in ABC
11.2:
First we normalize the CPI of XYZ and ABC
so we have :
XYZ ABC
clock frquency 100mhz 50 mhz 50 mhz
CPI A 2 2
CPI B 3 4
CPI C 4 3
so
11.2.1 Using the binary produced by its special compiler (and running it on both the CPUs), XYZ claims to be faster than ABC by a factor of 1.1
11.2.2 Using the binary produced by its special compiler (and running it on both the CPUs), XYZ claims to be faster than ABC by a factor of 1.1
11.2.3 (a) XYZ as there will be more cycles per instruction for XYZ
11.2.4 (a)Standard Compiler as there will be more cylces per instruction for standard compiler
11.2.5 (a)Standard Compiler as there will be more cylces per instruction for standard compiler