performance of single-cycle processor - performance

Question

Given:
Operation time required by:
memory units: 200 ps
ALU and adders: 100 ps
Register file: 50 ps
other units and wires: no delay
Instruction mix and operation time in ps:
25% loads (600 ps)
10% stores ( 550 ps)
45% ALU instructions (400 ps)
15% branches (350 ps)
5% jumps (200 ps)
Every instruction executes in 1 clock cycle
Two implementations: fixed length and variable length
Which implementation would be faster and by how much?
Solution
Reference Table
Rule: CPU execution time: IC * CPI * CCT
Since CPI = 1...
CPU execution time: IC * CCT
My questions are:
What does it mean when an implementation has variable / fixed length?
How were the values for CPU execution timesingleclock calculated?

Answer 1

What does it mean when an implementation has variable / fixed length?
Fixed length clock means that each clock cycle has the same period, irrespective of the instruction being execution. Variable length clock means that different clock cycles may have different periods, depending on the instruction being executed.
So in a fixed clock design, the clock cycle has to be at least 600 ps, which is the longest time any instruction would take to execute (the load instruction). In a variable clock design, we can calculate the average clock cycle as follows:
Average CPU clock cycle = 600*25% + 550*10% + 400*45% + 350*15% + 200*5% = 447.5 ps
How were the values for CPU execution timesingleclock
calculated?
To determine which implementation is faster, you need to measure speedup, which is defined as:
Speedup = CPU execution time(single) / CPU execution time(variable)
Using the definition of CPU execution time we get (note that the number of instructions is the same):
Speedup = CPU execution time(single) / CPU execution time(variable)
= (Instruction count * Clock cycle time(single)) / (Instruction count * Clock cycle time(variable))
= Clock cycle time(single) / Clock cycle time(variable)
= 600 / 447.5 = 1.34
Se the variable clock design is 1.34 faster.
Regarding CPU execution timevariable
CPU execution timevariable is technically equal to the sum of the individual clock cycle times of each executed instruction. But we used the average clock cycle time instead to calculate speedup. Will we get the same result either way? Let's find out!
Assume there are N executed instructions and let C1, C2, ..., CN denote the cycle times of each of them, respectively. Hence:
CPU execution time(variable) = C1 + C2 + ... + CN
= 600*25%*N + 550*10%*N + 400*45%*N + 350*15%*N + 200*5%*N
= N * average CPU clock cycle
So they are the same.