The number of instructions a program has is the static instruction count. The number of instructions actually executed (e.g. due to some instructions being repeated, as in a loop) is called the dynamic instruction count. Consider the following code:
li $t0, 99
li $t1, 44
loop: addi $t0, $t0, -1
lw $t1, 0($t1)
bne $t0, $zero, loop
add $v0, $t0, $zero
a. What is the static instruction count of this code? What is the dynamic instruction count? (Assume any pseudoinstructions are counted as 1 instruction).
b. How many times is the instruction memory accessed? How many times is the data memory accessed?
c. Assuming this runs on a single cycle processor (with CPI = 1) running at 100 KHz, how long will the code take to execute?
Execution Time = (CPI * IC) / Clock Rate
The Attempt at a Solution
a. Static Instruction Count: 6 (Since there are only 6 instructions in the code.)
Dynamic Instruction Count: (99 * 3) - 1 + 3 = 299 (The two li instructions and the add instruction only executed once. The loop itself executes 99 times but bne instructions executes only 98 times since at the 99th iteration, $t0 = 0 and bne instruction won't executed.)
b. Instruction Memory Accessed: 299 (Number of times program counter updated)
Data Memory Accessed: 99 (Number of times lw or load instruction was executed)
c. Given CPI = 1, IC = 299 (from part a), Clock Rate = 100000 (Converted KHz to Hz)
Execution Time = (1 * 299) / 100000 = 2.99 x 10-3 milliseconds
I am not sure if I calculated the Dynamic Instruction Count correctly and if the Instruction Memory Accessed is correct or not. Any input would greatly appreicated.