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mov r0, #0
Loop: cmp r0, #1000 <----+
beq Done |
instr 1 | 100 instructions
instr 2 |
... | Each instruction is store
add r0, r0, #1 | in 1 memory location (for simplicity)
b Loop <----+
Done: ...
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The loop is executed for 1000 times
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# instructions fetched = 1000 × 100
Time needed to fetch 1 instr = 50 nsec
Total time needed to fetch insr = 50 × 1000 × 100
= 5,000,000 nsec
Avg time = 5,000,000 / 100000 = 50 nsec per instruction
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First time throught the loop:
# instructions fetched = 1 × 100
Time needed to fetch 1 instr = 50 nsec (from memory)
Total time needed = 50 × 1 × 100
= 5000 nsec
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