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The instruction pipeline of a RISC processor has the following stages: Instruction Fetch $(IF)$, Instruction Decode $(ID)$, Operand Fetch $(OF)$, Perform Operation $(PO)$ and Writeback $(WB)$, The $IF$, $ID$, $OF$ and $WB$ stages take $1$ clock cycle each for every instruction. Consider a sequence of $100$ instructions. In the $PO$ stage, $40$ instructions take $3$ clock cycles each, $35$ instructions take $2$ clock cycles each, and the remaining $25$ instructions take $1$ clock cycle each. Assume that there are no data hazards and no control hazards.

The number of clock cycles required for completion of execution of the sequence of instruction is _____.
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Best answer
147 votes
147 votes
Total Instruction $= 100$
Number of stages $= 5$
In normal case total cycles $= 100 +5 -1 = 104$ cycles

Now, For PO stage $40$ instructions take $3$ cycles, $35$ take $2$ cycles and rest of the $25$ take $1$ cycle.
That means all other stages are perfectly fine and working with $CPI$ (Clock Cycle Per Instruction)$ = 1$

PO stage:
$40$ instructions take $3$ cycles i.e. these instructions are suffering from $2$ stall cycle,
$35$ instructions take $2$ cycles i.e. these instructions are suffering from $1$ stall cycle,
$25$ instructions take $1$ cycles i.e. these instructions are suffering from $0$ stall cycle,

So, extra cycle would be $40*2 + 35*1 + 25*0 = 80+35 = 115$ cycle.

Total cycles = $104 + 115 = 219$
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46 votes
46 votes
Total Instruction =100
Instruction Fetch (IF). Instruction Decode (ID). Operand Fetch (OF), and Write back (WB) performed in 1 cycle.
PO stage:
40 instructions takes 3 cycle
35 instructions take 2 cycles
25 instructions take 1 cycle
Average number of cycles required is = (40 x 3+35 x 2+25 x 1)/100 = 2.15 cycles.
On an average first instruction completed in 1+1+1+1+2.15 cycles = 6.15;
Remaining 99 instruction will takes 99 × 2.15 = 212.85 cycles
Total number of cycles is 6.15+212.85 = 219 cycles.
31 votes
31 votes
We have 100 instructions.We only focus on PO stages.Becoz of pipeline,instructions execute like this...

1st instruction takes (1+1+1+3+1)=7 Clock cycles.

2nd to 40th instructions PO stages takes=39*3 clock cycles.

41th to 75th instructions PO stages takes=35*2 clock cycles.

76th to 100th instruction PO stages takes=25*1 clock cycles.

So, total=7+(39*3)+(35*2)+(25*1).

219 clock cycles required.
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8 votes
8 votes
7 + 39×3 + 35×2 +25×1 = 219 clock cycle
Answer:

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