Computer System Architecture - Pipeline and Vector Processing

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6. We have two designs D1 and D2 for a synchronous pipeline processor. D1 has 5 pipeline stages with execution times of 3 nsec, 2 nsec, 4 nsec and 3 nsec while the design D2 has 8 pipeline stages each with 2 nsec execution time. How much time can be saved using design D2 over design D1 for executing 100 instructions?

  • Option : B
  • Explanation : Total execution time = (k + n – 1) * maximum clock cycle
    Where k = total number of stages and n = total number of instructions
    For D1 :
    k = 5 and n = 100
    Maximum clock cycle = 4ns
    Total execution time = (5 + 100 – 1) * 4 = 416
    For D2 :
    k = 8 and n = 100
    Each clock cycle = 2ns Total execution time = (8 + 100 – 1) * 2 = 214
    Thus, time saved using D2 over D1
    = 416 – 214
    = 202
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7. A nonpipelined single cycle processor operating at 100 MHz is converted into a synchronous pipelined processor with five stages requiring 2.5 nsec, 1.5 nsec, 2 nsec, 1.5 nsec and 2.5 nsec, respectively. The delay of the latches is 0.5 nsec. The speedup of the pipeline processor for a large number of instructions is

  • Option : C
  • Explanation : For non pipelined system time required
    = 2.5 + 1.5 + 2.0 + 1.5 + 2.5 = 10
    for pipelined system
    = Max (stage delay) + Max(Latch delay)
    = 2.5 + 0.5 = 3.0
    speedup = time in non-pipelined system/time in pipelined system
    = 10/3 = 3.33
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8. Which of the following are NOT true in a pipelines processor?
1. Bypassing can handle all Raw hazards.
2. Register renaming can eliminate all register carried WAR hazards.
3. Control hazard penalties can be eliminated by dynamic branch prediction.

  • Option : A
  • Explanation : In pipelined processor bypassing can handle all RAW hazards and register remaining can eliminate all register carried WAR hazard.
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Delayed branching can help in the handling of control hazardous

9. For all delayed conditional branch instructions, irrespective of whether the condition evaluates to true or false, A

  • Option : B
  • Explanation : For all delayed conditional branch instructions the first instruction in the fall through path is executed
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Delayed branching can help in the handling of control hazardous

10. The following code is to run on a pipelines processor with one branch delay slot
I1 : ADDR2 ⇽ R7 + R8
I2 : SIN R4 ⇽ R5 – R6x
I3 : ADD R1 ⇽ R2 + R3
I4 : STORE Memory [R4] ⇽ R1
BRANCH to Label if R1 == 0
Which of the instruction 11, 12, 13 or 14 can legitimately occupy the delay slot without any other program modification?

  • Option : B
  • Explanation : In delay slot I2 will be there
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