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F9.5 CPU performance

Logic and computer systems · GCSE · OCR J277 1.1.2, AQA 8525 3.4.5 · about 15 min

BugBotLab

What this lesson is about

Clock speed, cores and cache, measured with models.

Questions 5 marks in all

  1. [1 mark]What does a clock speed of 3 GHz mean?

    1. A3 billion clock pulses each second
    2. B3 million instructions in total
    3. C3 cores
    4. D3 GB of cache
    Answer: A. Hertz is per second; giga is a billion.

  2. [1 mark]Why might doubling the cores not halve the time a program takes?

    1. ASome of the work cannot be split across cores
    2. BCores share one clock pulse
    3. CMore cores use less cache
    4. DThe operating system can only use one core
    Answer: A. Steps that depend on the last answer must run one after another.

  3. [1 mark]Why does more cache usually make a CPU faster?

    1. AMore data is kept in fast memory, so fewer slow trips to RAM are needed
    2. BCache increases the clock speed
    3. CCache adds more cores
    4. DCache stores files when the power is off
    Answer: A. A cache hit avoids fetching from slower main memory.

  4. [1 mark]A job takes 60 seconds on 1 core and all of it can be split. How many seconds on 4 cores?

    Answer: 15. 60 ÷ 4.

  5. [1 mark]What is a drawback of a higher clock speed?

    1. AThe processor gets hotter and uses more power
    2. BIt needs more RAM
    3. CIt makes cache slower
    4. DIt reduces the number of cores
    Answer: A. Heat and battery life limit clock speed.

The task: the cache model

Use run_cache(requests, size) on the request list [5, 6, 5, 7, 5, 6, 8, 5, 6, 7], for cache sizes 1 to 6. Print one line for each size in the form size <n>: <hits> hits, then print best size: <n> for the smallest cache that gets the most hits, working it out with a loop.

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

def run_cache(requests, size):
    cache = []
    hits = 0
    for address in requests:
        if address in cache:
            hits = hits + 1
            cache.remove(address)
        elif len(cache) == size:
            cache.pop(0)
        cache.append(address)
    return hits, len(requests) - hits

requests = [5, 6, 5, 7, 5, 6, 8, 5, 6, 7]

The hint students can ask for: Call the model once for each size in the range, printing as you go. Track the best size by keeping it only when a size beats the best hits so far, so the smallest of equally good sizes wins.

A solution

from bugbot import *
connect()
def run_cache(requests, size):
    cache = []
    hits = 0
    for address in requests:
        if address in cache:
            hits = hits + 1
            cache.remove(address)
        elif len(cache) == size:
            cache.pop(0)
        cache.append(address)
    return hits, len(requests) - hits

requests = [5, 6, 5, 7, 5, 6, 8, 5, 6, 7]
best = 1
best_hits = -1
for size in range(1, 7):
    hits, misses = run_cache(requests, size)
    print(f"size {size}: {hits} hits")
    if hits > best_hits:
        best = size
        best_hits = hits
print("best size:", best)

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.