GATE CS Mastery
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GATE CS 2019 · Computer Organization and Architecture

Question 55

NATPrinted as CS Q.452 marksNo negative markingMedium (estimate)

Question

A certain processor deploys a single-level cache. The cache block size is 8 words and the word size is 4 bytes. The memory system uses a 60-MHz clock. To service a cache miss, the memory controller first takes 1 cycle to accept the starting address of the block, it then takes 3 cycles to fetch all the eight words of the block, and finally transmits the words of the requested block at the rate of 1 word per cycle. The maximum bandwidth for the memory system when the program running on the processor issues a series of read operations is __________ ×106 bytes/sec\times 10^6\text{ bytes/sec}.

Official answer and solution
Official answer
160
From
Official GATE 2019 answer key

Solution

Calculate the total cycle time per cache block transfer and compute memory bandwidth in bytes per second.

We are given:

  • Cache block size = 8 words8\text{ words}
  • Word size = 4 bytes4\text{ bytes}, so each block is 8×4=32 bytes8 \times 4 = 32\text{ bytes}
  • Clock frequency = 60 MHz=60×106 cycles/sec60\text{ MHz} = 60 \times 10^6\text{ cycles/sec}

The cycles required to service one cache block read miss are:

  1. Accept starting address: 1 cycle1\text{ cycle}
  2. Fetch all 8 words from memory: 3 cycles3\text{ cycles}
  3. Transmit words at 1 word/cycle: 8×1=8 cycles8 \times 1 = 8\text{ cycles}
    Total cycles per block miss:
Tblock=1+3+8=12 cyclesT_{\text{block}} = 1 + 3 + 8 = 12\text{ cycles}

The rate at which blocks can be serviced is:

Rate=60×106 cycles/sec12 cycles/block=5×106 blocks/sec\text{Rate} = \frac{60 \times 10^6\text{ cycles/sec}}{12\text{ cycles/block}} = 5 \times 10^6\text{ blocks/sec}

The maximum bandwidth in bytes per second is:

Bandwidth=(5×106 blocks/sec)×(32 bytes/block)=160×106 bytes/sec\text{Bandwidth} = (5 \times 10^6\text{ blocks/sec}) \times (32\text{ bytes/block}) = 160 \times 10^6\text{ bytes/sec}

Thus, the required value is 160160.

Final answer

160

Common trap

Counting transmission as 1 cycle instead of 8 cycles (1 word per cycle for 8 words).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2019 Computer Science and Information Technology (CS)
Question
55 of 65, printed as CS Q.45
Section
Computer Science and Information Technology

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