Question

1) We have a word addressable 64MbByte main memory and a 128KByte cache. A word has 2 bytes, and a block has 16 words.
What would the address formats be if we used direct, associative or 2-way set associative?
2) For example, let’s say we have a single disk that rotates at 10,000 RPM and that there are 2048 sectors on each track, 4096 bytes per sector, and there are 8192 tracks. Assume a seek time of 2ms and an overhead of 1ms.
a) What’s the rotational latency?
b) How much data is on the disk?
c) How much time does it take to read 21 Mb from the disk?
3) Consider the following single-platter magnetic disk.
• Each sector is 1024 bytes and there are 2048 tracks.
• The seek time (including overhead) is 3 ms (thousands of a second).
• The rotational latency is 2.00 ms and the average access time (seek + overhead + latency + transfer time) for 64 Kbytes (from the same track) is 5.5 ms.
a) How fast does the disk spin in RPM (revolutions per minute)?
b) How many sectors are on each track?
c) How large is the disk in Mbytes?
4) Assume we are using 32 bit floating point representation with an 8 bit biased exponent field. With a simple diagram, explain the limitations of this format in terms of underflow and overflow. In general terms, what is the impact of expanding the exponent field by 2 ?

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We recall the formula for average disk read/write time. It is equal to the sum between: seek time, overhead, rotational latency and transfer time.
Seek time =2 ms
Overhead = 1ms
Rotational latency = 3ms
It is needed to compute the transfer time.
In one rotation are read 8 Mbytes. We need 21 Mbytes to read. This task is performed within 21/8 = 2.625 rotations.
The necessary time to perform 2.625 rotations = 2.625* 6ms = 15.75 ms...

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