Question

Question 12

Consider a magnetic disk drive with 10 double sided platters, 512(=2^9) tracks per surface, and 1024 (=2^10) sectors per track. Sector size is 2KBytes (=2^10 Bytes ). Assume the average seek time 0.02 ms, the seek time from a track to its adjacent track can be ignored. The drive rotates at 5400 rpm. Successive tracks in a cylinder can be read without head movement. Furthermore, disks can be handled independently, so multiple tracks in a cylinder can be read at the same time. Estimate the time required to transfer a 5-MB (=2^25 Bytes) file. If the file is stored sequentially.  Provide details regarding how you get the answer. (Data is stored at sequentially sectors on a track and then sequential tracks on a cylinder, if a cylinder capacity is not enough, it will store sequentially in the same manner on the adjacent cylinder.)

Hint:

1. Determine how many cylinder needs to read
2. Determine how long it takes to read a cylinder (seek time+rotational delay+transfer time)
3. Based on the previous result get the time needed

Question 13 (10 points)

For a given computer system, the main memory is 4096 Mbyte (2^32 byte); word size is 8 bytes (2^3 bytes); block size (aslo the cache line size) is 128 bytes (2^7 bytes); cache size is 128 Kbytes (2^17 bytes). For the hexadecimal main memory addresses FFFF6666, show the following information: Tag, set and word offset for a eight-way set-associative cache(eight-way set means: each set can hold 8 lines) . Provide details regarding how you get the answer. Show in binary, no conversion to hex is needed.

Answer 1

Answer 12). The following details are given:

1. No. of platters = 10 (doubles sided)
2. No. of tracks per surface = 2⁹
3. No. of sectors per track = 2K Bytes which is given in braces as  2¹⁰, so considering 2¹⁰ sector/track
4. sector size is 2¹⁰ Bytes
5. Average seek time = 0.02 ms
6. Rotational Speed of disk = 5400 rpm

Now, we need to find time taken to transfer a 5-MB as 2²⁵ Bytes of data.

5400 revolution in 60 sec

1 revolution in 60/5400 = 1/90 sec

In one revolution it will cover entire track, 2¹⁰ sector

In one revolution total data it will cover = 2¹⁰ * 2¹⁰   Bytes = 2²⁰Bytes

Data transfer rate = 2²⁰/ (1/90) = 90 * 2²⁰ Bytes/sec

Only transfer for 2²⁵ Bytes of data, time consumed will be 2²⁵ / (90 * 2²⁰) = 0.35555 sec = 355.555 ms

Average seek time = 0.02 msec

Average rotational delay = (1/2) * 60/5400 = 0.055555 sec = 55.555 ms

Effective total time taken will be = Tranfer rate + seek time + rotational delay

= 355.555 + 0.02 + 55.555 = 411.13 ms

To find the number of cylinder need, we need to find total number of track required to read 2²⁵ Bytes of data.

Since 1 track has 2²⁰ Bytes, therefore cylinder required = 2²⁵/2²⁰ = 32

Answer 13). Given

1. Main memory size = 2³² Bytes
2. Word size. = 2³ Bytes
3. Block size = 2⁷ Bytes
4. Cache size = 2¹⁷ Bytes

Also it is 8-way set associativity

Main memory has 32 bytes that will take 32 bit to represent a instruction  

Tag

Set Bit

Word Offset

<--------------------------------- 32 bit --------------------------------------------------------------------->

We first to need number of cache line, which will be calculated as

No. of cache line. = Cache size / Block size = 2¹⁷ / 2⁷ = 2¹⁰

No of set = Total cache line / P-way set associative = 2¹⁰ / 8 = 2⁷ which will be represented by 7 bit.

Also it is given that word size is 8 bytes, means 3 bit will take to consider word offset address.

therefore,

instruction will be seems like as

Tag bit

Set bit ( 7 bit)

Word offset (3 bit)

<------------------------- 32 bit ------------------------------------------------------------------------------>

Now tag bit can be calculated as 32 bit - 7 bit - 3 bit = 22 bit

Given hexadecimal main memory address FFFF6666, in binary as

1111 1111 1111 1111 0110 0110 0110

and this instruction will be as

1111 1111 1111 0110 01

10 01101

110

<-------Tag bit -------------------------------------------------> <---- Set bit---------------><-- word offset bit-->

For the given memory address,

Tag position is (1111 1111 1111 0110 01)₂

Set position (10 01101 )₂

Word offset position is (110)₂

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