1. Imagine a 4 Terabyte Disk Drive, formatted into 512 byte disk blocks. How many disk blocks will be on the disk, and how many bits are needed for block numbers (Each disk block needs a unique...

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1. Imagine a 4 Terabyte Disk Drive, formatted into 512 byte disk blocks. How many disk blocks will be on the disk, and how many bits are needed for block numbers (Each disk block needs a unique (integer) identifier - how many bits is it?) [This question has two answers] 2. How many disk block numbers can you store, tightly packed in a single disk block? [Hint: how many bits are in 512 bytes]. How many disk block numbers can you store in a disk block, assuming we don't overlap sector addresses in shared bytes? [This question has two answers, on one line.] 3. Now assume the disk is formatted into 4KB sectors. Assuming block addresses occupy a whole number of bytes, how many sector numbers can fit in a disk block? 4. Imagine a 10 GB file in our 4TB disk drive. How many disk blocks does the file occupy? 5. Imagine you want to access bytes starting at address 1 billion, and that these blocks holding data are allocated consecutively (like data in an array). You can calculate which disk block holds these bytes by dividing by 512 (very easy in binary) and using the quotient as the number of disk blocks beyond the first. Which component of disk access delay remains as a cost for this access? 6. You must know the maximum size of the file when it is created, and it's very expensive to shift bytes over if you want to insert data in the middle. So we consider each disk block having a disk address pointer to the next block. (Without using a FAT style structure), how long does it take to perform a random access fetch of data at location 1 billion, if each disk access is 5 msec? (Estimate, that is pretend that 512 is about 500) 7. So we put a linked list in memory mimicking the locations of the sectors holding the file (the FAT). Assume the CPU is 5 million times faster than the disk, for stepping to the next element of a linked list, How long will it take to randomly access data at location 1 billion As in Data Structures, contiguous arrays constrain insertion and deletion, and linked structures deny random access, so we go for trees. Since a binary tree (degree 2) is relatively deep, these trees are degree 100 or more. See Figure 5-11. (Also notice that Fig. 5-2(c) exploits the tree structure in the data to implementation tree structure for random access. For this question, estimate, and use KB, MB, GB, TB as appropriate, in your answers. A Single Indirection block names a sequence of blocks holding actual data. Assuming 512 byte sectors, how much data does a Single Indirect block encompass? How much data does a Double Indirect block encompass? How much data does a Triple Indirect block encompass? What is the maximum size of a file? [This questions has 4 answers] 8. How much data does a Single Indirect block encompass? How much data does a Triple Indirect Block encompass?
Answered Same DayApr 23, 2021

Answer To: 1. Imagine a 4 Terabyte Disk Drive, formatted into 512 byte disk blocks. How many disk blocks will...

Gopiraju answered on Apr 25 2021
138 Votes
1) Ans:-
Disk size is = 4 TB
Block size is = 512 Byte
No.of disk blocks = 4 TB/ 512 B
= 4*2^40/2^9
= 2^32
=> No.of disk blocks= 2^32.
=> No.of bits needed for identifying blocks is 32.
=> No.of disk blocks on the disk = 8G ( i.e.., 1G=2^30 ) Blocks.
2) Ans:-
No.of blocks can be stored in a disk block = 512*8/32
= 2^9*2^3/2^5
= 2^7
= 128
=> No.of blocks can be stored in a disk block = 128.

=> No.of bits need to identify unique block is 7
=> No.of block numbers store in one block = 512*8/7
= 585
So in one block we can store almost 585 block numbers.
3) Ans:-
Sector size = 4 KB
No.of Sectors = 4 TB/4 KB
= 4*2^40/4*2^10
= 1G
= 2^30 ( i.e.., 1G=2^30 )
=> No.of sectors = 2^30.
To identify one sector we need 30 bits.
No.of sectors...
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