Question

CTR Mode (a) Suppose that we are using the 3-DES in CTR mode for encrypting 8TB...

CTR Mode

(a) Suppose that we are using the 3-DES in CTR mode for encrypting 8TB of capacity. (1TB, terabyte, has 240 bytes) What is the maximum length of the IV which can be used? Justify your answer.

(b) We are using the AES in CTR mode for encrypting a disk with 8TB capacity. What is the maximum length of the IV? Justify your answer.

Homework Answers

Answer #1

(a) As we know,

1 byte = 8 bits

1 TB = 240 bytes

Disk capacity = 8 TB = (8 * 8 * 240) bits = 64 * 240 bits

As we know, the block size in 3-DES = 64 bits [Assuming the default value]

So, number of blocks = (64 * 240)/64 = 240 blocks

So, we will need 40 bits to represent these 240 blocks.

So, these 40 bits will be used for counter.

So, maximum length of the IV = (64 - 40) = 24 bits

(b) As we know,

1 byte = 8 bits

1 TB = 240 bytes

Disk capacity = 8 TB = (8 * 8 * 240) bits = 64 * 240 bits

As we know, the block size in AES = 128 bits [Assuming the default value]

So, number of blocks = (64 * 240)/128 = 239 blocks

So, we will need 39 bits to represent these 239 blocks.

So, these 39 bits will be used for counter.

So, maximum length of the IV = (128 - 39) = 89 bits

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