Question

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.

Answer #1

**(a)** As we know,

1 byte = 8 bits

1 TB = 2^{40} bytes

Disk capacity = 8 TB = (8 * 8 * 2^{40}) bits = 64 *
2^{40} bits

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

So, number of blocks = (64 * 2^{40})/64 = 2^{40}
blocks

So, we will need 40 bits to represent these 2^{40}
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 = 2^{40} bytes

Disk capacity = 8 TB = (8 * 8 * 2^{40}) bits = 64 *
2^{40} bits

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

So, number of blocks = (64 * 2^{40})/128 =
2^{39} blocks

So, we will need 39 bits to represent these 2^{39}
blocks.

So, these 39 bits will be used for counter.

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

**Please comment in case of any doubt.
Please upvote if this helps.**

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