Question

We consider the long-term security of the Advanced Encryption Standard (AES) with a key length of 128-bit with respect to exhaustive key-search attacks.

The AES is perhaps the most widely-used symmetric cipher at this time. It got standardized after 4 years of scrutiny by the research community.

Assume that an attacker has a special purpose application specific integrated circuit (ASIC) which checks 5×10⁸ keys per second, and she has a budget of $1 million. One ASIC costs $50, and we assume 100% overhead for integrating the ASIC (manufacturing the printed circuit boards, power supply, cooling, etc.).
How many ASICs can we run in parallel with the given budget? How long does an average key search take? Relate this time to the age of the Universe, which is about 10¹⁰ years.

Answer 1

speed of palatalized machine = 5 x 10^ 12 key /second

The length of time to find a key in the average case = 1.08×10^18 years

Step-by-step explanation:

By the number specified ,each machine cost $100

$1,000,000 ÷ $100 = 10,000 machine

Each of our 10,000 machine check $5 * 10^8 per seconds , we can calculate the speed of our palatalized machine as;

5 x 10^8 x 10^4 = 5 x 10^ 12 key /second

On average, we’ll find the correct key halfway through our search of the key space (2^128), so the average case will be 2^127 checks.

To calculate the length of time to find a key in the average case:

2^127 keys / 5×10^12 keys/second ≈ 3.4×10^25 seconds

This comes out to:

1.08×10^18 years

This is approximately 10^8 times longer than the elapsed age of the universe.