Memory Interfacing – Solved PYQs

Summary

In this educational session, the presenter introduces numerical problems related to memory interfacing, crucial for competitive exams. The session focuses on calculating average access times for cache and main memory, emphasizing practical application over theoretical knowledge. Two methods are discussed for solving such problems, with the second method, considering level-wise organization, being the correct approach as seen in an ISRO CS 2017 exam question. The presenter also covers a GATE CS 2015 question, illustrating the importance of understanding cache hit and miss scenarios. The session concludes with advice on careful problem-solving during exams.

Takeaways

• 😀 The session introduces types of numerical problems related to memory interfacing in competitive exams.
• 🔍 It's emphasized that theoretical knowledge needs to be complemented with practical application.
• ⏱️ Question 1 discusses calculating the average access time of CPU with given cache and main memory access times and a hit ratio.
• 💡 Two approaches are presented for solving the problem: simultaneous connection and level-wise organization.
• 📊 The first approach assumes simultaneous access, leading to an average access time of 54 nanoseconds.
• 📈 The second approach, considering level-wise organization, results in an average access time of 60 nanoseconds.
• 🎯 The correct answer for the competitive exam (ISRO CS 2017) is the second approach, with 60 nanoseconds.
• 🤔 The importance of understanding the organization type (level-wise or simultaneous) is highlighted for problem-solving.
• 📝 Question 2 from GATE CS 2015 involves calculating average read access time with given cache hit and miss times.
• 🧮 The average read access time for Question 2 is calculated as 14 nanoseconds, using the level-wise organization method.
• 📚 The session concludes with advice to be careful while solving numerical problems, especially during exams.

Q & A

What is the average access time of the CPU for a cache memory with 30 nanoseconds access time and main memory with 150 nanoseconds access time, given a hit ratio of 80 percent?

The average access time can be calculated using two approaches. In the first approach, assuming simultaneous connection to the processor, the average access time is calculated as 0.8 * 30ns + (1 - 0.8) * 150ns, which equals 54 nanoseconds. In the second approach, considering level-wise organization, the average access time is calculated as 0.8 * 30ns + (1 - 0.8) * (30ns + 150ns), which equals 60 nanoseconds.

What is the significance of the hit ratio in calculating the average access time of a CPU?

The hit ratio, which is the probability of finding the required data in the cache, plays a crucial role in determining the average access time of a CPU. A higher hit ratio means more data is found in the cache, reducing the need to access the slower main memory, thus improving the average access time.

How does the size and organization of cache memory affect its access time compared to main memory?

Cache memory is typically smaller and more quickly accessed than main memory due to its closer proximity to the processor and optimized organization. This results in a faster access time for cache memory compared to main memory.

What is the difference between the two approaches to calculating the average access time as described in the script?

The first approach assumes that cache and main memory are simultaneously connected to the processor, while the second approach considers the level-wise organization where the cache access time is added to the main memory access time in case of a miss. The second approach results in a higher average access time due to this additional consideration.

Why is it important to consider the organization of memory when calculating average access time?

The organization of memory, whether it is simultaneous or level-wise, affects how the processor accesses data. This, in turn, influences the average access time calculation, as different organizations have different implications for how data is retrieved in case of a cache miss.

In the context of the script, what does 't_cache' and 't_mm' represent?

In the script, 't_cache' represents the access time of the cache memory, and 't_mm' represents the access time of the main memory.

What was the question asked in the ISRO CS 2017 paper regarding cache memory and main memory access times?

The question in the ISRO CS 2017 paper asked for the average access time of the CPU given a cache memory access time of 30 nanoseconds, main memory access time of 150 nanoseconds, and a hit ratio of 80 percent.

How does the level-wise organization of memory affect the calculation of average read access time?

In a level-wise organization, the average read access time includes the access time of the cache and the main memory in case of a miss. This is because the processor must access both levels of memory to retrieve the required data.

What was the average read access time for a certain processor as discussed in the script, given a cache hit time of 5 nanoseconds and a cache miss time of 50 nanoseconds with an 80 percent hit ratio?

The average read access time for the given scenario is calculated as 0.8 * 5ns + (1 - 0.8) * 50ns, which equals 14 nanoseconds.

Why is it advised to look at the options first when solving numerical problems in competitive examinations as suggested in the script?

Looking at the options first can provide clues about the expected answer format and helps in selecting the correct approach to solve the problem, especially when there are multiple ways to calculate the same quantity.