Memory Interfacing – Solved PYQs
Summary
TLDRIn 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.
Outlines
💡 Introduction to Numerical Problems in Memory Interfacing
This paragraph introduces the session focused on numerical problems related to memory interfacing, which are commonly featured in competitive exams. The speaker emphasizes the importance of not only understanding theoretical concepts but also gaining practical experience. The session begins with a question about calculating the average access time of a CPU given the access times of cache and main memory and a hit ratio. Two approaches are discussed: one assuming simultaneous connection of cache and main memory to the processor, and the other considering a hierarchical memory system. The first approach calculates the average access time as 54 nanoseconds, while the second, which considers the access time of the previous level in case of a miss, results in 60 nanoseconds. The speaker notes that the second approach is correct for the question asked in the ISRO CS 2017 paper and advises students to look at the options first and then solve the problem, as the question usually refers to hierarchical organization unless specified otherwise.
🔍 Detailed Calculation of Average Read Access Time
The second paragraph delves into a specific numerical problem from the GATE CS 2015 paper, which involves calculating the average read access time given the time taken for a read request on a cache miss and a cache hit, along with a hit ratio. The speaker clarifies that the question explicitly mentions the level-wise organization, which should be considered for solving the problem. The calculation involves adding the access times for cache hits and misses, weighted by the hit ratio. The speaker advises to be careful during calculations, not to overthink, and to simply add the relevant times together. The final answer is calculated as 14 nanoseconds, and the speaker concludes the session by reminding viewers to be extra careful while solving numerical problems, especially during exams.
Mindmap
Keywords
💡Cache Memory
💡Main Memory
💡Access Time
💡Hit Ratio
💡Miss Ratio
💡Level-wise Organization
💡Read Request
💡Cache Hit
💡Cache Miss
💡Average Access Time
💡Competitive Examinations
Highlights
Introduction to types of numerical problems based on memory interfacing in competitive exams.
The importance of hands-on experience in addition to theoretical knowledge.
Question 1: Cache memory access time is 30 nanoseconds, main memory is 150 nanoseconds.
Assumption of simultaneous connection of cache and main memory to the processor.
Cache memory is smaller and faster than main memory.
Hit ratio of 80 percent (0.8) is given.
Average access time calculation: 0.8 * 30 + 0.2 * 150 nanoseconds.
Result of the first approach: 54 nanoseconds.
Second approach considers level-wise organization.
Access time includes cache and main memory in case of a miss.
Result of the second approach: 60 nanoseconds.
The second approach is the correct answer for the question asked in ISRO CS 2017.
Advice on smart work and considering options before answering.
Question 2: Read request takes 50 nanoseconds on a cache miss, 5 nanoseconds on a cache hit.
Observation: 80 percent of processor read requests result in cache hit.
Average read access time calculation: 0.8 * 5 + 0.2 * 50 nanoseconds.
Result of Question 2: 14 nanoseconds.
Emphasis on careful calculation and not overthinking during problem-solving.
Closing remarks and anticipation for the next session.
Transcripts
hello everyone
welcome to today's session today we will
be introduced to the types of the
numerical problems
based on memory interfacing that are
generally asked
in various competitive examinations
clearly
having only the lucid theoretical
conceptions are not enough
in order to concrete that knowledge
foundation we essentially need
hands-on experience in the applications
too so
let's get to learning let's begin with
the question number one
a cache memory needs an access time of
30 nanoseconds
and main memory 150 nanoseconds
what is the average access time of the
cpu assuming heat ratio is 80 percent
now there are two possible solutions to
this question
we will see them one by one first
approach
assume both the cachet and the main
memory are simultaneously connected to
the processor
the question states that the access time
of cache that is t
cache is 30 nanoseconds also tmm or the
access time of the main memory
is 150 nanoseconds now this is a
believable information
as the main memory is way bigger in size
than the cache
also due to its different organization
the access time
is bound to be greater than the cache
now coming to the hit ratio
it's already been mentioned in the
question itself 80
that is 0.8 now let's find out
the average access time usually termed
as t
average for this specific organization
h cache is the possibility of finding
the required word inside the cache
and t cache indicates the access time
now considering the chances of not
finding the word inside the cache that
is 1 minus h cache
that is the miss ratio of the cache we
will look for the word inside the main
memory
and tmm is the access time of the main
memory
substituting the values 0.8 into 30
plus 1 minus 0.8 into 150
and the unit remains the same
nanoseconds
which equates to 24 plus
0.2 into 150 which finally produces the
result as
54 nanoseconds now coming to the next
approach
i presume you already guessed it indeed
we are talking about the level wise
organization
and the only difference in this is the
way we calculate
the average memory access time here in
case of a miss the access time of the
previous level
is also considered due to the
organization so substituting the values
0.8 into 30 plus 1 minus 0.8
but instead of considering the access
time of the main memory only
we are to consider the t cache as well
because
not being able to find the needed word
inside the cachet
the processor opted for the main memory
level wise
which finally results in 60 nanoseconds
now this question was asked in isro cs
2017 paper
and these were the options judging the
options we can say
the second approach produces the correct
answer the examiners were looking for
so option a 60 is the correct choice for
this question
now you might be thinking having two
ways of solving the same question
is bit of a confusing thing but let me
tell you here you
need to do smart work look for the
options first and then
answer moreover if questions are asked
like this one
generally they refer to the level wise
organization unless mentioned otherwise
but worry not there are questions which
clearly mention the exact organization
which should consider
for problem solving let me read out the
question number two
assume that for a certain processor a
read request takes 50 nanoseconds on a
caching miss
and five nanoseconds on a cache hit
suppose
while running a program it was observed
that eighty percent of the processor rid
requests
result in cache hit we are to find out
the average read access time that 2 in
nanoseconds for this question
now it was asked in gate cs 2015 paper
and it's clearly talking about the level
was organization
how is so let's consider the access
times
a read request takes 50 nanoseconds
during cache miss
that means total time to access both the
cache
and the main memory must be 50
nanoseconds
because during cache hit it's only
taking 5 nanoseconds
here the question itself is specifying
the organization to be considered
and regarding hit ratio it's 80 percent
that is 0.8 similar to the previous one
now let's figure out the average read
access time
or t average substituting the values
0.8 into 5 plus 1 minus 0.8
into 50 nanoseconds one more thing
during calculations
be a bit more careful don't overthink
and just add
five more to this 50. here 50 is t
cash a plus tmm altogether so finally we
will have the answer as
14 nanoseconds all alright folks
that was all for this session remember
always be extra careful
while solving the numerical problems
especially during examinations
so i hope to see you in the next one
thank you all for watching
[Music]
you
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