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COE 768 Sample Midterm 1

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Sample Midterm 1 The following partial specifications are given: Question 1.1 To determine the miss penalty (time to replace the block in cache) for the worst-case scenario (D-bit for the block to be replaced always = ) calculate the following: • Bus clock period • Bus transfer time • Block replacement time (if D-bit = 1) • Miss penalty

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Sample Midterm 1
The following partial specifications are given:




Question 1.1

To determine the miss penalty (time to replace the block in cache) for the worst-case sce
block to be replaced always = 1) calculate the following:

• Bus clock period

• Bus transfer time

• Block replacement time (if D-bit = 1)

• Miss penalty

This question is just a matter of knowing the formula.

• The bus clock period is the reciprocal of the bus clock frequency.

1 1
tbus = = 6
≈ 7.5 × 10−9 s or 7.5 ns
fbus 133.33 × 10 Hz

Make sure you know the SI conversions, most notably for mega (M) and nano (n), as these will be those yo
tests.

• The block transfer time is the product of the number of words being transferred and the bus clock period.

Tbt = # of words × tbus = 32 words × 7.5 ns = 240 ns

• The block replacement time is given by the following formula.

, This is because it allows two data-words during one clock cycle period, so the block transfer gets cut in half. The r
the same.

Tmiss = Tbr = (80 ns + 120 ns)(2) + 10 ns = 410 ns

The system with DDR-SDRAM main memory has a miss penalty of 410 ns.

Question 1.3
Taking in account that average access time for the cache is equal to: Hit rate × Hit time +
penalty. Find the best variant of systems organization (with minimum average data acces
following options:

1. Block size = 32 words and Hit rate = 98% & SDRAM based main memory

2. Block size = 16 words and Hit rate = 96% & SDRAM based main memory

3. Block size = 32 words and Hit rate = 98% & DDR-SDRAM based main memory

4. Block size = 16 words and Hit rate = 96% & DDR-SDRAM based main memory

Calculate the value for the average data access time for each option. Fill in the Table 1.1
variant with min{Tav }.

For this question, we are already given the formula for average access time, which is:

Tav = (Rhit ⋅ Thit ) + (Rmiss ⋅ Tmiss )

The miss rate is given by Rmiss = 1 − Rhit .

Tav = (Rhit ⋅ Thit ) + ((1 − Rhit ) ⋅ Tmiss )

We already know how to calculate Tmiss from the previous question, which is:

Tmiss = Tbr + Thit

Substituting the formula for block replacement Tbr and block transfer Tbt :

• For SDRAM based main memory:

Tmiss = (Taddr + # of words × tbus ) (2) + Thit

• For DDR-SDRAM based main memory:

# of words × tbus
Tmiss = (Taddr + ) (2) + Thit
2

Using the formula for:

• Option 1

Tmiss = (80 ns + (32 words × 7.5 ns)) (2) + 10 ns = 650 ns

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