GATE CS 2024 Set 1 — Question 56

NAT+2 / -0MediumCache NumericalsMemory Hierarchy & CacheComputer Organization & Architecture

Computer Organization & Architecture → Memory Hierarchy & Cache → Cache Numericals

Last updated 20 August 2026

Question

A given program has 25% load/store instructions. Suppose the ideal CPI (cycles per instruction) without any memory stalls is 2. The program exhibits 2% miss rate on instruction cache and 8% miss rate on data cache. The miss penalty is 100 cycles. The speedup (rounded off to two decimal places) achieved with a perfect cache (i.e., with NO data or instruction cache misses) is _________

Correct answer

3 to 3

Solution

To calculate the speedup achieved with a perfect cache, we compare the execution time (or CPI) of the program with the actual cache to the execution time with a perfect cache.

1. Identify Given Parameters:

Ideal CPI ( $CPI_{ideal}$ ) = 2
Percentage of load/store instructions = 25% = 0.25
Instruction cache miss rate ( $m_i$ ) = 2% = 0.02
Data cache miss rate ( $m_d$ ) = 8% = 0.08
Miss penalty ( $P$ ) = 100 cycles

2. Calculate Stalls per Instruction for the Actual Cache:

Instruction Fetch Stalls: Every instruction requires a fetch from the instruction cache.

\text{Stalls}_{I} = 1 \times m_i \times P = 1 \times 0.02 \times 100 = 2 \text{ cycles}

Data Access Stalls: Only load/store instructions access the data cache.

\text{Stalls}_{D} = 0.25 \times m_d \times P = 0.25 \times 0.08 \times 100 = 2 \text{ cycles}

Total Stalls per Instruction:

\text{Total Stalls} = \text{Stalls}_{I} + \text{Stalls}_{D} = 2 + 2 = 4 \text{ cycles}

3. Calculate Actual CPI ( $CPI_{actual}$ ):

CPI_{actual} = CPI_{ideal} + \text{Total Stalls} = 2 + 4 = 6

4. Calculate Perfect Cache CPI ( $CPI_{perfect}$ ):
A perfect cache has a 0% miss rate, meaning there are no memory stalls.

CPI_{perfect} = CPI_{ideal} = 2

5. Calculate Speedup:

\text{Speedup} = \frac{\text{Execution Time}_{actual}}{\text{Execution Time}_{perfect}} = \frac{CPI_{actual}}{CPI_{perfect}}

\text{Speedup} = \frac{6}{2} = 3.00

The speedup achieved with a perfect cache is 3.00.

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