Alternative 1:A small D-cache with a hit rate of 94% and a hit access time of 1 cycle (assume that no additional cycles on top of the baseline CPI are added to the execution on a cache hit in this case).Alternative 2: A larger D-cache with a hit rate of 98% and the hit access time of 2 cycles (assume that every memory instruction that hits into the cache adds one additional cycle on top of the baseline CPI). a)[10%] Estimate the CPI metric for both of these designs and determine which of these two designsprovides better performance. Explain your answers!CPI = # Cycles / # InsnLet X = # InsnCPI = # Cycles / XAlternative 1:# Cycles = 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150)CPI= 0.50*X*2 + 0.50*X(0.94*2 + 0.06*150) / X1= X(0.50*2 + 0.50(0.94*2 + 0.06*150) ) / X= 0.50*2 + 0.50(0.94*2 + 0.06*150)= 6.44Alternative 2:# Cycles = 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150)CPI= 0.50*X*2 + 0.50*X(0.98*(2+1) + 0.02*150) / X2= X(0.50*2 + 0.50(0.98*(2+1) + 0.02*150)) / X= 0.50*2 + 0.50(0.98*(2+1) + 0.02*150)= 3.97Alternative 2 has a lower CPI, therefore Alternative 2 provides better performance.
Answer:
True
Explanation:
If a module or code is not ready then the unit test can use the Stubs to simulate a called upon function to test the process. On the other hand if the main unit is not ready the test can use Drivers to simulate the calling of said function to test the rest of the modules.
Therefore, a unit test will use either Drivers or Stubs at a given moment for testing but not both simultaneously.
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17, 12.5, 6, and 10. HOPE THIS HELPS
Answer:
0.
Explanation:
Given
int x;
x=3/(int)(4.5+6.4)
Required
What is x?
The first line of the code segment declares x as integer. This means that, it will only hold non decimal numbers.
With the above explanation, options (a), (c) and (d) can not be true.
Solving further:
x=3/(int)(4.5+6.4)
The computer evaluates the denominator as:
x=3/(int)(10.9)
The denominator is then converted to an integer. So, we have:
x = 3/10;
3/10 = 0.3 but
Recall that: <em>x will only hold non decimal numbers.</em>
So:
x = 0;
