Int sum = 0;
int item = 0;
do
{
item;
sum = item;
if (sum > 4)
break;
} while (item < 5);
You will generate a compile error, as having (item;) as a statement on its own is invalid.
T<span>he factors affecting the purchasing decision for dbms software are :
</span><span>a) Cost
b) DBMS features and tools
c) Underlying model
d) Portability
e) DBMS hardware requirements</span>
Answer 515be559d883f8eab00011e5
head is the tag at the top of your page containing your meta-tags, styles, scripts and title. headings are the h1, h2, h3 etc tags that allow you to size your text. a is an anker usable in the url for navigating to other pages or as in-page navigation.
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:
2^11
Explanation:
Physical Memory Size = 32 KB = 32 x 2^10 B
Virtual Address space = 216 B
Page size is always equal to frame size.
Page size = 16 B. Therefore, Frame size = 16 B
If there is a restriction, the number of bits is calculated like this:
number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]
where
physical memory size = 32KB which is the restriction
n bit machine = frame size = 16
Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11
