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expeople1 [14]
3 years ago
6

Assume that the 100,000 element integer array that you allocated starts at address 0x50000000 in memory, the size of an integer

is 4 bytes and the D-cache is initially empty. As you read the integers in the array one-by-one, starting at index 0, how many D-cache misses will you see for reading the first 40 integers when the cache block size is (to be answered without running pin):a) 4 bytes:b) 32 bytes:c) 64 bytes:
Computers and Technology
1 answer:
Vanyuwa [196]3 years ago
6 0

Answer:

40, 5 and 3

See explaination for the details

Explanation:

A)

Consider the data.

Assume D-cache is empty

An integer size =4bytes

The cache block size =4bytes

Therefore, the number of D-cache misses for reading the first 40 integers is,

= (40×4)/4

= 160/4

= 40

b)

Consider the data.

Assume D-cache is empty

An integer size =4bytes

The cache block size =32bytes

Therefore, the number of D-cache misses for reading the first 40 integers is,

= (40×4)/32

= 160/32

= 5

c)

Consider the data.

Assume D-cache is empty

An integer size =4bytes

The cache block size =64bytes

Therefore, the number of D-cache misses for reading the first 40 integers is,

= (40×4)/64

= 160/64

= 3

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julsineya [31]

Answer:

The correct answer for the given question is option(b) i.e " No subject line"

Explanation:

In the e-mail subject is the important part which describe the information which kind of particular mail is send or received .if we do not give any subject line in the e-mail header then most of the chances that email will not read by recipients due to this suspicion will be caused.

Unknown Sender and Multiple recipients do not cause suspicion so the correct answer is option(b) i.e No subject line".

3 0
3 years ago
What would be the result of sending the string AAAAAAAAAAAAAAAAAAAA into a variable that has allocated space for 8 bytes
DerKrebs [107]

The result of sending the string above into a variable that has allocated space for 8 bytes is that there will be Buffer Overflow Attacks.

<h3>What occurs when a string of data  sent to buffer is large?</h3>

A buffer overflow is also known as buffer overrun. This  is known to be a process that occurs when or if  the size of data is known to be more than the storage capacity of any kind of memory buffer.

Note that due to the overload, the program that is trying to write the data to the buffer will therefore overwrites to the adjacent memory locations.

Learn more about string from

brainly.com/question/26427527

4 0
3 years ago
Which of the following events happened first?
horrorfan [7]

Answer:

D. ARPANET was developed

Explanation:

ARPANET is an acronym for Advanced Research Projects Agency Network and it was established in 1969 by the Advanced Research Projects Agency (ARPA) of the United States Department of Defense.

ARPANET was the first wide area packet switching network for transmitting electronic data and communications between computers on a single network with a distribution functionality. The TCP/IP protocol was first implemented on the Advanced Research Projects Agency Network (ARPANET), as well as some other standard protocols such as the NCP and the 1822 protocol.

The development of ARPANET happened first in the evolution of the internet.

E-mail is an acronym for electronic mail and it was invented by Ray Tomlinson in 1971; World Wide Web was created by Tim Berners-Lee in 1990; Web 2.0 evolved in 1999.

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3 years ago
A 1 MB digital file needs to transmit a channel with bandwidth of 10 MHz and the SNR is 10 dB. What is the minimum amount of tim
hodyreva [135]

Answer:

A 1 MB digital file needs 0.23 seconds to transfer over a channel with bandwidth 10 MHz and SNR 10 dB.

Explanation:

We can calculate the channel capacity using Shannon's Capacity formula:

C = B + log₂ (1 + SNR)

Where C = Channel Capacity

           B = Bandwidth of the Channel

           SNR = Signal to Noise Ratio

We are given SNR in dB so we need to convert it into a ratio.

SNR_{dB} = 10log₁₀ (SNR)

10 = 10log₁₀ (SNR)

1 = log₁₀ (SNR)

SNR = 10¹

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So, using Shannon Channel Capacity formula:

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C = 34.5 MHz = 34.5 Mb/s

Time required = 8Mb/34.5Mb/s = 0.23 seconds

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8 0
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ddd [48]
Indeed, this is to be true as software is to programs and applications(apps) hardware is to physical computers such as hard drive, SSD, motherboard, ram, graphics card, ect...
8 0
3 years ago
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