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3 years ago

You discover memory is corrupted, what would be an indication of a software vs. a hardware issue?

1 answer:

8 0

Hey there!

Memory (RAM) is considered to be hardware since it is a physical component that makes up the computer and is accessed through the CPU rather than internal code that makes the computer run (which is software). Therefore, if you are having issues with the memory, it is likely a problem with a memory chip, making it a hardware issue.

Hope this helped you out! :-)

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Direct Mapped Cache. Memory is byte addressable. Fill in the missing fields based upon the properties of a direct-mapped cache.

xz_007 [3.2K]

The question given is incomplete and by finding it on internet i found the complete question as follows:

Direct Mapped Cache.

Memory is byte addressable

Fill in the missing fields based upon the properties of a direct-mapped cache. Click on "Select" to access the list of possible answers Main Memory Size Cache Size Block Size Number of Tag Bits 3 1) 16 KiB 128 KiB 256 B 20 2) 32 GiB 32 KiB 1 KiB 3) 64 MiB 512 KiB 1 KiB Select] 4 KiB 4) 16 GiB 10 Select ] Select ] 5) 10 64 MiB [ Select ] 6) Select] 512 KiB 7

For convenience, the table form of the question is attached in the image below.

Answers of blanks:

1. 3 bits

2. 20 bits

3. 64 MB

4. 16 MB

5. 64 KB

6. 64 MB

Explanation:

Following is the solution for question step-by-step:

Part 1:

No. of Tag bits = No. of bits to represent

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 128 KB = $2^7 * 2^{10} = 2^{17}$

Cache Memory = 16 KB = $2^4 * 2^{10}= 2^{14}$

Putting values in A:

Tag bits = 17 - 14 = 3 bits

Part 2:

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 32 GB = $2^5 * 2^{30} = 2^{35}$

Cache Memory = 16 KB = $2^5 * 2^{10}= 2^{15}$

Putting values in A:

Tag bits = 35 - 15 = 20 bits

Part 3:

Given:

Tag bits = 7

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

So from equation A

7 = Main Memory size - 19

Main Memory = 7 + 19

Main memory = 26

Main Memory = $2^6 * 2^{20} = 64 MB$

Part 4:

Given that:

Main Memory Size = $2^4 * 2^{30} = 2^{34}$

Tag bits = 10

Cache Memory Bits = 34 - 10 = 24

Cache Memory Size = $2^4 * 2^{20} = 16 MB$

Part 5:

Given that:

Main Memory Size = 64 MB = $2^6 * 2^{20}$

Tag bits = 10

Cache Memory Bits = 26 - 10 = 16

Cache Memory Size = $2^{16} = 2^6 * 2^{10} = 64 KB$

Part 6:

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

Tag Bits = 7

Main Memory Bits = 19 + 7 = 26

Main Memory size = $2^{26} = 2^6 * 2^20 = 64 MB$

i hope it will help you!

6 0

3 years ago

Mark works as a Network Administrator for NetTech Inc. The company has a Windows Server 2008 domain-based network. The network c

mart [117]

Hello This Answer Is Not Real This Is All A Simulation

5 0

3 years ago

Weak Induction

slega [8]

Answer:

Following are the answer to this question:

Explanation:

In option 1:

The value of n is= 7, which is (base case)

$\to 3^7$

when n=k for the true condition:

$\to 3^k$

when n=k+1 it tests the value:

$\to 3^{(k+1)}= 3^k,3\\\to < (k!) 3 \ substituting \ equation \\\to$

since k>6 hence the value is KH>3 hence proved.

In option 2:

when:

for n=1:(base case)

$\log(1!)$

0<=0 \\ condition is true

when the above statement holds value n=1

when n=k

$\log(k!)$

when n=k+1

$\log(k+1)!=\log(k!)+\log(k+1)\\$

$\because k \log k$ $[\therefore KH>K \Rightarrow \log(KH>\loK)]$

In option 3:

when n=1:

$A_1 \cup B=A_1 \cup B\\\\$

when n=k

$\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a \\\\$

hence n=k+1 is true.

7 0

3 years ago

Expressions provide an easy way to perform operations on data values to produce other data values. True False

Papessa [141]

Expressions provide an easy way to perform operations on data values to produce other data values, True.

<h3>What is an Expression? </h3>

An expression is a combination of one or more operands (Constant, Variable, Array element, Function), operators(Multiplication,Division, Subtraction etc) to be interpreted by a programming language following rules of precedence or association to produce other data values.

Three kinds of expressions includes:

An arithmetic expression

A character expression

A logical or relational expression

Therefore, it is true that Expressions provide an easy way to perform operations on data values to produce other data values.

6 0

2 years ago

Create the following matrix M: 1 7 13 19 25 ?-3 9 15 21 27 5 11 17 2329By writing one command and using the colon to address ran

zimovet [89]

Answer:

Matlab code is:

>> $M=reshape(1:2:29, 3,5)$

>> $\% (a)$

>> $Va=M(3,:)$

>> $\% (b)$

>> $Vb=M(:,4)$

>> $\% (c)$

>> $Vc=[M(2,:) M(:,3)']$

Explanation:

>> $\% \ Making\ the\ Matrix\ With\ required\ terms\ 1\ to\ 29\ with\ spaces\ of\ 5\ in\ three\ rows$

>> $M=reshape(1:2:29, 3,5)$

$M =$

1 7 13 19 25

3 9 15 21 27

5 11 17 23 29

>> $\% (a)$

>> $\%\ Slicing\ M\ \ from\ third\ row\ till\ end$

>> $Va=M(3,:)$

$Va =\\\ 5\ 11\ 17\ 23\ 29$

>> $\% (b)$

>> $\% \ Slicing\ the\4th\ column\ of\ M\ matrix\$

>> $Vb=M(:,4)$

$Vb =\\ 19\\ 21\\ 23\\$

>> $\% (c)$

>> $\% \ slicing\ the\ 2nd\ row\ and\ 3rd\ column\ of\ M\ Matrix\ and\ combining\ them\ to\ make\ a\ row\ matrix$

>> $Vc=[M(2,:) M(:,3)']$

$Vc =\\ 3\ 9\ 15\ 21\ 27\ 13\ 15\ 17$

The code is tested and is correct. If you put a semi colon ' ; ' at the end of every statement then your answer will be calculated but matlab doesn't show it until you ask i.e. typing the variable (Va, Vb, Vc )and press enter.

6 0

3 years ago