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Ludmilka [50]
3 years ago
12

True or false questions. If a statement is false, you must explain the reason why it is false: a). Suppose that TCP is being run

over IPsec with the AH protocol. If TCP retransmits the same packet, then the two packets will have the same sequence number in the AH header. b). Suppose certifier.com creates a certificate for foo.com. Typically the entire certificate would be encrypted with certifier.com's public key.
Computers and Technology
2 answers:
Dima020 [189]3 years ago
8 0

Answer:

a. True

b. False

Explanation:

b. False

At the initial stage, the certificate needs to be encrypted not by the public key of certifier.com but by the by the private key of certifier.com.

After which, the certificate which contains lots information, (i.e there's more information in the certificate than just the public key to be certified). The certificate also the CA’s information, the certificate expiration date, etc.

Only this information’s digital fingerprint is encrypted by the certifier.com’s private key.

DIA [1.3K]3 years ago
7 0

Answer:

Explanation:

The TCP protocol uses the OPTIONS field to exchange information like maximum segment size accepted between the TCP layers on the two hosts. The flags currently defined are

• URG Urgent pointer field is valid

• ACK Acknowledgement field is valid

• PSH This segment requests a push

• RST Reset the connection

• SYN Synchronize sequence numbers

• FIN Sender has reached end of its byte stream

OFFSET

This integer indicates the offset of the user data within the segment. This field is only required as the number of bits used in the OPTIONS field can vary

URGENT POINTER

This field can be initialized to point to a place in the user data where urgent information such as escape codes etc. are placed. Then the receiving host can process this part immediately when it receives the segment.

TCP/IP (or rather; the TCP/IP protocol suite) is not limited to the TCP and IP protocols, but consist of a multitude of interrelated protocols that occupy the upper three layers of the ARPA model. With the exception of the Point-to-Point Protocol (PPP) which resides in the upper half of the Network Interface Layer, the TCP/IP suite generally does not include the Network Interface layer, but merely depends on it for access to the medium.

The Network Interface layer

The Network Interface layer is responsible for transporting frames between hosts on the same physical network. It is implemented in the Network Interface Card or NIC, using hardware and firmware (i.e. software resident in Read Only Memory).

The NIC employs the appropriate medium access control methodology, such as CSMA/CA, CMSA/CD, token passing or polling, and is responsible for placing the data received from the upper layers within a frame before transmitting it. The frame format is dependent on the system being used, for example Ethernet or Frame Relay, and holds the hardware address of the source and destination hosts as well as a checksum for data integrity.

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Consider the classes below, declared in the same file: class A { int a; public A() { ​ a = 7; } } class B extends A { int b; pub
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Answer:

2. <em>A reference of type A can be treated as a reference of type B</em> - False

Base class or its objects are not related to their derived class (or its objects).

Explanation:

class A {

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///////////////////////////////////////////////////////////////////////////////////////////////////////

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gulaghasi [49]

Answer:

Following is given the answer step by step:

Explanation:

First of all we will write a program using MARIE that will support the statement: Sum = (A + B) - (C + D). All the necessary comments are given in front of each statement:

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Add     B                  # B will be added to A

Store   Temp1         # A + B will be stored in Temp1

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Load    Temp1         # Temp1 will be loaded in the memory

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We can see from above program that memory is accessed 9 times. While if C + D get executed first than memory accesses will be reduced to 7.

Above same program could be written using an architecture of 4 registers:

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Load   R1 , A     #A  will be loaded into R1

Load   R2 , B    # B will be loaded into R2

Add     R1 , R2    # R2 gets added to R1 and the result is stored in R1 (A + B)

Load   R3 , C      # C loaded into R3

Load   R4 , D     # D loaded into R4

Add    R3 , R4   # Value in R4 gets added into R3 and R3 becomes (C + D)

                          #no memory accesses required for this operation

Subt   R1 , R4     #R4 (C + D) gets subtracted from R1 (A + B)

                         #no memory accesses required for this operation

Store  Sum        # The recent value will be stored into Sum

Here memory is accessed 5 times in total.

<h2>I hope it will help you!</h2>
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