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4 years ago

Host Y sends the first TCP ACK message for the transaction? a. true b. false

Computers and Technology

1 answer:

kupik [55]4 years ago

8 0

Answer: True

Explanation:

TCP is the transmission control protocol which and TCP ACK is the transmission control handshake method which are used by TCP. It basically indicates the next sequence number in the method when the flag is set.

Firstly, the ACK (Acknowledge) send by the each end for the initial sequence number itself but it does not contained any type of data. In the TCP segment header ACK contain 32 bit field.

The acknowledgement is just a proof to clients that ACK is a specific to the SYN when the clients initiate.

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Answer:

The following are the answer to the given points:

Explanation:

In point (a):

Calculating the long-distance call cost:

$= 4 \times 5 \times 0.40 \\= 20 \times 0.40\\= 8$

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$= 10 \times 3 \times 0.05\\ = 30 \times 0.05\\= 1.5$

Calculating the overall cost of PSTN:

$= 25 + (4 \times 5 \times 0.40) + (10 \times 3 \times 0.05) + 2000 + 275 \\= 25 + 8 + 1.5 + 2000 + 275 \\= 2309.5$

In point (b):

Calculating the call rate per second and the average arrival rate:

$\to (\lambda) = 0.2$

The call average length:

$\to (T_s) = - 8 \\\\ = (8 \times 60) \ seconds \\\\ = 480 \ seconds$

The complete agent number:

$\to (m) = 9$

The strength of traffic:

$\to u = \lambda \times T_s \\\\ = (0.2 \times 480) \\\\ = 96$

The occupancy of the agent:

$\to p = \frac{u}{m} \\\\$

$= \frac{96}{9} \\ \\= 10.66$

Calculation of obtained:

$= \frac{(\frac{um}{m!})}{(\frac{um}{m!})} + (1-p) \sum {m-1} _{k=0} \ \frac{uk}{k!}$

We get = 0.329 to substitute values.

In point (c):

The rate of blocking = $5 \%$

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In point (d):

Calculating the number of link, which is required:

$= \frac{275}{5} \\\\ =55$

In point (e):

Calculating the Line Number:

$= \frac{275}{5} \\\\ =55$

PSTN line number:

$= (\frac{2000}{55}) \\\\ = 36.3636\\\\= 37$

In point (f):

The gross design expense = $ 2309. 5

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4 years ago

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