Answer:
Follows are the solution to this question:
Explanation:
In point a:
N = Size of window = 4
Sequence Number Set = 1,024
Case No. 1:
Presume the receiver is k, and all k-1 packets were known. A window for the sender would be within the range of [k, k+N-1] Numbers in order
Case No. 2:
If the sender's window occurs within the set of sequence numbers [k-N, k-1]. The sender's window would be in the context of the sequence numbers [k-N, k-1]. Consequently, its potential sets of sequence numbers within the transmitter window are in the range [k-N: k] at time t.
In point b:
In the area for an acknowledgment (ACK) would be [k-N, k-1], in which the sender sent less ACK to all k-N packets than the n-N-1 ACK. Therefore, in all communications, both possible values of the ACK field currently vary between k-N-1 and k-1.
I believe there were not nicknames for first generation computers.
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Sorry!
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Answer:
Answer is provided in the explanation section
Explanation:
Given data:
Bandwidth of link = 10* 106 bps
Length of packet = 12* 103 bits
Distance of link = 40 * 103m
Transmission Speed = 3 * 108 meters per second
Formulas:
Transmission Delay = data size / bandwidth = (L /B) second
Propagation Delay = distance/transmission speed = d/s
Solution:
Transmission Delay = (12* 103 bits) / (10* 106 bps) = 0.0012 s = 1.2 millisecond
Propagation Delay = (40 * 103 meters)/ (3 * 108mps) = 0.000133 = 0.13 millisecond
Explanation:
Mainframes typically run on large boxes with many processors and tons of storage, as well as high-bandwidth busses. PCs are desktop or mobile devices with a single multi-core processor and typically less than 32GB of memory and a few TBs of disk space. Second, a mainframe OS usually supports many simultaneous users.
Hi!
The answer is True because a typical audience does not have a large attention span so using short phrases and images is the best way to get your message across to them.
