Explanation:
There are only two inputs to the circuit, and one output. There is an internal signal, not an input, between the inverter and the AND gate.
Here, it is convenient to label the top input A, the inverter output B, and the bottom input C. Of course, the output is labeled D.
Then B is the logical inverse of A, and D is the AND of B and C.
_____
<em>Check for you</em>
D will only be 1 for A=0 and C=1. It will be 0 otherwise.
B hope this helps, can you make me brainliest if it is right?
The answer to your question is:
A. Have the key in your hand.
Answer:
a. 2100
b. 2199
Explanation:
GIven that:
The file size = 1000 MB
The payload size is = 100 bytes
The negotiated window size is = 1000 bytes.
This implies that the sliding window can accommodate maximum number of 10 packets
The sender receives an ACK 1200 from the receiver.
Total byte of the file is :
1000 MB = 1024000000 bytes
a.
Sender receives an ACK 1200 from the receiver but still two packets are still unacknowledged
=1200 + 9 * 100
= 1200 + 900
= 2100
b. What is the last byte number that can be sent without an ACK being sent by the receiver?
b. Usually byte number starts from zero, in the first packet, the last byte will be 99 because it is in 1000th place.
Thus; the last byte number send is :
= 1200 + 10 *100 -1
= 1200 + 1000-1
= 1200 + 999
= 2199
