1177 Words5 Pages

Homework 5
Chapter 5
Question P4.
a.)
Answer.
Lets represent the decimal numbers into the binary first
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
10 = 1010
Lets take 16 bits and calculate the check sum
So we have,
Checksum = 1’s complement of (0000000100000010 + 0000001100000100 + 0000010100000110 + 0000011100001000 + 0000100100001010)
Checksum = 1’s complement of (00011001 00011110)
Checksum = 11100110 11100001
b.)
Answer. Lets represent the ascii values from B to K in binary
B = 66 = 01000001
C = 67 = 01000010
D = 68 = 01000011
E = 69 = 01000100
F = 70 = 01000101
G = 71 = 01000110
H = 72 = 01000111
I = 73 = 01001000
J = 74 = 01001001
K = 75 = 01001010
Checksum = 1’s complement of (0100000101000010 + 0100001101000100*…show more content…*

So now if one of the AP is on channel 1 and the other one is on channel 2, then there won’t be any collision and they can now transmit the frame independently. Question P6. Answer. If there are two wireless station trying to transmit, let’s say if A is transmitting data D1. While this transmission is going say another wireless station B wants to transmits its data D2, but as B detects that there is another transmission ongoing it won’t transmit its data D2. As soon as A finishes transmitting data it has two option either wait for some time and then transmit again or select a backoff value. If it chooses a backoff value it will give an opportunity to station B to transmit its data. And hence to give an equal opportunity to transmit. Question P7. Answer. The station wants to transmit data of 1000 bytes i.e. 8000bits. Let us consider maximum transmission rate of 802.11b i.e. 11Mbps. Here the header is of 256 bits. So the entire frame will be of header plus data i.e. 256 + 8000 = 8256 bits. So the time required to transmit the frame is 8256bits/11Mbps i.e. 751 microseconds. And time required to transmit the control frame is 256 bits/11 Mbps i.e. 23 microseconds. So the time required to transmit and receive the acknowledgement would

So now if one of the AP is on channel 1 and the other one is on channel 2, then there won’t be any collision and they can now transmit the frame independently. Question P6. Answer. If there are two wireless station trying to transmit, let’s say if A is transmitting data D1. While this transmission is going say another wireless station B wants to transmits its data D2, but as B detects that there is another transmission ongoing it won’t transmit its data D2. As soon as A finishes transmitting data it has two option either wait for some time and then transmit again or select a backoff value. If it chooses a backoff value it will give an opportunity to station B to transmit its data. And hence to give an equal opportunity to transmit. Question P7. Answer. The station wants to transmit data of 1000 bytes i.e. 8000bits. Let us consider maximum transmission rate of 802.11b i.e. 11Mbps. Here the header is of 256 bits. So the entire frame will be of header plus data i.e. 256 + 8000 = 8256 bits. So the time required to transmit the frame is 8256bits/11Mbps i.e. 751 microseconds. And time required to transmit the control frame is 256 bits/11 Mbps i.e. 23 microseconds. So the time required to transmit and receive the acknowledgement would

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