计算机网络自顶向下方法第七版第六章答案,《计算机网络 自顶向下方法》（第7版）答案（第八章）（五）...

《计算机网络 自顶向下方法》(第7版)答案(第八章)(五)

《计算机网络 自顶向下方法》(第7版)答案(第八章)(五)

P22

a) F

b) T

c) T

d) F

P23

If Trudy does not bother to change the sequence number, R2 will detect the duplicate when checking the sequence number in the

ESP header. If Trudy increments the sequence number, the packet will fail the

integrity check at R2.

P24

a) Since IV = 11, the key stream is 111110100000 ……….

Given,

m = 10100000

Hence,

ICV = 1010 XOR 0000 = 1010

The

three fields will be:

IV: 11

Encrypted

message: 10100000 XOR 11111010 = 01011010

Encrypted

ICV: 1010 XOR 0000 = 1010

b) The receiver extracts the IV (11) and generates the

key stream 111110100000 ……….

XORs

the encrypted message with the key stream to recover the orig