For the address 172.30.173.68/23, what is the binary conversion of the broadcast address?
A) 10101100 00011110 10101100 11111111
B) 10101100 00011110 10101101 11111111
C) 10101100 00011110 10101110 11111111
D) 10101100 00011110 10101111 11111111
B
Explanation: B) The address 172.30.173.68 is a Class B address, which means the first two bits are 10 and the next 14 are reserved for network bits. The other 16 bits are used for hosts. With the mask of /23, or 255.255.254.0, the host portion is subdivided to provide 7 host bits to be used for subnet and the other 9 host bits to be used for host space. This means the first 23 bits will be the same for all the devices within a subnet, with the highest number being used for the broadcast address. The address 172.30.173.68 converts to 10101100 00011110 10101101 01000100 in decimal. The first 23 bits don't change; the rest of the bits change to 1s to get the binary broadcast address 10101100 00011110 10101101 11111111.
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For each of the sequence w = e1,...,elast below, determine whether they are subsequences of the following data sequence: {A, B}{C, D}{A, B}{C, D}{A, B}{C, D} subjected to the following timing constraints: mingap = 0 (interval between last event in ei and first event in ei+1 is > 0) maxgap = 2 (interval between first event in ei and last event in ei+1 is ? 2) maxspan = 6 (interval between first event in e1 and last event in elast is ? 6) ws = 1 (time between first and last events in ei is ? 1) (a) w = {A}{B}{C}{D} (a) w = {A}{B}{C}{D} (c) w = {A}{B, C, D}{A} (d) w = {B,C}{A, D}{B,C} (d) w = {B,C}{A, D}{B,C}