Question 36

Destination IP Address:
131.23.151.76

• First octet: $131$.
• Second octet in binary: $23 = 00010111_2$.

Evaluate Routing Table Entries:

• Prefix 1: 131.16.0.0/12 (length 12, interface 3)
  The first 12 bits consist of all 8 bits of octet 1 (131) and the first 4 bits of octet 2.
  For 16, binary is $00010000_2$, so first 4 bits are 0001.
  For 23, first 4 bits are 0001.
  Since 0001 matches 0001, this prefix matches. (Prefix length = 12)

• Prefix 2: 131.28.0.0/14 (length 14, interface 5)
  Requires matching first 6 bits of octet 2.
  For 28, binary is $00011100_2$ -> first 6 bits are 000111.
  For 23, first 6 bits are 000101.
  000111 != 000101, so no match.

• Prefix 3: 131.19.0.0/16 (length 16, interface 2)
  Requires matching all 8 bits of octet 2 ($19 = 00010011_2$).
  Packet has $23 \ne 19$, so no match.

• Prefix 4: 131.22.0.0/15 (length 15, interface 1)
  Requires matching first 7 bits of octet 2.
  For 22, binary is $00010110_2$ -> first 7 bits are 0001011.
  For 23, binary is $00010111_2$ -> first 7 bits are 0001011.
  Since 0001011 matches 0001011, this prefix matches! (Prefix length = 15)

Longest Prefix Match Rule:
Between the matching prefixes /12 (interface 3) and /15 (interface 1), the longest prefix is /15.
The associated interface identifier is 1.