Bonus Quiz Question: Given an IP address of 191.90.0.0 and the fact that you have 1000 subnets created, find the following information of subnet number 100 (i.e. S100): (1) Number of bits borrowed (s): (2) Configured Given Subnet Mask (GSM) in dot decimal format: (3) Number of host bits available (h): (4)Subnet ID (SID) in dot decimal format: (5) Broadcast address (BA) in dot decimal format: (6)1st Host IP address in dot decimal format: (7) Last Host IP Address in dot binary format: Answers: S=1000 subnets S=2^s s=10 ClassBDSM/16 GSM/26 => 255.255.255.192 h=32-GSM =32-26 = 6 SID & GSM GIP: 10111111.01011010.(00000000.00) 000000 GSM: 11111111.11111111.(11111111.11) 000000 SID: 191. 90. 0. .0 SID (100) = 11111111.11111111. (00011001.00) 000000 (100 is 01100100) SID (100) = 191.90.25.0 BA(100) = 11111111.11111111. (00011001.00) 111111 = 191.90.25.63 1st HIP: SID+1 = 191.90.25.1 Last HIP = BA-1 = 191.90.25.62 = 10111111.01011010.00011001.00111110|