What range of addresses would be in the network 192.168.23.128 with a 255.255.25

Question

What range of addresses would be in the network 192.168.23.128 with a 255.255.255.1927 9. 10. What netmask should be used for a network to allow just 16 addresses in the network? 11. What is the network address for a computer with the IP address 192.168.48.75 and a netmask of 255.255.248.0? 12. Write a routing table for a router that will direct packets to two different networks with the addresses 192.168.2.0 and 192.168.3.0. Both networks are connected to the ethi Ethernet interface. The router's etho interface is connected to its local network which has the address 192.168.1.0. What IP address should be assigned to the router? IP address of the router: Gateway nterface Net Mask Network Address

Explanation / Answer

9. Since Netmask is 255.255.255.192 and Network address is 192.168.23.128 . 192 in binary form as 110000000 where subnet is 2 bits and host is 6 bits so the range of addresses is 256-192 = 64 bits where subnet bits are 2 where remaining . 64-2=62 bits are host addresses and the range of subnets would be 64,128,192 where 192.168.23.128 class C address belongs to 128 subnet where the broadcast is from 128 to 191

10.To allow 16 addresses in the Netmask then 256-16 = 240 so the Netmask will be 255.255.255.240

11.The binary form of 248 is 11111000.00000000 so the subnet bits are 2^5-2 is 30 bits and hosts are 2^16-30= 2046 bits so now for 192.168.48.75 the network address would be 192.168.55.255 as for every 8 subnets network address are 8,16,24,32,40,48,56...... 192.168.48.75 lies in 48 subnet so broadcast is 55.255 so broadcast address is 192.168.55.255

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