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Your previous Network Administrator had subnetted your network into many subnets. Only info you have is that one of the hosts IP was 200.252.144.33 and the mask was 255.255.255.240.

Which class network is it ?

How many bits are being subnetted ?

What is the broadcast address ?

a. B, 28, 255.255.255.47

b. C, 20, 255.255.255.47

c. C, 28, 255.255.255.47

c. C, 28, 255.255.255.32



Your previous Network Administrator had subnetted your network into many subnets. Only info you have..

Answer / indu

C

explanation:

We see that the first byte is 200. That puts it in class C.

Mask's last byte 240 (128+64+32+16), that 4 bits. N=4 (see subnetting

formulas above). That subnetted bits = 8+8+8+4 = 28.

There should be 14 (2^4-2) hosts in the subnet.

The highest Ip in the subnet will be 200.252.144.33+13 = 200.252.144.46

The broadcast address is always 1 higher than the highest Ip in the range.

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