1 CS 250-Computer Networks Fundamentals Spring 2018 Quiz # 4 Name a) 209.165.200
ID: 3701153 • Letter: 1
Question
1 CS 250-Computer Networks Fundamentals Spring 2018 Quiz # 4 Name a) 209.165.200.32/26 b) 209.165.200.190/26 c)209.165.200.32/27 d) 209.165.200.80/27 e) 209.165.200.190/28 t 209.165.200.80/28 2. Consider network 193.154.62.64 27 and answer the following questions a) What subnet mask would be used for this network? b) What broadcast address would be used for this network? c) Which of the following IP address are on that network (more than one)? 193.154.62.32 193.154.62.92 - 193.154.62.84 - 193.154.62.97Explanation / Answer
Answer 1) Correct answer are option (c) and (f)
Explanation
(a) No
IP Address 209.165.200.32
11010001.10100101.11001000.00100000
Netmask 255.255.255.192 = 26
11111111.11111111.11111111.11000000
Therefore Network address after taking '&' operation will be
Network address
209.165.200.0/26
11010001.10100101.11001000.00000000
(b) No
IP Address
209.165.200.190
11010001.10100101.11001000.10111110
Netmask
255.255.255.192 = 26
11111111.11111111.11111111.11000000
Network address is
209.165.200.128/26
11010001.10100101.11001000.10000000
IP address is not same as Network Id.
(c) Yes
IP Address
209.165.200.32
11010001.10100101.11001000.00100000
Netmask
255.255.255.224 = 27
11111111.11111111.11111111.11100000
Network address-
209.165.200.32/27
11010001.10100101.11001000.00100000
In this case Ip address is same as Network ID.
(c) No
IP Address
209.165.200.80
11010001.10100101.11001000.01010000
Netmask
255.255.255.224 = 27
11111111.11111111.11111111.11100000
Network address
209.165.200.64/27
11010001.10100101.11001000.01000000
(e) No
IP Address
209.165.200.190
11010001.10100101.11001000.10111110
Netmask
255.255.255.240 = 28
11111111.11111111.11111111.11110000
Network Address is
209.165.200.176/28
11010001.10100101.11001000.10110000
(f) Yes
Address
209.165.200.80
11010001.10100101.11001000.01010000
Netmask
255.255.255.240 = 28
11111111.11111111.11111111.11110000
Network address is
209.165.200.80/28
11010001.10100101.11001000.01010000
In this case ip address is same as network address.
Answer 2) (a)Subnet of 193.154.62.64/27 is
255.255.255.224 = 27
11111111.11111111.11111111.11100000
(b) Broadcast address
193.154.62.95
11000001.10011010.00111110.01011111
27 is network id and 5 is host id then for broadcast address make all bits of host id to 1.
(c) 193.154.62.92 and 193.154.62.84 are on that number
First IP on that network
193.154.62.65
11000001.10011010.00111110.01000001
Last IP on that network
193.154.62.94
11000001.10011010.00111110.01011110
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