🔒 Closed Patulong naman po sa networking na subject

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Matsiako_07

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1. Given the network address of 112.44.0.0 and the network mask of 255.255.0.0. Would the two
stations (computer) with addresses 112.44.22.19/16 and 112.44.23.2/16 be on the same
network? Elaborate your explanation.
2. Workstations with addresses 172.16.22.1/22 and 172.16.23.9/22 share what network and broadcast
address? Elaborate your explanation.
Additional question for number 2:
 What is the host bit?
 What is the subnet mask?
 How many host per subnet?
 What is the borrowed bit?
 What is the magic number?
3. How many network devices can be supported on a single network using network mask of 255.255.240.
0? Show solution.
4. Which of the following devices share the same network? Show solution and elaborate your answer
answer.
A 192.168.78.25/29
B 192.168.78.23/29
C 192.168.78.33/29
D 192.168.78.38/29
E 192.168.78.41/29

Diko po kasi magets yung turo ng prof ko.
 
1.no.
pag ginwan ng range kc yan magging
ntwrk add. host range (usable) broadcst add.
112.44.0.0 - 112.44.0.1-112.44.0.254 - 112.44.0.255 /16
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . /16
112.44.22.0 - 112.44.22.1 - 112.44.22.254 - 112.44.22.255 /16 (si 112.44.22.19 ay nandto)
112.44.23.0 - 112.44.23.1 - 112.44.23.254 - 112.44.23.255 /16 (si 112.44.23.2 ay dto)

2.172.16.22.1/22 and 172.16.23.9/22
/22(255.255.252.0) means 4 increment. (172 is class B nnnn.nnnn.hhhh.hhhh)
ntwrk add. host range (usable 1022) broadcst add.
172.16.0.0 - 172.16.0.1 - 172.16.3.254 172.16.0.3.255 /22
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . /22
172.16.20.0 - 172.16.20.1 - 172.16.23.254 172.16.23.255 /22

3.
255.255.255.240 convert to cidr prefix = /28 /28 ay meron 16 IP add. (16 - 2(netwrk add and broadcst add.)= 14)
16 IP with 14 usable Host.

4.
A 192.168.78.25/29
B 192.168.78.23/29
C 192.168.78.33/29
D 192.168.78.38/29

E 192.168.78.41/29

192.168.78.0 - 192.168.78.1 - 192.168.78.6 - 192.168.78.7
192.168.78.8 - 192.168.78.9 - 192.168.78.14 - 192.168.78.15
192.168.78.16 - 192.168.78.17 - 192.168.78.22 - 192.168.78.23 (andto si 192.168.78.23/29)
192.168.78.24 - 192.168.78.25 - 192.168.78.30 - 192.168.78.31 (andto si 192.168.78.25/29)
192.168.78.32 - 192.168.78.33 - 192.168.78.38 - 192.168.78.39 (anjan sila o .33 at .38)
192.168.78.40 - 192.168.78.41 - 192.168.78.46 - 192.168.78.47 (andto si 192.168.78.41/29)


PS: mabilisang sagot lang po yan pachck m nlng dn sa iba.
 
Assignment ba ito? Simple.

1) YES. Dahil pasok sa 3rd octet ang either 22 or 23.
valid 1st address: 112.44.0.1/16
valid last address: 112.44.255.254/16

2) YES. Both subnets belong to parent subnet 172.168.20.0/16
Subnet A: 172.16.22.1/22 belongs to 172.16.

3) 4094 valid IP addresses (excluding network and broadcast).

4) 192.168.78.33/29, 192.168.78.38/29, 192.168.78.41/29 all belong to the same subnet 192.168.78.24/29
192.168.78.25/29 belongs to its own subnet 192.168.78.24/29
192.168.78.23/29 is actually a broadcast IP for the subnet 192.168.78.16/29

EZ
 
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