Soal dan Jawaban tugas Prak.netadmin

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1.Diketahui :
Untuk host =300 2n- 2 ≥300, maka n =9 29 - 2 =510 Sehingga subnet mask menjadi : 11111111.11111111.11111110.0000000
Jawaban: C. 255.255.254.0

2 . Refer to the exhibit. Host A is being manually configured for connectivity to the LAN.
Which two addressing scheme combinations are possible configurations that can be applied
to the host for connectivity? (Choose two.)
a. Address - 192.168.1.14
Gateway - 192.168.1.33
b. Address - 192.168.1.45
Gateway - 192.168.1.33
c. Address - 192.168.1.32
Gateway - 192.168.1.33
d. Address - 192.168.1.82
Gateway - 192.168.1.65
e. Address - 192.168.1.63
Gateway - 192.168.1.65
f. Address - 192.168.1.70
Gateway - 192.168.1.65


Jawaban: B dan F
Karna pada Address - 192.168.1.45, Gateway - 192.168.1.33 merupakan dalam suatu jaringan dimana terletak pada jaringan (subnet 192.168.1.32).
Pada Address - 192.168.1.70, Gateway - 192.168.1.65 merupakan dalam suatu jaringan dimana terletak pada jaringan (subnet 192.168.1.64).


3. IP address = 172.31.192.166
Subnet Mask = 11111111.11111111.11111111.11111000
Host Id = 23-2= 6
Net Id = 25-2 =30
Net id range broadcast
172.31.192.0 172.31.192.1- 172.31.192.6 172.31.192.7
172.31.192.8 172.31.192.9- 172.31.192.14 172.31.192.15
...................
172.31.192.160 172.31.192.161- 172.31.192.166 172.31.192.167
Jawaban :   E
172.31.192.166 -> 10101100.00011111.11000000.10100110
255.255.255.248->11111111.11111111.11111111.11111000
172.31.192.160 -> 10101100.00011111.11000000.10100000



4.Which subnet masks would be valid for a subnetted Class B address? (Choose two.)
a. 255.0.0.0
b. 255.254.0.0
c. 255.224.0.0
d. 255.255.0.0
e. 255.255.252.0
f. 255.255.255.192
Jawaban: D dan E
Karena batas dari submask untuk kelas B adalah 255.255.0.0 – 255.255.254.0


5. Net id range broadcast 172.16.128.0 172.16.159.255 172.16.159.255 172.16.160.0 …….
Net id :160-128 = 32 = 25 Subnetmask : 11111111.11111111.11111000.00000000
Jawaban : D. 172.16.128.0 and 255.255.224.0

6. Which type of address is 223.168.17.167/29?

Jawaban: D
karna address 223.168.17.167/29 menempati pada submask 255.255.255.248
maka memiliki blok ip 8. Sedangkan alamat 223.168.17.167 merupaka broadcast dari jaringan 223.168.17.160.




7. IP address : 192.168.99.0/29 (kelas C)
Subnetmask : 11111111.11111111.11111111.11111000
Host Id : 23-2 = 6 host/subnet
Net Id : 25-2 = 30 subnet
Jawaban : C. 30 networks / 6 hosts

 8. IP address : 192.168.4.0 (kelas C)
subnetmask : 255.255.255.224 = 11111111.1111111111.1111111.11100000
Host Id : 25-2 = 30 host/subnet
*Dikurangi 2 untuk pemakaian broadcast dan localhost/loopback
Jawaban : C. 30

9. 27 host /subnet = 2n-2 ≥27 , n = 5 2^5-2=30
jumlah host id = 30 /subnet, maka subnet mask = 11111111.11111111.11111111.11100000
Jawaban : C. 255.255.255.224

10.An IP network address has been subnetted so that every subnetwork has 14 usable host IP
addresses. What is the appropriate subnet mask for the newly created subnetworks?


Jawaban: C
karna -2 = 14, yaitu host yang mecukupi sehingga submasknya adalah 11111.11111.11111.11110000 (255.255.255.240).
 


11. Pada kelas B ,membutuhkan 100 networks
2n -2≥100, n = 7
Subnetmask : 11111111.11111111.11111111.10000000 = 255.255.255.128
Jawaban : F. 255.255.255.128

12. IP Address = 172.32.65.13(Kelas B)
Default Mask = 255.255.0.0
Jawaban :C. 172.32.0.0

13. IP address of 172.16.210.0/22 Subnetmask : 11111111.11111111.11.0000000
Host Id : 210-2= 1022
Net id range broadcast
172.16.0.0 172.16.1.0 – 172.16.2.0 172.16.3.0
172.16.4.0 172.16.5.0 – 172.16.6.0 172.16.7.0
……
172.16.208.0 172.16.209.0 - 210.0 172.16.211.0
Jawaban : C. 172.16.208.0

14. Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose
three.)


jwb. A dan F
Submask 11111111.11111111.11111100.00000000, maka subnetnya  -> 64 dan blocknya adalah 4. Sehinga subnetwork yang tepat adalah 115.64.8.32 dan 115.64.12.128

15. IP address 200.10.5.68/28
Subnetmask : 11111111.11111111.11111111.11110000
Host Id : 24-2 =14
Net id range broadcast
200.10.5.0 200.10.5.1 – 200.10.5.4 200.10.5.15
……
200.10.5.64 200.10.5.65 – 200.10.5.78 200.10.5.79
Jawaban : C. 200.10.5.64

16. 172.16.0.0/19
Subnetmask : 11111111.11111111.11100000.00000000
Net Id : 23-2 = 6
Host Id: = 213 = 8190 /subnet
Jawaban : E. 8 subnets, 8190 hosts each

17. Kelas B, 500 subnet, setiap subnet digunakan 100 host, mask??
Subnet = 2N > 500, N = 9 (bit “1”)
Subnetmask = 11111111.11111111.11111111.10000000 = 255.255.255.128
Jawaban : B. 255.255.255.128

18 . What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
Jawaban C
submask 11111111.11111111.11111000.00000000 (255.255.248.0), maka blocknya adalah 256-248 = 8. Sedangkan IP address 172.16.66.0 terdapat pada subnet 172.16.64.0


19. Kelas B, 100 subnet dan 500 host/subnet
Subnet = 2N > 100, N = 7 (bit “1”)
Subnetmask = 11111111.11111111.11111110.00000000
Jawaban : B. 255.255.254.0

20. IP address 192.168.19.24/29
Subnetmask : 11111111.11111111.11111111.11111000 = 255.255.255.248
Host Id = 23-3 =6host /network
Net id range broadcast
192.168.19.0 192.168.19.1 - 192.168.19.6 192.168.19.7
….
192.168.19.24 192.168.19.25 - .30 192.168.19.31
Jawaban : C. 192.168.19.26 255.255.255.248

21. need a minimum of 300 subnets with a maximum of 50 hosts per subnet. Which of
the following masks will support the business requirements? (Choose two.)

Jawaban: E
Banyak host 50, maka -2 = 62 sehinggasubmasknya adalah 11111111.11111111.11111111.11000000 (255.255.255.192)


22. IP address 172.16.112.1/25
Subnetmask : 11111111.11111111.11111111.10000000 = 255.255.255.128
Host Id = 27- 2 =126
Net id range broadcast
172.16.112.0 172.16.112.1- .126 172.16.112.127
172.16.112.128 ………..
Jawaban : A. 172.16.112.0

23. Jumlah host yang ada = 3350
Host 2n - 2 > 3350, n = 12
Subnetmask : 11111111.11111111.11111000.00000000 = 255.255.248.0
Jawaban : C. 255.255.248.0

24. Subnet 172.16.17.0/22
Subnetmask : 11111111.11111111.11111100.00000000 = 255.255.255.252
Host per blok : 256-252 = 4
Jawaban : E. 172.16.18.255 255.255.252.0

25. Your router has the following IP address on Ethernet0: 172.16.112.1/20. How many hosts
can be accommodated on the Ethernet segment?
Jawaban:. C
submask 11111111.11111111.11110000.00000000 (255.255.240.0), maka banyakny ahost yang tersedia adalah -2 = 4094

26. Prefix /27 Class C
11111111.11111111.11111111.11100000
Subnet 23-2=6
Host Id 25-2 = 30 subnet
Jawaban : D,E,F

27. Kelas B, 450 host/subnet
Host Id 2n - 2 > 450, n = 9
Subnetmask = 11111111.11111111.11111110.00000000 = 255.255.254.0
Jawaban : C. 255.255.254.0

28. Host A is connected to the LAN, but it cannot connect to the Internet. The host
configuration is shown in the exhibit. What are the two problems with this configuration?
(Choose two.)
a. The host subnet mask is incorrect.
b. The host is not configured for subnetting.
c. The default gateway is a network address.
d. The default gateway is on a different network than the host.
e. The host IP address is on a different network from the Serial interface of the router.
Jawaban  A dan  E

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