question on ip address

[kelvinwgr]

Hi can anyone explain to me on how to tackle this qs:

Which 2 addresses can be assigned to a host with subnet mask of 255.255.254.0? (Choose two.)

A) 113.10.4.0
B) 186.54.3.0
C) 175.33.3.255
D) 26.35.2.255
E) 17.35.36.0

[bpman57]

I have been practice testing this type of subnetting today, so here goes:

    The network portion of the address is the first 23 bits as indicated by the subnet mask 255.255.254.0 or in binary 11111111.11111111.11111110.00000000.

the first address 113.10.4.0  we know 113.10 is a network portion of address by 255.255

so the 3rd octet needs to be analyzed since 3rd octet of mask is 254 as follows

                                          binary
                            113.10.  00000100.    0
                            113.10.  0000010 | 0.00000000
                                                      ^
                                                this is the boundary btwn network and host.
                                                to the right you see all zeros so this is a network address with no
                                                host specified and can therefore not be assigned to a host.

the second address    186.54.3.0  analyze 3rd octet

                              186.54.  00000011.    0
                              186.54.  0000001 | 1.00000000
                                                      ^
                                                  boundary -  so  the 1.00000000 to the right is a valid host of the
                                                                    186.54.2.0  network
                              186.54.3.0  is therefore a valid host address


the third address        175.33.3.255  analyze 3rd octet

                              175.33.  00000011.  11111111
                              175.33.  0000001 | 1.11111111
                                                        ^
                                                  boundary  -  here you see all ones to the right (host portion) which
                                                                    is broadcast address for network 175.33.2.0 and       
                                                                    therefore 175.33.3.255 is NOT a valid host address.


the fourth address        26.35.2.255    analyze 3rd octet

                                26.35.  0000001 | 0.11111111
                                                        ^
                                                              here 0.11111111 is a valid host address for network
                                                              26.35.2.0

the fifth address            17.35.36.0    check 3rd

                                  17.35.    0010010 | 0.00000000
                                                          ^
                                                              here all zeros to the right = 17.35.36.0  255.255.254.0 is a
                                                                                                    network with no host specified
                                                                                                    Not a valid host address




                                    I hope this helps ------  study on!
                                 

[kelvinwgr]

Hi there,

MILLIONS THANKS TO YOU. Appreciate.

Just that I wonder how am I going to convert the decimal 36 to binary 00100100 for the last option e) where I can bring nothing into the ccna exam room. I can't possibly memorise the table =(

[bpman57]

Hey Kelvin,

    I am testing Thursday (3rd attempt) for ICND2. I don't think I'm giving anything away if I advise you to do this. When you sit for exam, you have about 15 minutes to view a tutorial of the exam function. I highly recommend you view this. It won't take anywhere near the 15 minutes to view, so with the extra minutes you have left over write a small table on your plastic 'white sheet' of the powers of 2 up to at least 256 and the corresponding mask values (from memory).

    example -    128    64    32    16    8    4    2    1

                      128    192  224    240  248  252  254  255


This was very helpful to me for ip addressing VLSM networks

I wish I had discovered how2pass sooner as this site is the MOST thorough and difficult practice test I have used. If you work through this site for the 640-802 or the 640-822 / 640-816 test paths you will be very prepared. I have been at this for six months. I passed the 640-822 test on first attempt (after failing the 802 test twice  :-[ ) and I have used lab simulators and have rented rack time with real equipment.
From my experience this how2pass is the best final prep, as they ask you the hard 'trick' questions that will trip you up on the real exam.

  Cisco means business with their certifications and you will need to be well prepared to know AND use all of your knowledge and be able to do so quickly!


P.S.  you can also view exam tutorial on Cisco's website in the Learning center


                        Best Wishes for Success!

[kelvinwgr]

Hi bpman57,

Thanks for the advice. I admire your perseverance. I will try my best for my 640-802. Way to go my friend..... Cheers

[bpman57]

Passed Thurday 640-816 ICND2 w/ 946!  How2 pass got me over the 800 hump! 

[kelvinwgr]

Wow that's great. Congrats to you my friend. Cheers!!!

[fanren]

i want to help you, but i have no idea. you can check it on google.

[megathumpzilla]

Dudes please.

255.255.254.0 gives you 2 3rd octet addresses. the first pair is 0- 1.2-3.4-5.
Even # 3rd octet can use the  x.x.even.255 for a host, and x.x.odd.o for a host.
Odd 254s/23s have a broadcast address at 255.
if it isn't a reserved address and meets these criteria, you are correct.

Usually I figure out how many networks there are. Then I divide the whatever octet by that number to get my network address closest to my target.

There are only 128,64,32,8,4,2,1,0 per octet. 32 and 64 might get tricky, but that's just 4 and 8 possible networks. If it is beyond midpoint of 64 128 190 255, it's easy to figure out if it is +/- 32 away. 101 is closer to 128-32 than it is to 64(boundary 96=32+64)
Blamo.
128 is over/under, and if I can't divide a 3 digit number by 8 in my head I should do something else.

Don't forget, the reason we don't base everything off 256 is because 0 is a number not a void. anything that hits on a boundary(32 64 96 128 etc in this case) is really hitting the first number of the next network because of the 0(33rd,65th,97th,129th) available address.. And just before that is the broadcast. Broad cast is actually the 32nd address but is numbered 31.

Oh, so masks make sense now... Where is the start of a network? The number of the last network number is where you start The actual 32nd  address is what you are looking for, 31.

Only bad thing about google is it won't work when your network can't get to the internet. If you need a tool Boson is a good one.


I can't and on my fingers, too many steps, but I can nail an ip subnet zero network faster than I can find and open a calculator.

[chas]

Got this question on my ICND1 exam the other day.  The answer is B and D. Here is how I got my answer.

Which 2 addresses can be assigned to a host with subnet mask of 255.255.254.0? (Choose two.)

A) 113.10.4.0
B) 186.54.3.0
C) 175.33.3.255
D) 26.35.2.255
E) 17.35.36.0

Look at the 3rd Octet, 255.255.254.0

256 - 254 = 2

3rd Octet is increment by 2

Networks are
0
2
4
6 etc.. (even number)

In this example:

Subnetwork 26.35.2.0
First valid host 26.35.2.1
Last valid host 26.35.3.254
Broadcast 26.35.3.255
Next subnetwork 26.35.4.0  (remember 3rd Octet is increment by 2)

A) 113.10.4.0  (Subnetwork)
B) 186.54.3.0  (Valid host)
C) 175.33.3.255 (Broadcast)
D) 26.35.2.255  (Valid host)
E) 17.35.36.0  (Broadcast)

Answer... B and D