About 70-291

Discussion in 'MCSA' started by RWH, Aug 17, 2006.

  1. RWH

    RWH Guest

    Studying for 70-291 using the Microsoft Press textbook by Mackin and MacLean
    (2nd ed.), and need some clarification. The answer to a practice question in
    Chapter 2 is given as follows:

    Original Internally Number of
    Number of
    Address Configured Available
    Available Hosts
    Block Subnet Mask Subnets
    Per Subnet

    208.147.66.0/25 255.255.255.248 8 6

    I am having a heck of a time getting a grip on this topic and am wondering,
    is this a misprint or am I just stuck on "Stupid"?

    The formula in the book for figuring the number of subnets is 2^n, where
    n=the number of bits in the subnet mask. OK, so far so good.

    So the original mask would be 255.255.255.128, right? (/25)

    And the newly configured mask is 255.255.255.248, as stated above, which
    would be /29, am I right?

    Applying what is presented in the expository section in the text to this
    question, the number of bits in the Subnet Mask would be 29 - 25, which is 4.

    The number of subnets can be calculated as 2^(number of bits in the Subnet
    Mask) = 2^4=16.

    So, should the correct answer be "16" for the number of available subnets in
    this example?

    This chapter and topic has me a little bewildered, as you can tell, and the
    misprints in the book make it even more confusing. There is a web page to
    list corrections in the second edition, but so far nothing has been posted.

    Thanks in advance. And, yes, go ahead and take some shots if this is a DUMB
    QUESTION!! But if anyone agrees with me and corroborates my answer, you will
    win a new Caddy. (Just kidding.)
     
    RWH, Aug 17, 2006
    #1
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  2. RWH

    RWH Guest

    Rats. Those columns don't look too hot. It should be:

    Original Address Block: 208.147.66.0/25
    Internally Configured Subnet Mask: 255.255.255.248
    Number of Available Subnets: 8
    Number of Available Hosts per Subnet: 6

    Sorry about that.

    "RWH" wrote:

    > Studying for 70-291 using the Microsoft Press textbook by Mackin and MacLean
    > (2nd ed.), and need some clarification. The answer to a practice question in
    > Chapter 2 is given as follows:
    >
    > Original Internally Number of
    > Number of
    > Address Configured Available
    > Available Hosts
    > Block Subnet Mask Subnets
    > Per Subnet
    >
    > 208.147.66.0/25 255.255.255.248 8 6
    >
    > I am having a heck of a time getting a grip on this topic and am wondering,
    > is this a misprint or am I just stuck on "Stupid"?
    >
    > The formula in the book for figuring the number of subnets is 2^n, where
    > n=the number of bits in the subnet mask. OK, so far so good.
    >
    > So the original mask would be 255.255.255.128, right? (/25)
    >
    > And the newly configured mask is 255.255.255.248, as stated above, which
    > would be /29, am I right?
    >
    > Applying what is presented in the expository section in the text to this
    > question, the number of bits in the Subnet Mask would be 29 - 25, which is 4.
    >
    > The number of subnets can be calculated as 2^(number of bits in the Subnet
    > Mask) = 2^4=16.
    >
    > So, should the correct answer be "16" for the number of available subnets in
    > this example?
    >
    > This chapter and topic has me a little bewildered, as you can tell, and the
    > misprints in the book make it even more confusing. There is a web page to
    > list corrections in the second edition, but so far nothing has been posted.
    >
    > Thanks in advance. And, yes, go ahead and take some shots if this is a DUMB
    > QUESTION!! But if anyone agrees with me and corroborates my answer, you will
    > win a new Caddy. (Just kidding.)
    >
     
    RWH, Aug 17, 2006
    #2
    1. Advertising

  3. RWH

    M D Guest

    AFAIK you're right (plesase don't kill me If I'm not ;) )

    /25 (255.255.128.0) is 10000000 in the 3rd octet -> only 1 bit in the
    network ID
    /29 (255.255.248.0) is 11111000 in the 3rd octet -> 5 bits in the network ID

    => 2^(5-1) = 2^4 = 16 subnets

    BTW: when you've got an original /25 subnet mask, you can set up to
    2^(32-25) (-2) = 2^7 (-2) = 128 (-2) hosts;
    with 2^3 (-2) = 8 (-2) hosts per subnet you can create up to 128/8 = 16
    subnets
    as you said.

    I'm studying 70-291 too in the same MS Press book but 1st edition... I can
    assure that misprintings there are even worse...
    Good luck with your studies
    MD






    "RWH" <> ha scritto nel messaggio
    news:...
    > Rats. Those columns don't look too hot. It should be:
    >
    > Original Address Block: 208.147.66.0/25
    > Internally Configured Subnet Mask: 255.255.255.248
    > Number of Available Subnets: 8
    > Number of Available Hosts per Subnet: 6
    >
    > Sorry about that.
    >
    > "RWH" wrote:
    >
    >> Studying for 70-291 using the Microsoft Press textbook by Mackin and
    >> MacLean
    >> (2nd ed.), and need some clarification. The answer to a practice question
    >> in
    >> Chapter 2 is given as follows:
    >>
    >> Original Internally Number of
    >> Number of
    >> Address Configured Available
    >> Available Hosts
    >> Block Subnet Mask Subnets
    >> Per Subnet
    >>
    >> 208.147.66.0/25 255.255.255.248 8
    >> 6
    >>
    >> I am having a heck of a time getting a grip on this topic and am
    >> wondering,
    >> is this a misprint or am I just stuck on "Stupid"?
    >>
    >> The formula in the book for figuring the number of subnets is 2^n, where
    >> n=the number of bits in the subnet mask. OK, so far so good.
    >>
    >> So the original mask would be 255.255.255.128, right? (/25)
    >>
    >> And the newly configured mask is 255.255.255.248, as stated above, which
    >> would be /29, am I right?
    >>
    >> Applying what is presented in the expository section in the text to this
    >> question, the number of bits in the Subnet Mask would be 29 - 25, which
    >> is 4.
    >>
    >> The number of subnets can be calculated as 2^(number of bits in the
    >> Subnet
    >> Mask) = 2^4=16.
    >>
    >> So, should the correct answer be "16" for the number of available subnets
    >> in
    >> this example?
    >>
    >> This chapter and topic has me a little bewildered, as you can tell, and
    >> the
    >> misprints in the book make it even more confusing. There is a web page to
    >> list corrections in the second edition, but so far nothing has been
    >> posted.
    >>
    >> Thanks in advance. And, yes, go ahead and take some shots if this is a
    >> DUMB
    >> QUESTION!! But if anyone agrees with me and corroborates my answer, you
    >> will
    >> win a new Caddy. (Just kidding.)
    >>
     
    M D, Aug 18, 2006
    #3
  4. RWH

    RWH Guest

    Thanks, MD. You just saved me a lot of time and frustration. That's how I was
    trying to figure it, but just could not understand why my answer didn't match
    the one given in the book.

    I know what you mean about the first edition's typos. The Second Edition has
    improved somewhat and Amazon's got it. I had hoped all the misprints were
    fixed, but evidently, it still has a few.

    Thanks again and good luck to you, too.
    RWH

    "M D" wrote:

    > AFAIK you're right (plesase don't kill me If I'm not ;) )
    >
    > /25 (255.255.128.0) is 10000000 in the 3rd octet -> only 1 bit in the
    > network ID
    > /29 (255.255.248.0) is 11111000 in the 3rd octet -> 5 bits in the network ID
    >
    > => 2^(5-1) = 2^4 = 16 subnets
    >
    > BTW: when you've got an original /25 subnet mask, you can set up to
    > 2^(32-25) (-2) = 2^7 (-2) = 128 (-2) hosts;
    > with 2^3 (-2) = 8 (-2) hosts per subnet you can create up to 128/8 = 16
    > subnets
    > as you said.
    >
    > I'm studying 70-291 too in the same MS Press book but 1st edition... I can
    > assure that misprintings there are even worse...
    > Good luck with your studies
    > MD
    >
    >
    >
    >
    >
    >
    > "RWH" <> ha scritto nel messaggio
    > news:...
    > > Rats. Those columns don't look too hot. It should be:
    > >
    > > Original Address Block: 208.147.66.0/25
    > > Internally Configured Subnet Mask: 255.255.255.248
    > > Number of Available Subnets: 8
    > > Number of Available Hosts per Subnet: 6
    > >
    > > Sorry about that.
    > >
    > > "RWH" wrote:
    > >
    > >> Studying for 70-291 using the Microsoft Press textbook by Mackin and
    > >> MacLean
    > >> (2nd ed.), and need some clarification. The answer to a practice question
    > >> in
    > >> Chapter 2 is given as follows:
    > >>
    > >> Original Internally Number of
    > >> Number of
    > >> Address Configured Available
    > >> Available Hosts
    > >> Block Subnet Mask Subnets
    > >> Per Subnet
    > >>
    > >> 208.147.66.0/25 255.255.255.248 8
    > >> 6
    > >>
    > >> I am having a heck of a time getting a grip on this topic and am
    > >> wondering,
    > >> is this a misprint or am I just stuck on "Stupid"?
    > >>
    > >> The formula in the book for figuring the number of subnets is 2^n, where
    > >> n=the number of bits in the subnet mask. OK, so far so good.
    > >>
    > >> So the original mask would be 255.255.255.128, right? (/25)
    > >>
    > >> And the newly configured mask is 255.255.255.248, as stated above, which
    > >> would be /29, am I right?
    > >>
    > >> Applying what is presented in the expository section in the text to this
    > >> question, the number of bits in the Subnet Mask would be 29 - 25, which
    > >> is 4.
    > >>
    > >> The number of subnets can be calculated as 2^(number of bits in the
    > >> Subnet
    > >> Mask) = 2^4=16.
    > >>
    > >> So, should the correct answer be "16" for the number of available subnets
    > >> in
    > >> this example?
    > >>
    > >> This chapter and topic has me a little bewildered, as you can tell, and
    > >> the
    > >> misprints in the book make it even more confusing. There is a web page to
    > >> list corrections in the second edition, but so far nothing has been
    > >> posted.
    > >>
    > >> Thanks in advance. And, yes, go ahead and take some shots if this is a
    > >> DUMB
    > >> QUESTION!! But if anyone agrees with me and corroborates my answer, you
    > >> will
    > >> win a new Caddy. (Just kidding.)
    > >>

    >
    >
    >
     
    RWH, Aug 18, 2006
    #4
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