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Editorial Team
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Asked: 2026-06-15T01:10:41+00:00

Consider a router that interconnects three subnets: Subnet1, Subnet2, and Subnet3. Suppose all of

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Consider a router that interconnects three subnets: Subnet1, Subnet2, and Subnet3.
Suppose all of the interfaces in each of these three subnets are required to have the prefix
223.1.17/24. Also suppose that Subnet #1 is to support at least 60 interfaces
Subnet #2 is to support at least 90 interfaces
Subnet #3 is to support at least 12 interfaces. Provide
three network addresses (of the form a.b.c.d/x) that satisfy these constraints.

What I thought is
Subnet 1: 223.1.17.0/26
Subnet 2: 223.1.17.64/25
Subnet 3 : 223.1.17.192/28

Is this correct?

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  1. Editorial Team

    No. You have the size calculations done correctly but subnet 1 and 2 will overlap. A /25 network must start at either 0 or 128.

    The requirements for each network is:
    60 rounded up to the nearest power of two is 64 => we need 6 bits
    Subnet #2 needs 90 addresses so again, rounding up to 128 => we need 7 bits
    Subnet #3 only has 12 adresses, rounded gives us 16 => 4 bits.

    You can fit this in one of two ways (you can permutate subnet 1 and three in more). 

    subnet 1 223.1.17.128/26 
subnet 2 223.1.17.0/25
subnet 3 223.1.17.192/28

    or

    subnet 1 223.1.17.0/26 
subnet 2 223.1.17.128/25
subnet 3 223.1.17.64/28
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