You have a DHCP server named Server1 that runs Windows Server 2016.
You have a single IP subnet.
Server1 has an IPv4 scope named Scope1. Scope1 has an IP address range of 10.0.1.10 to 10.0.1.200 and a
length of 24 bits.
You need to create a second logical IP network on the subnet. The subnet will use an IP address range of
10.0.2.10 to 10.0.2.200 and a length of 24 bits.
What should you do?
A.
Create a second scope, and then create a superscope.
B.
Create a superscope, and then configure an exclusion range in Scope1.
C.
Create a new scope, and then modify the IPv4 bindings.
D.
Create a second scope, and then run the DHCP Split-Scope Configuration Wizard.
Explanation:
References: https://technet.microsoft.com/en-us/library/cc958938.aspx
Agree
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Correct Answer: A
The only thing you need to do to meet the question criteria is Create a second scope. However there would be no need to split the scope which would rule out D.
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You cannot create a second scope as the first scope will be in the form of 10.0.0.0/8. If you try to create another scope with 10.0.2.200 you will be prompted with a warning saying that the address range and mask are in conflict with an existing scope. The way you should go about this is change the first scope’s end IP address from 10.0.1.200 to 10.0.2.200 and it will look like 10.0.1.10-10.0.2.200
A. Create a second scope, and then create a superscope – not working because of the aforementioned warning that doesn’t let you continue
B. Create a superscope, and then configure an exclusion range in Scope1. – in order to create a superscope you need both scopes configured first, which you can’t because of the conflict explained above
C. Create a new scope, and then modify the IPv4 bindings. This would be the most accurate answer but not entirely correct. You don’t need to create a new scope, you can modify the IPv4 bindings of the existing scope.
D. Create a second scope, and then run the DHCP Split-Scope Configuration Wizard. – A split scope will require a secondary DHCP server and is not fit for purpose in this specific example.
I’d go for C. Try it in your LAB
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