Q: 15
Which statement is correct about IS-IS?
Options
Discussion
Option C for sure. IS-IS runs as a link-state protocol-just like OSPF, it builds a complete topology map, not just next hops. It isn't distance vector or path vector, and classful/classless doesn't really apply since IS-IS doesn't carry IP subnet info in the same way RIP or EIGRP might. Pretty confident but correct me if you see conflicting Juniper docs.
C . IS-IS is always considered link-state, never distance or path vector. Juniper materials are clear about that. Disagree?
B is tempting since IS-IS path calculation might sound like path vector, but that's more for BGP. Not super confident though, since IS-IS does have some path characteristics.
A is wrong, C. IS-IS uses link-state updates, not distance vectors. Juniper always lists it as link-state in their docs.
C, IS-IS is always classified as link-state, not distance or path vector. Pretty sure that's what Juniper will expect.
Pretty simple, C. IS-IS is a link-state protocol, not distance or path vector. Open to pushback but I'm confident here.
C makes the most sense here. IS-IS uses SPF just like OSPF, so it's classic link-state not distance vector. If anyone has more insight on why A is tempting, let me know but I'm pretty confident.
C or B? IS-IS is generally classed as link-state, so C should be it. But I get why B might trip people up, since it calculates paths too. Still, I'm pretty sure Juniper sticks with link-state for IS-IS. Someone double-check me if that's out of date.
B is off, C is right. IS-IS is definitely a link-state protocol.
Nah, I don't think A or D fit here. IS-IS is definitely a link-state protocol, so C makes sense since it uses the SPF algorithm like OSPF. Pretty sure, unless Juniper changed something recently.
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