I got it all correct but it only shows 90% in my score.
networktut
October 17th, 2020
@Got it all correct: Thanks for your information, we have just fixed it!
Rust
November 18th, 2020
Hi. I checked and found that right answer for question number 4 is A.
R1 will send to R2 by RIP route 10.1.1.0/24 and AD 120, but when on R2 will come same route from R3 with AD 110, in will create loop. That’s why if you decrease AD from 120 to 100 this route will appear in RIB.
B wrong answer, because OSPF on R3 doesn’t know about 10.1.1.0/24. R3 learned 10.1.1.0/24 from EIGRP. That’s why when you will redistribute OSPF to EIGRP this route will no appear on R4.
Getafix
November 25th, 2020
Hi Rust,
I thought about your answer but it seems to be incorrect (atleast as i see it). There are 2 redistributions happening.
RIP–>EIGRP = R2 will inject 10.1.1.0/24 into EIGRP and prefix will reach R3 via EIGRP and AD 90
EIGRP –> OSPF = R3 will inject 10.1.1.0/24 into OSPF and prefix will reach R2 via OSPF and AD 110
OSPF–> EIGRP = R3 will inject 10.1.1.0/24 into EIGRP and prefix will reach back R2 via EIGRP and AD 90
R2 now has 3 routes to choose from for 10.1.1.0/24 (RIP=120, OSPF= 110, EIGRP=90)
Hence 90 route gets installed in the routing table. This AD 90 route is only received because it was being redistributed from OSPF back into EIGRP unfiltered.
The correct answer would be to filter the routes from OSPF –> EIGRP
I hope this makes sense
Thanks
Retake
March 30th, 2022
@Rust is correct. Answer is A.
I’ve labbed it. Answer A corrects the issue which results in all PC on R2, R3, and R4 being able to reach PC1.
@Getafix: The AD of EIGRP external routes is 170.
@networktut:
Please kindly review and update the correct answer for Q4. Thank you.
I got it all correct but it only shows 90% in my score.
@Got it all correct: Thanks for your information, we have just fixed it!
Hi. I checked and found that right answer for question number 4 is A.
R1 will send to R2 by RIP route 10.1.1.0/24 and AD 120, but when on R2 will come same route from R3 with AD 110, in will create loop. That’s why if you decrease AD from 120 to 100 this route will appear in RIB.
B wrong answer, because OSPF on R3 doesn’t know about 10.1.1.0/24. R3 learned 10.1.1.0/24 from EIGRP. That’s why when you will redistribute OSPF to EIGRP this route will no appear on R4.
Hi Rust,
I thought about your answer but it seems to be incorrect (atleast as i see it). There are 2 redistributions happening.
RIP–>EIGRP = R2 will inject 10.1.1.0/24 into EIGRP and prefix will reach R3 via EIGRP and AD 90
EIGRP –> OSPF = R3 will inject 10.1.1.0/24 into OSPF and prefix will reach R2 via OSPF and AD 110
OSPF–> EIGRP = R3 will inject 10.1.1.0/24 into EIGRP and prefix will reach back R2 via EIGRP and AD 90
R2 now has 3 routes to choose from for 10.1.1.0/24 (RIP=120, OSPF= 110, EIGRP=90)
Hence 90 route gets installed in the routing table. This AD 90 route is only received because it was being redistributed from OSPF back into EIGRP unfiltered.
The correct answer would be to filter the routes from OSPF –> EIGRP
I hope this makes sense
Thanks
@Rust is correct. Answer is A.
I’ve labbed it. Answer A corrects the issue which results in all PC on R2, R3, and R4 being able to reach PC1.
@Getafix: The AD of EIGRP external routes is 170.
@networktut:
Please kindly review and update the correct answer for Q4. Thank you.