Fortinet NSE8_812: What is the Forward Error Correction Behavior for 500 Mbps Download with 8% Packet Loss?

Discover the Forward Error Correction (FEC) behavior for a Fortinet SD-WAN network with 500 Mbps download and 8% packet loss. Learn how to configure FEC mappings and understand the correct ratio of redundant to base packets in this scenario. Prepare for the Fortinet NSE8_812 certification exam with this detailed explanation.

Question

Review the VPN configuration shown in the exhibit.

config vpn ipsec fec
edit "fecprofile"
config mappings
edit 1
set base 8
set redundant 2
set packet-loss-threshold 10
next
edit 2
set base 9
set redundant 3
set bandwidth-up-threshold 450000
next
edit 3
set base 5
set redundant 3
bandwidth-bi-threshold 5000000
next
end
next
end

config vpn ipsec phase1-interface
edit "vdl-pl"
set fec-health-check "1"
set fec-mapping-profile "fecprofile "
set fec

What is the Forward Error Correction behavior if the SD-WAN network traffic download is 500 Mbps and has 8% of packet loss in the environment?

A. 1 redundant packet for every 10 base packets
B. 3 redundant packet for every 5 base packets
C. 2 redundant packet for every 8 base packets
D. 3 redundant packet for every 9 base packets

Answer

C. 2 redundant packet for every 8 base packets

Explanation

The VPN configuration in the exhibit shows three FEC mapping profiles:

1. base 8, redundant 2, packet-loss-threshold 10
2. base 9, redundant 3, bandwidth-up-threshold 450000
3. base 5, redundant 3, bandwidth-bi-threshold 5000000

In the given scenario, the SD-WAN network traffic download is 500 Mbps, which is below the bandwidth-bi-threshold of 5,000,000 Kbps (5 Gbps) specified in mapping 3. Therefore, mapping 3 does not apply.

The packet loss in the environment is 8%, which is below the packet-loss-threshold of 10% specified in mapping 1. As a result, mapping 1 will be applied.

Mapping 1 specifies that for every 8 base packets, 2 redundant packets will be added. This means that the Forward Error Correction behavior in this scenario will be 2 redundant packets for every 8 base packets.

Therefore, the correct answer is C. 2 redundant packets for every 8 base packets.

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