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message fragments transmitted on the VL. The right hand side is the maximum frame
frequency feasible with the given BAG.
If the inequation is disobeyed, the network cannot cope with the transmit schedule,
meaning that some messages will never be transmitted or cannot always be
transmitted as per their transmit interval. Aperiodically transmitted messages
(queuing port messages) are affected first, since they are lower prioritized as
periodically transmitted messages (sampling port messages).
⎡ ⎤
VL
n
msg
msg
msg VL
MTC BAG
PL MTU 1
1
≤ Σ
=
Formula 1: Bandwidth constraints of messages on the same VL
Transmit Scheduling Example
Formula 2 shows the calculations for an example of two messages, msg1 and msg2
being transmitted on the same VL. Since the payload of msg1 exceeds the MTU, the
message is divided into two fragments. The payload of msg2, however, fits the MTU
so it will not get fragmented. When inserting into Formula 1, it can be deduced that
the inequation is disobeyed, hence the transmit schedule is not feasible.
To get a working transmit schedule, its necessary to change one or more of the four
properties used in the inequation. Often it's easier and better to change the message
properties than the VL properties, since changing the latter affect all messages on that
particular VL. For example, the transmit schedule would work if MTCmsg2 is changed
to 10 ms. Alternatively, MTCmsg1 and MTCmsg2 can be changed to 8 ms and 4 ms,
respectively, which would also ensure that the formula is adhered to. In both cases the
AFDX Network Configuration
28/30 􀁟 700008_TUT-AFDX-EN_1000
left side of the inequation will exactly equal the right side, meaning that the VL
bandwidth is fully exploited.
⎡ ⎤ ⎡ ⎤
⎡ ⎤ ⎡ ⎤
!!
2
1
15
11
2
1
3
100 100
5
200 100
1
100 , 3
200 , 5
2 , 100
2
2
1
1
2 2
1 1
Inequation disobeyed
MTC BAG
PL MTU
MTC
PL MTU
PL bytes MTC ms
PL bytes MTC ms
BAG ms MTU bytes
msg VL
msg VL
msg
msg VL
msg msg
msg msg
VL VL
⇓
≤
⇓
+ ≤
⇓
+ ≤
= =
= =
= =
Formula 2: Example using Formula 1 for two messages on the same VL
Bandwidth Allocation
AFDX® / ARINC 664 Tutorial 􀁠 29/30
Bandwidth Allocation
As previously mentioned, the purpose of the switch is to police that the network’s
maximum bandwidth is not exceeded. The network load can be calculated by the sum
of the load of each message on the network. Since a message consists of a protocol
part as well as a payload part, it is necessary to take both parts into consideration
when performing the calculation. The AFDX protocol overhead is shown in Table 1.
Table 1: AFDX protocol overhead including leading and trailing bytes
In other words, every time an AFDX message is transmitted, 67 bytes alone is
protocol information which is a significant amount considering that the smallest
allowable AFDX payload size is merely 17 bytes.
With this information it is now possible to establish a formula for the network load of
an AFDX message as shown in Formula 3.
( )
( ) Mbps
MTC
PL
Message Network Load
bytes ms
MTC
PL
Message Network Load
msg
msg
msg
msg
1000
67
8
/
67
∗
+
=
+
=
c
Formula 3: AFDX message network load
Protocol Type Protocol Element Size (bytes)
IEEE802.3 Preamble 7
IEEE802.3 SFD (Start Frame Delimiter) 1
IEEE802.3 MAC (Media Access Controller) Destination 6
IEEE802.3 MAC Source 6
IEEE802.3 Ethernet Type 2
RFC IP (Internet Protocol) Header 20
RFC UDP (User Datagram Protocol) Header 8
AFDX RSN (Redundancy Sequence Number) 1
IEEE802.3 FCS (Frame Check Sequence) 4
IEEE802.3 IFG (Inter Frame Gap) 12
Total: 67
AFDX Network Configuration
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To establish a formula for calculating the maximum allowable total network load, it is
necessary to take into consideration that messages can be fragmented as was also the
case in Formula 1. Formula 4 shows the inequation which the network configuration
must adhere to in order not to exceed the maximum network bandwidth. The left hand
side of the inequation is the sum of the network loads caused by the transmitted
message fragments each having a payload of MTU bytes. The right hand side is the
　
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