Internet Engineering Task Force                       Christian E. Hopps
INTERNET-DRAFT                                             Merit Network
Expires July 1999                                        26 January 1999

             Analysis of an Equal-Cost Multi-Path Algorithm

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   Equal-cost multi-path (ECMP) is a routing technique for routing
   packets along multiple paths of equal cost.  The forwarding engine
   identifies paths by next-hop.  When forwarding a packet the router
   must decide which next-hop (path) to use. This document gives an
   analysis of one method for making that decision.  The analysis
   includes the performance of the algorithm and the disruption caused
   by changes to the set of next-hops.

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1.  Hash-Threshold

   One method for determining which next-hop to use when routing with
   ECMP can be called hash-threshold.  The router first selects a key by
   performing a hash (e.g., modulo-K where K is large, or CRC16) over
   the packet header fields that identify a flow. The N next-hops have
   been assigned unique regions in the key space. The router uses the
   key to determine which region and thus which next-hop to use.

   As an example of hash-threshold, upon receiving a packet the router
   performs a CRC16 on the packet's header fields that define the flow
   (e.g., the source and destination fields of the packet), this is the
   key.  Say for this destination there are 4 next-hops to choose from.
   Each next-hop is assigned a region in 16 bit space (the key space).
   For equal usage the router may have chosen to divide it up evenly so
   each region is 65536/4 or 16k large.  The next-hop is chosen by
   determining which region contains the key (i.e., the CRC result).

2.  Analysis

   There are a few concerns when choosing an algorithm for deciding
   which next-hop to use. One is performance, the computational
   requirements to run the algorithm.  Another is disruption (i.e., the
   changing of which path a flow uses).  Balancing is a third concern;
   however since the algorithm's balancing characteristics are directly
   related to the chosen hash function this analysis does not treat this
   concern in depth.

   For this analysis we will assume regions of equal size.  If the hash
   function is uniformly distributed the distribution of flows amongst
   paths will also be uniform.

2.1.  Performance

   The performance of the hash-threshold algorithm can be broken down
   into three parts: selection of regions for the next-hops, obtaining
   the key and comparing the key to the regions to decide which next-hop
   to use.

   Since regions are restricted to be of equal size the calculation of
   region boundaries is trivial.  The boundaries can be calculated as

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              i = 0;
              regionssize = keyspace.size / #{ next-hops }
              for n in { next-hops }
                   n.start =  i * regionsize;
                   n.end = n.start + regionsize;
                   i = i + 1

   This calculation is O(N).  Furthermore the calculation can be done
   when the next-hops are added to or removed from the destination

   The algorithm doesn't specify the hash function used to obtain the
   key.  Its performance in this area will be exactly the performance of
   the hash function.  It is presumed that if this calculation proves to
   be a concern it can be done in hardware parallel to other operations
   that need to complete before deciding which next-hop to use.

   Finally the next-hop must be chosen. This is done by determining
   which region contains the key.  The time required to do this is
   dependent on the way the next-hops are organized in memory. The
   problem reduces to a search.  For example if the next-hops are stored
   as a linked list the time is O(N) as the router must traverse the
   list comparing each next-hop's region boundaries against the key.  If
   the next-hops are stored as an ordered array a binary search can be
   used yielding O(logN).

   As [1] notes if the number of next-hops is limited to a fixed maximum
   the comparison can be done in parallel in hardware, thus O(1).

2.2.  Disruption

   Protocols such as TCP perform better if the path they flow along does
   not change while the stream is connected.  Disruption is the
   measurement of how many flows have their paths changed due to some
   change in the router.  We measure disruption as the fraction of total
   flows whose path changes in response to some change in the router.

   Some algorithms such as round-robin (i.e., upon receiving a packet
   the least recently used next-hop is chosen) are disruptive regardless
   of any change in the router.  Clearly this is not the case with hash-
   threshold.  As long as the region boundaries remain unchanged the
   same next-hop will be chosen for a given flow.

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   Because we have required regions to be equal in size the only reason
   for a change in region boundaries is the addition or removal of a
   next-hop.  In this case the regions must all grow or shrink to fill
   the key space.  The analysis begins with some examples of this.

             |   1   |   2   |   3   |   4   |   5   |
             |    1    |    2    |    4    |    5    |

              Figure 1. Before and after deletion of region 3

   In figure 1. region 3 has been deleted.  The remaining regions grow
   equally and shift to compensate.  In this case 1/4 of region 2 is now
   in region 1, 1/2 (2/4) of region 3 is in region 2, 1/2 of region 3 is
   in region 4 and 1/4 of region 4 is in region 5.  Since each of the
   original regions represent 1/5 of the flows, the total disruption is
   1/5*(1/4 + 1/2 + 1/2 + 1/4) or 3/10.

   Note that the disruption to flows when adding a region is equivalent
   to that of removing a region.  That is, we are considering the
   fraction of total flows that changes regions when moving from N to
   N-1 regions, and that same fraction of flows will change when moving
   from N-1 to N regions.

             |   1   |   2   |   3   |   4   |   5   |
             |    1    |    2    |    3    |    5    |

              Figure 2. Before and after deletion of region 4

   In figure 2. region 4 has been deleted.  Again the remaining regions
   grow equally and shift to compensate.  1/4 of region 2 is now in
   region 1, 1/2 of region 3 is in region 2, 3/4 of region 4 is in

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   region 3 and 1/4 of region 4 is in region 5.  Since each of the
   original regions represent 1/5 of the flows the, total disruption is

   To generalize, upon removing a region K the remaining N-1 regions
   grow to fill the 1/N space.  This growth is evenly divided between
   the N-1 regions and so the change in size for each region is
   1/N/(N-1) or 1/(N(N-1)).  This change in size causes non-end regions
   to move.  The first region grows and so the second region is shifted
   towards K by the change in size of the first region. 1/(N(N-1)) of
   the flows from region 2 are subsumed by the change in region 1's
   size. 2/(N(N-1)) of the flows in region 3 are subsumed by region 2.
   This is because region 2 has shifted by 1/(N(N-1)) and grown by
   1/(N(N-1)).  This continues from both ends until you reach the
   regions that bordered K.  The calculation for the number of flows
   subsumed from the Kth region into the bordering regions accounts for
   the removal of the Kth region.  Thus we have the following equation.

                           K-1              N
                           ---    i        ---  (i-K)
             disruption =  \     ---    +  \     ---
                           /   (N)(N-1)    /   (N)(N-1)
                           ---             ---
                           i=1            i=K+1

   We can factor 1/((N)(N-1)) out as it is constant.

                                /  K-1         N        \
                          1     |  ---        ---       |
                     =   ---    |  \    i  +  \   (i-K) |
                       (N)(N-1) |  /          /         |
                                \  ---        ---       /
                                     1        i=K+1

   We now use the the concrete formulas for the sum of integers.  The
   first summation is (K)(K-1)/2.  For the second summation notice that
   we are summing the integers from 1 to N-K, thus it is (N-K)(N-K+1)/2.

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                             (K-1)(K) + (N-K)(N-K+1)
                           = -----------------------

   Considering the summations, one can see that the least disruption is
   when K is as close to half way between 1 and N as possible.  This can
   be proven by finding the minimum of the concrete formula for K
   holding N constant.  First break apart the quantities and collect.

                            2K*K - 2K - 2NK + N*N + N
                          = -------------------------

                             K*K - K - NK      N + 1
                          = --------------  + -------
                               (N)(N-1)        2(N-1)

   Since we are minimizing for K the right side (N+1)/2(N-1) is constant
   as is the denominator (N)(N-1) so we can drop them.  To minimize we
   take the derivative.

                             -- (K*K - (N+1)K)

                             = 2K - (N+1)

   Which is zero when K is (N+1)/2.

   The last thing to consider is that K must be an integer.  When N is
   odd (N+1)/2 will yield an integer, however when N is even (N+1)/2
   yields an integer + 1/2.  In the case, because of symmetry, we get
   the least disruption when K is N/2 or N/2 + 1.

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   Now since the formula is quadratic with a global minimum half way
   between 1 and N the maximum possible disruption must occur when edge
   regions (1 and N) are removed.  If K is 1 or N the formula reduces to

   Thus to minimize disruption we recommend adding new regions to the
   center rather than the ends.

3.  Comparison to other algorithms

   Other algorithms exist to decide which next-hop to use.  These
   algorithms all have different performance and disruptive
   characteristics.  Of these algorithms we will only consider ones that
   are not disruptive by design (i.e., if no change to the set of next-
   hops occurs the path a flow takes remains the same).  This will
   exclude round-robin and random choice.  We will look at modulo-N and
   highest random weight.

   Modulo-N is a simpler form of hash-threshold.  Given N next-hops the
   hash function includes a final modulo-N which directly maps the
   result to one of the next-hops.  This operation is the fastest of the
   three we consider, however if a next-hop is added or removed the
   disruption is (N-1)/N.

   Highest random weight (HRW) is another comparative method similar to
   hash-threshold.  The router seeds a pseudo-random number generator
   with the packet header fields which describe the flow and the next-
   hop to obtain a weight.  The next-hop which receives the highest
   weight is selected.  The advantage with using HRW is that it has
   minimal disruption (i.e., disruption due to adding or removing a
   next-hop is always 1/N.)  The disadvantage with HRW is an only
   slightly more complex function to choose the next-hop.  A description
   of HRW along with comparisons to other methods can be found in [1].
   Although not used for next-hop calculation an example usage of HRW
   can be found in [2].

   If the complexity of HRW's next-hop selection processes is acceptable
   we think it should be considered as an alternative to hash-threshold.

4.  Security Considerations

   This document is an analysis of an algorithm used to implement an
   ECMP routing decision.  This analysis does not directly effect the

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   security of the Internet Infrastructure.

5.  References

[1]  Thaler, D., and C.V. Ravishankar, "Using Name-Based Mappings to
     Increase Hit Rates", IEEE/ACM Transactions on Networking, February

[2]  Estrin, D., Farinacci, D., Helmy, A., Thaler, D., Deering, S.,
     Handley, M., Jacobson, V., Liu, C., Sharma, P., and L. Wei,
     "Protocol Independent Multicast-Sparse Mode (PIM-SM): Protocol
     Specification", RFC 2362, June 1998.

6.  Author's Address

     Christian E. Hopps
     Merit Network
     4251 Plymouth Road, Suite C.
     Ann Arbor, MI  48105
     Phone: +1 734 936 0291

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