Connection Preemption in MultiClass Networks

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Connection Preemption in Multi-Class Networks

• Fahad Rafique Dogar
• Work done while at LUMS, Pakistan

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Agenda

• Preemption Problem
• Earlier Work
• Our Contribution
• Conclusion

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Problem Scenario
7. Preemption decision for R4-gtR8
1. New connection request (R1,R8,bw,class)
5. Preemption decision for R6-gtR7
2. Makes an admission control decision If enough
bandwidth is available then accept the request
otherwise reject the request A third
possibility accept the request by preempting
lower priority connections
6. Preemption decision for R7-gtR4
3. Makes a constraint based routing decision Say
route R1-gtR6-gtR7-gtR4-gtR8

• Makes a preemption decision for R1-gtR6

We consider the problem of which connections to
preempt!!!
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Preemption Problem Constraint and Objectives

• What is the constraint while making the
  preemption decision?
• available bw preempted bw bw of new
  request
• Some possible objectives?
• Minimize the number of preempted connections
• Minimize the preempted bandwidth
• Minimize the priority of preempted connections
• We consider 1 and 2, in that order

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Earlier Work

• MinnConn Peyravian et al. Infocom99
• Enhanced version of our problem
• Considers priority as the third objective, so
  tries to achieve objectives 1,2, and 3, in that
  order
• Lets assume that priority of preemptable
  connections is the same i.e., we only consider
  bronze class traffic for preemption. So
  MinnConnOur Problem
• Authors claim MinnConn solves the problem
  optimally in polynomial time
• Lets verify the above claims!

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MinnConn
Bandwidth demand of the new request
Available bandwidth
While bw of new request is greater than available
bw
Bandwidth required for preemption
Finding the minimum bw connection that is greater
than the bw required for preemption --- i?0 if
such connection is found
If no SINGLE connection can fulfill the
preemption request
Finding the connection with the largest bw
Removing the chosen connection
Including it in the preemption set
Updating the available bandwidth
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MinnConn contd.

• Does it run in polynomial time?
• Inner loops (steps 4 and 11) run k times (where k
  is the number of connections in the preemptable
  set)
• Outer loop can also run a maximum of k times
  since in every iteration at least one connection
  is chosen for preemption
• So complexity is O(k2)
• Is it optimal?
• Consider C70,50,50,20 Bp 100 and aj 0
• MinnConn result 70,50 while optimal
  result50,50
• Greedy approach of selecting the largest
  connection is sub-optimal

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Our Contribution

• We show that solving this problem optimally in
  polynomial time is highly unlikely
• Prove that this problem is NP-complete by
  reducing it to the subset sum problem
• Propose exact and approximate algorithms to solve
  this problem
• Exact algorithm is optimal and runs in
  exponential time
• Polynomial time approximation algorithm gives a
  bounded difference from the optimal

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NP-completeness Proof

• Subset Sum (SS) Problem
• Given a set Va1,,an of n positive integers
  and a number t, is there any subset S of V, such
  that
• How is it different from our problem?
• Yes/No problem (rather than finding a set)
• Sum is made equal to threshold (rather than
  overshoot)
• No restriction on the cardinality of the solution
  subset
• This difference is the key to reducing our
  problem to the subset sum problem

3 differences
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Proof (Contd.)

• How to solve the subset sum problem using the
  solution to our problem? Basic idea
• Introduce n dummy connections, each corresponding
  to a valid connection
• Choose n connections from 2n options either an
  actual connection or its dummy counterpart (but
  not both) is selected
• Dummy values are selected for those connections
  that are not part of the actual solution
• From the resulting set of cardinality n, discard
  the dummy values
• if solution sum equal to threshold (and not
  greater) then a subset exists whose sum is equal
  to threshold otherwise no subset exists

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Proof (Contd.)

• Examples (threshold100 in each case V and S
  are the input and output of our algorithm
  respectively)
• V30,60,80,10,D1,D2,D3,D4 S30,60,10,D3 SS
  solution --- YES
• V100,110,130,150,D1,D2,D3,D4
  S100,D2,D3,D4 Solution ---YES
• V50,60,80,10,D1,D2,D3,D4 S50,60,D3,D4
  Solution --- NO
• Challenge
• How to modify the input and the threshold value
  such that the solution to our problem can be used
  (as described above) to solve SS problem

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Proof (Contd.)

• SS Input Va1,,an and t
• We construct Vc1,b1 ,cn,bn and t

Ensures that those cis are chosen that minimize
the overshoot from the threshold
Ensures that exactly n elements are chosen
Ensures that either ci or the corresponding bi
(but not both) is selected
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Proof --- Putting it together

• Given SS Input Va1,,an and t
• Check whether the sum of all elements exceed
  threshold (if not then no solution subset exists)
• Construct Vcis, bis and t and pass it to our
  problem
• Discard the dummy elements from the solution set
• Keep the l most significant bits of cis
• If their sum equals threshold then a subset
  exists whose sum is equal to the threshold else
  no subset exists
• Steps 1,3,4,and 5 can be performed in polynomial
  time
• If our problem solver in step 2 can run in
  polynomial time then subset sum problem can be
  solved in polynomial time as well

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Exact Algorithm (V,t,K)

• In any iteration i, the length of L can be as
  long as 2i
• So algorithm has exponential complexity

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Approximate Algo.

• Similar to the exact algorithm but uses a trim
  function to reduce the length of L in each
  iteration
• Trimming
• If two values are quite close (within some factor
  (1 d)) then we can keep the larger one and
  discard the smaller value
• Keeping the larger value ensures that our
  solution is feasible though not optimal
• But solution is within (1 d)K of the optimal
• simulation results show that actual difference is
  much less

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Conclusion

• Our contribution
• Proof of NP-Completeness
• Exact algorithm
• Approximate Algorithm
• Other applications of this problem
• Process preemption in OS
• Job preemption in scheduling systems
• Take home message
• Dont blindly trust INFOCOM papers ?