statistical multiplexing end-to-end principle ee122 tas 9/7/12
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Statistical Multiplexing End-to-End Principle
EE122 TAs9/7/12
Statistical MultiplexingSharing of a single link over multiple flows on
demand, allocating only for the average bandwidth needed
Statistical MultiplexingSharing of a single link over multiple flows on
demand, allocating only for the average bandwidth needed
Packet Switching or Circuit Switching?Packet Switching or Circuit Switching?
Statistical MultiplexingSharing of a single link over multiple flows on
demand, allocating only for the average bandwidth needed
Packet Switching or Circuit Switching?Packet Switching or Circuit Switching?
Hotel Telephone Operator
Colin, Andrew, Panda, Thurston and Scott check into Hotel Durant, which accommodates 5
Once in a while, one of them makes a call using the hotel telephone
Sometimes two of them call at the same time…Sometimes three of them call at the same time…
Hotel Telephone Operator Kay
Hey Joshua, did you report your lecture participation to
[email protected] yet?
Hotel Telephone Operator Kay
Hey Joshua, did you report your lecture participation to
[email protected] yet?
Yeah, the HW looks long, so get started
early. Hint hint.
Hotel Telephone Operator Kay
Hey Joshua, did you report your lecture participation to
[email protected] yet?
Yeah, the HW looks long, so get started
early. Hint hint.
Mmm… Cadbury chocolate
Hotel Telephone Operator
Sometimes four of them call at the same time…Sometimes all five call at the same time…
Let’s say each phone conversation lasts 5 minutes (300 seconds) on average.
There are 86400 seconds in a day… what are the chances these calls will overlap?
Not very high...
Hotel Telephone Operator
How many lines do we need?Do we need all 5?Let’s say we allocate 3.
What will happen if all 5 call at the same time?
Hotel Telephone Operator Kay
GO SPAIN!!!
ITALY!!!
Mmm… Hershey’s
SPAIN!!!
GO ITALY!
Hotel Telephone Operator
How many lines do we need?Do we need all 5?Let’s say we allocate 3.
What will happen if all 5 call at the same time?Not everyone can be serviced, and there will be some dropped calls.But the probability of this happening is very low.
A Similar Problem
The WorldShared Link
Alice, Bob, Eve and Mallory each have a cat videoThey want to broadcast this to the worldBut they share a single link to the world
A Similar Problem
The World
First they tried sharing the link by dividing the available bandwidth evenly
TOO SLOW!!
Hope For A Solution
• They could go one at a time– We decided this is bad
• But notice they are not really sending at the same time
• The chance of someone sending is a random variable
• Let us try and apply the law of large numbers
Statistical Multiplexing
The World
Better: Everyone gets more bandwidthLess wasted bandwidth, but not reserving a circuit
Statistical Multiplexing
• Works because everyone sends at random times
• What is the expected bandwidth used?• What would we need if we reserved
bandwidth for all?• When does this fail?
• TADA INTERNET
The Law of Large Numbers
The average value taken by a large set of observations of a random variables approaches the
mean of the random variable
• For our purposes– The sum of a set of random variables is
Sum = Mean × Number of Variables• Note this is smaller than
Maximum × Number of Variables
The Law of Large Numbers
• Example:– 30 end hosts, each want to send at 100kbps– But each host only sends 10% of the time
How much bandwidth do we need to guarantee that everyone gets 100kbps no matter what?
30 * 100kbps = 3Mbps
The Law of Large Numbers• Example:– 30 end hosts, each want to send at 100kbps– But each host only sends 10% of the time
How much bandwidth does the Law of Large Numbers say we need?
100kbps * 10% * 30 = 300kbps
Statistical Multiplexing says we only need a link with 300kbps, by the Law of Large Numbers.
End-to-End Principle
• Packets sometimes get lost• Have reliable links, each hop resends packets
that were lost
Host A Host B
Reliable Links
Host A Host B
Unreliable Links
End-to-End Principle
• Packets sometimes get lost• Have reliable links, each hop resends packets
that were lost• Does this violate the end-to-end principle?• Is this ever a reasonable scheme?– Remember End-To-End has an exception
• Does this violate layering?
The 5 Layers
Application
Transport
Network
Datalink
Physical
L5
L1
End-to-End Principle
• One needs a more perfect mechanism for delivering cat video
• Introducing the LOLCat Switch• Produce a cat video at the switch– Given a description of the kind of cat, props, and
witty caption, network switches will assemble a cat video
– Really fast, don’t need to go beyond the first hop– Ever!!!
Host A Host B
LOLCat SwitchKind of CatDescription
Witty Captioncat253.avi
End-to-End Principle
• Does this violate the end-to-end principle?• Does it violate layering?• Why is this a bad idea?– Duplicating application functionality– Unnecessarily complicating network– Inflexible switch design
Probability Primer
Independent Events• Independent: occurrence of one event does not
affect the likelihood of the other event• Example of two independent events:– Flow 1 sends 5 packets in a particular frame– Flow 2 sends 2 packets in a particular frame
• Assume that:– Event A happens with probability pA
– Event B happens with probability pB
• What is probability both event A and B happen?• Answer: Probability = pApB 30
Consider Dice
• Probability that rolling a single die yields a 1? 1/6th
• Probability rolling two dice yields 1 1?1/36th
• Probability rolling two dice yields a 1 and a 2?– Write it down!
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Be Careful Counting Events
First die is 1 and second die is 2:1/36th
One die is 1 and one die is 2: 1/18th
• That’s because you could have 1 2 or 2 1
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Mutually Exclusive Events
• First roll of die is 1:1/6th
• First roll of die is 2:1/6th
• First roll of die is 3:1/6th
• First roll of die is 4:1/6th
• First roll of die is 5:1/6th
• First roll of die is 6:1/6th
• This set of events is complete (or exhaustive) in that one of them must be true
Total of probabilities = 133
Exclusive vs Independent Events
• First roll of die is 1, second roll of die is 3– Independent
• First roll of die is 1, first roll of die is 2– Exclusive
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Computing Averages
• Assume that x is some property of an event and that events A, B, C are mutually exclusive and complete (i.e., one of them happens)– E.g., x = number of packets sent in a particular frame
• Assume that:– Event A has x=5– Event B has x=2– Event C has x=10
• What is the average value of x? (denoted by <x>)– Write it down!
• Average <x> = 5pA + 2pB + 10pC
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THIS IS ALL YOU NEED TO KNOW!
• The problems are easier than you think…• …but think clearly before computing the
answer…
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