statistical multiplexing and end-to-end principle
TRANSCRIPT
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Statistical Multiplexing End-to-End Principle
EE122 TAs 9/7/12
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Statistical Multiplexing Sharing of a single link over multiple flows on
demand, allocating only for the average bandwidth needed
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Statistical Multiplexing Sharing 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?
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Statistical Multiplexing Sharing 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?
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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…
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Hotel Telephone Operator Kay
Hey Joshua, did you report your lecture participation to
[email protected] yet?
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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.
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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
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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...
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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?
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Hotel Telephone Operator Kay
GO SPAIN!!!
ITALY!!!
Mmm… Hershey’s
SPAIN!!!
GO ITALY!
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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.
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A Similar Problem
The World Shared Link
Alice, Bob, Eve and Mallory each have a cat video They want to broadcast this to the world But they share a single link to the world
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A Similar Problem
The World
First they tried sharing the link by dividing the available bandwidth evenly
TOO SLOW!!
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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
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Statistical Multiplexing
The World
Better: Everyone gets more bandwidth Less wasted bandwidth, but not reserving a circuit
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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
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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
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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
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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.
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End-to-End Principle
• Packets sometimes get lost
• Have reliable links, each hop resends packets that were lost
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Host A Host B
Reliable Links
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Host A Host B
Unreliable Links
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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?
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The 5 Layers
Application
Transport
Network
Datalink
Physical
L5
L1
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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!!!
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Host A Host B
LOLCat Switch
Kind of Cat Description
Witty Caption cat253.avi
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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
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Probability Primer
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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
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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 = 1
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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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