# Alternating Bit Protocol

Document Sample

```					    Sliding window protocol
last + w

(s, r)
next
S                  R      j

last
:                  (r, s)           }    accepted
messages

.
.
The sender continues the send action
without receiving the acknowledgements of at most
w messages (w > 0), w is called the window size.
Sliding window protocol
{program for process S}                   {program for process R}
define   next, last, w : integer;         define j :     integer;
initially next = 0, last = -1, w > 0      initially j = 0;

do (m[next],       next)     is      received   
do last+1 ≤ next ≤ last + w 
send (m[next], next); next := next + 1
if j = next  accept message;
 (ack, j) is received                                              send (ack, j);
if   j > last last := j                               j:= j+1
     j ≤ last     skip               j ≠ next  send (ack, j-1)
fi                                    fi;
 timeout (R,S)  next := last+1          od

{retransmission begins}
od
Why does it work?
Lemma. Every message is accepted exactly once.

Lemma. m[k] is always accepted before m[k+1].
(Argue that these are true.)

Observation. Uses unbounded sequence number.
This is bad. Can we avoid it?
Theorem

If the communication channels are non-FIFO, and the message
propagation delays are arbitrarily large, then using bounded
sequence numbers, it is impossible to design a window protocol
that can withstand the (1) loss, (2) duplication, and (3) reordering
of messages.
Why unbounded sequence no?

(m’’,k)   (m’, k)   (m[k],k)

New message           Retransmitted
using the same        version of m
seq number k

We want to accept m” but reject m’. How is that possible?
Alternating Bit Protocol
m[2],0   m[1],1   m[0],0   m[0],0

S                                              R

ack, 0

ABP is a link layer protocol. Works on FIFO channels only.
Guarantees reliable message delivery with a 1-bit sequence
number (this is the traditional version with window size = 1).
Study how this works.
Alternating Bit Protocol
program ABP;
{program for process S}
define sent, b : 0 or 1; next : integer;
initially next = 0, sent = 1, b = 0, and channels are empty;            S
do sent ≠b               send (m[next], b);
next := next+1; sent := b            m[2],0
 (ack, j) is received  if j = b  b := 1-b
 j ≠ b  skip
m[1],1
fi                                         a,0
timeout (R,S)              send (m[next-1], b)                m[0],0
od
{program for process R}
m[0],0
define j : 0 or 1; {initially j = 0};
do          (m[ ], b) is received 
if j = b  accept the message;
send (ack, j); j:= 1 - j                      R
 j ≠ b  send (ack, 1-j)
fi
od
How TCP works
Supports end-to-end logical connection between any two
computers on the Internet. Basic idea is the same as those of
sliding window protocols. But TCP uses bounded sequence
numbers!
It is safe to re-use a sequence number when it is unique. With a
high probability, a random 32 or 64-bit number is unique. Also,
current sequence numbers are flushed out of the system after a
time = 2d, where d is the round trip delay.
How TCP works
Sender                           Recei ver

SYN seq = x

SYN, seq=y, ack = x+1

ACK, ack=y+1

send (m, y+1)

ack (y+2)
How TCP works

• Three-way handshake. Sequence numbers are unique w.h.p.
• Why is the knowledge of roundtrip delay important?
• What if the window is too small / too large?
• What if the timeout period is too small / too large?
and control the window size to save its buffer space.
Distributed Consensus
Reaching agreement is a fundamental problem in distributed
computing. Some examples are

Commit or Abort in distributed transactions
Reaching agreement about which process has failed
Clock phase synchronization
Air traffic control system: all aircrafts must have the same view

If there is no failure, then reaching consensus is trivial. All-to-all broadcast
Followed by a applying a choice function … Consensus in presence of
failures can however be complex.
Problem Specification

input                        output

p0     u0                             v
p1     u1                             v
p2     u2                              v
p3     u3                              v

Here, v must be equal to the value at some input line.
Also, all outputs must be identical.
Problem Specification
Termination.   Every non-faulty process must eventually decide.
Agreement.     The final decision of every non-faulty process
must be identical.
Validity.      If every non-faulty process begins with the same
initial value v, then their final decision must be v.
Asynchronous Consensus

Seven members of a busy household decided to hire a cook, since they do not
have time to prepare their own food. Each member separately interviewed
every applicant for the cook’s position. Depending on how it went, each
member voted "yes" (means “hire”) or "no" (means “don't hire”).
These members will now have to communicate with one another to reach a
uniform final decision about whether the applicant will be hired. The process
will be repeated with the next applicant, until someone is hired.

Consider various modes of communication…
Asynchronous Consensus

Theorem.
In a purely asynchronous distributed system,
the consensus problem is impossible to solve
if even a single process crashes

Famous result due to Fischer, Lynch, Patterson
(commonly known as FLP 85)
Proof
Bivalent and Univalent states

A decision state is bivalent, if starting from that state, there exist
two distinct executions leading to two distinct decision values 0 or 1.
Otherwise it is univalent.

A univalent state may be either 0-valent or 1-valent.
Proof

Lemma.
No execution can lead from a 0-valent to a 1-valent
state or vice versa.

Proof.
Follows from the definition of 0-valent and 1-valent states.
Proof
Lemma. Every consensus protocol must have a bivalent initial state.

Proof by contradiction. Suppose not. Then consider the following input patterns:

s[0]       0 0 0 0 0 0 …0 0 0         {0-valent)
0 0 0 0 0 0 …0 0 1                             s[j] is 0-valent
0 0 0 0 0 0 …0 1 1                             s[j+1] is 1-valent
…         …      …         …                   (differ in jth position)
s[n-1]     1 1 1 1 1 1 …1 1 1         {1-valent}

What if process (j+1) crashes at the first step?
Proof
The system from reaching
Lemma.                                                                          consensus
bivalent
Q
In a consensus protocol,
starting   from    any    initial   bivalent                  bivalent         bivalent           bivalent

bivalent state, there must            S                   R                U                  T

exist a reachable bivalent                     action 0           action 1       action 0              action 1

state T, such that every                         R0               R1                T0                 T1
action taken by some process                   o-valent         1-valent           o-valent           1-valent

p in state T leads to either a
0-valent or a 1-valent state.                   Actions 0 and 1 from T must be
taken by the same process p. Why?

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 21 posted: 6/17/2012 language: English pages: 19