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Mohamed G. Gouda                                                CS 356
Spring 2009                                                 	Homework 2
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1. (4 points)
The connection protocols in Chapter 6 are designed under the assumption that
sent messages can be lost or reordered before they are received. Design a
simplex connection protocol, where a process p can establish a connection and
send up to m data messages to another process q, under the assumption that sent
messages are not corrupted, lost, nor reordered. Specify your designed protocol
using the AP notation.

Solution:

process p
inp m      : integer
var conp   : boolean     {init. false}
    cansnd : 0..m
    readyp : boolean     {init. true}
    
begin
    ~conp ^ readyp   -> send crqst to q; readyp:=false
[]  rcv crply from q -> conp,cansnd,readyp:=true,m,true  
[]  conp ^ readyp    -> if cansnd>0 -> send data to q;
                                       cansnd:=cansnd-1
                        []  true    -> send drqst to q;
                                       readyp:=false
                        fi
[]  rcv drply from q -> conp,readyp:=false,true
end

process q

var conq   : boolean    {init. false}

begin
    rcv crqst from p -> if true -> conq:=true; send crply to p
                        [] true -> conq:=false;send drply to p
                        fi
[]  conq             -> conq:=false; send drply to q
[]  rcv data from p  -> if conq          -> {accept data} skip
                        [] ~conq         -> {discard data} skip
                        fi
[]  rcv drqst from p -> if conq          -> conq:=false;
                                            send drply to p
                        [] ~conq         -> skip
                        fi                                                    
end
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2. (3 points) 
Extend the simplex protocol that you designed in solving problem 1 to a multi-
connection full-duplex protocol where each of two processes r and s can 
establish up to v connections, and send up to m data messages over each 
established connection, to the other process. Again the extended protocol is to
be designed under the assumption that sent messages are not corrupted, lost, 
nor reordered. Specify the extended protocol using the AP notation.

Solution:

process r
const v
inp   m             : integer
var   conp,conq     : array[0..v-1] of boolean      {init. false}
      cansnd        : array[0..v-1] of 0..m
      readyp        : array[0..v-1] of boolean      {init. false}
par   c             : 0..v-1

begin
    ~conp[c] ^ readyp[c] -> send crqst(c) to s; readyp[c]:=false
[]  rcv crply(c) from s  -> conp[c],cansnd[c],readyp[c]:=true,m,true  
[]  conp[c] ^ readyp[c]  -> if cansnd[c]>0 -> send data(c) to s;
                                              cansnd[c]:=cansnd[c]-1
                            []  true       -> send drqst(c) to s;
                                              readyp[c]:=false
                            fi
[]  rcv drply(c) from s  -> conp[c],readyp[c]:=false,true
[]  rcv crqst(c) from s  -> if true -> conq[c]:=true; send crply(c) to s
                            [] true -> conq[c]:=false;send drply(c) to s
                            fi
[]  conq[c]              -> conq[c]:=false; send drply(c) to p;
[]  rcv data(c) from s   -> if conq[c]            -> {accept data} skip
                            [] ~conq[c]           -> {discard data} skip
                            fi
[]  rcv drqst(c) from s  -> if conq[c]            -> conq[c]:=false;
                                                     send drply(c) to p
                            [] ~conq[c]           -> skip
                            fi
end
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3. (3 points)
Consider the corruption detection protocol that uses parity bits in Section 
8.1. This protocol is to be used to transmit blocks of 6000 data bits each, 
after augmenting each block with n parity bits. Assume that only the first
1000 bits and the last 1000 bits in an augmented block can be corrupted while
this block is being transmitted. What are the best values for the two constants
m and n in this protocol in this case? Explain your answer.  

Solution:

The smallest value of n that guarantees detecting every corruption
is 2000. Thus m = 3 and n = 2000. 
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