CIS 725 Data Link Layer

Flow Control Producer-consumer problem Sliding window protocol - Go Back N - Selective retransmission

Go Back N na x x x x ns S:: ns = 1; na = 1 do ns < na + W /\ avail(nl_buff) /\ sbuff[ns] = null  sbuff[ns] = nl_buff; R!(sbuff[ns],ns); ns++ [] sbuff[ns] != null  R ! (sbuff[ns],ns); ns++ [] R ? ack(s)  if s > na then na = s [] timeout  ns = na od

R:: nr = 1 do S ? (D,x)  if nr = x then deliver(D); nr++ S ! Ack(nr) od

Acks sent but not yet received nr na ns ns <= na + W outstanding data packets ns na 1 1 nr W = 3 d1 2 1 1 3 1 d2 2 d3 4 1 3 a2 a3 4 2 5 2 d4 d5

Cumulative Acks ns na 1 1 nr W = 3 d1 2 1 1 d2 3 1 2 d3 4 1 3 a2 a3 4 1 5 1 d4 d5

- Alternative: Timeout per message d,1 d,1 d,2 d,3 a,2 a,3 d,4 d,2 d,3 a,2 start timer start timer a,2 start timer timeout d,1 start timer d,5 d,2 - Alternative: Timeout per message

Go Back N S:: ns = 1; na = 1 do ns < na + W /\ avail(nl_buff) /\ sbuff[ns] = null  sbuff[n] = nl_buff; R!(sbuff[ns],ns); ns++; if ns = na + W then start timer [] sbuff[ns] != null  R ! (sbuff[ns],ns); ns++ [] R ? ack(s)  if s > na then na = s; cancel timer [] timeout  ns = na od

A ------------------B Bandwidth =108 bytes/sec RTD = 2 msecs 2msecs ---- 2* 105 bytes 1 packet = 1000 bytes W = 200 W will also depend on buffer space at the receiver d,2 a,2 Timeout interval = 2msecs d,W

Selective Retransmission Retransmit only those packets the lost packets

Selective Retransmission S:: ns = 1; na = 1 do ns < na + W /\ avail(nl_buff) /\ sbuff[ns] = null  sbuff[n] = nl_buff; R!(sbuff[ns],ns); ns++ [] R ? nack(s, bitmap)  na = s; for (0 <= i < W) if bitmap[i] then R! (sbuff[s+i], s+i); [] timeout  R ! (sbuf[na], na) od

R:: nr = 1 do S ? (D,x)  rbuff[x] = D; recd[x] = true; if x != nr then S !nack(nr, bitmap) [] recd[nr]  deliver(rbuff[nr]); nr++ od

R:: nr = 1 do S ? (D,x)  rbuff[x] = D; recd[x] = true; if x != nr then S !nack(nr, bitmap) [] recd[nr]  deliver(rbuff[nr]); nr++ od

Receiver timeout R:: nr = 1 do S ? (D,x)  rbuff[x] = D; recd[x] = true; [] timeout  S !nack(nr, bitmap) [] recd[nr]  deliver(rbuff[nr]); nr++ od

Receiver based Windows S:: ns = 1; na = 1 do : [] R ? Ack(s,x)  W = x; if s > na then na = s od

R:: nr = 1 do S ? (D,x)  if nr = x then deliver(D); nr++; compute W; S ! ack(nr, W); od