1. CIS 725
Data Link Layer

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

3. 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

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

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

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

7. - 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

8. 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

9. A ------------------B Bandwidth =108 bytes/sec RTD = 2 msecs
2msecs * 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

12. Selective Retransmission
Retransmit only those packets the lost packets

13. 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

14. 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

15. 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

16. 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

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

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