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dharmesh
Posts: 13
Posted 15:05 Aug 01, 2015 |

I have a question regarding the home-work solution of question 1.

So basically after doing a bit of goggling for question 1, I found that DIFS is only set once, when we have to transfer a single frame of data. Even if we send the frame using the RTS/CTS we will be having DIFS only once.

So my question is, even if we use RTS/CTS to reserve the channel, then also we will have only one occurrence of DIFS, but in the solution we have 2 occurrence of DIFS. So can you please correct me if I am wrong.

Also at some place I found that, if we want to transfer a data for say 1000 bytes, then we use the frame without the data as 32 bytes instead of 34 bytes. Again please correct me if I am wrong. Attached is the screenshot of 1DIFS and 3SIFS.

Thanks

 

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raylongma1018
Posts: 81
Posted 22:19 Aug 01, 2015 |

I believe from our lectures notes in page 29 Difs means the waiting time space we used by the sender only when sender wishes to send rts or data when it senses a channel idle , and page 30 tells sifs is the waiting time used by the receiver. so 2 difs and 2 sifs is the correct answer. there are many different version from website, i think we should stick on the one from our lecture notes.

also for control frame size, just check page34, you can easily find the control frame size without data by adding all the bytes for each part 2+2+6+6+6+2+6+4 = 34

max data size for data for each frame is 2312 since the data from homework 1 we used only contain 1200 byte which is less than 2312 so we only send the data once. if the data we send is greater than 2312 say 3312 than we send the first frame with size 2312+34 and then send next frame with 1000+34 on the next data frame. plus additional difs and sifs for ack, and data frame.

I don't know if we need to memorize the max data frame size(2312)  for the midterm

 

dharmesh
Posts: 13
Posted 22:22 Aug 01, 2015 |

I guess I got it now. Thanks

plakhan
Posts: 37
Posted 09:27 Aug 02, 2015 |

Hi Raylong,

Thank you for the explanation. So if we have case where the sender sends a frame size of 500 bytes first without reserving the channel and then send 800 bytes with RTS/CTS. In this case what would be the calulation. Can you please explain .

Thanks

plakhan
Posts: 37
Posted 09:29 Aug 02, 2015 |

In Particular, how many DIFS and SIFS should be there in the transmission?

krishnarajen
Posts: 16
Posted 10:52 Aug 02, 2015 |

On my understanding it should be 3 DIFS & 3 SIFS 

raylongma1018
Posts: 81
Posted 11:47 Aug 02, 2015 |

from page 29 the lecture notes states in the wireless network, the sender wishes to send a frame, then he must use "csma/ca". unlike "csma/cd", "csma/ca" once transmit a frame it will transmit the frame no matter what. so sender should always send a rts and wait for cts from the reciever in order to send a packet. Notice if the sender wishes to send 2 or more frames, it will send the next data frame after reciever ack from the previous data frame, and sender do not need to create rts/cts again. 

I believe sender should always reserve a channel when using csma/ca, otherwise it will have potential collision with other sender.

there is 1 difs for sender sends a rts, 1 difs for sending 1 data, and 1 difs for sending 2nd data so 3 difs

1 sifs for reciever sending cts, 1 sifs for ack of 1 data , 1 sifs ack for 2nd data so 3sifs