Modeling and Simulation NETW 707

Modeling and Simulation NETW 707
Lecture 7 ARQ Modeling: Modeling Error/Flow Control (2) Course Instructor: Dr.-Ing. Maggie Mashaly C3.220

Modeling of Go Back N Protocol

Limitation of Stop-and–Wait Protocol
Machine A Machine B Physical Channel First Frame - Bit enters Channel Last Frame arrives at B Last ACK Arrives at A Processing Time Channel is Idle Inefficiency of Stop-and-Wait because the channel is left idle without use until an acknowledgement is received Potential Solution: Allowing the transmitter to continue sending enough frames so that the channel is kept busy while the transmitter is waiting for an acknowledgement

Go-Back-N ARQ Protocol Example Go Back 4 An outstanding frame:
Sender Receiver Fr_0 Fr_1 ACK_1 Fr_2 ACK_2 Fr_3 ACK_3 An outstanding frame: It is a frame that has been transmitted and is yet to be acknowledged Fr_4 Fr_5 Out of Sequence Frames. So, do not accept them At this Point there are WS=4 Outstanding Frames. So, GO-BACK-4 Fr_6 Fr_3 Fr_4 Fr_5 Fr_6 Fr_7 Fr_8 ACK_4 ACK_5 ACK_6 ACK_7 ACK_8

Go-Back-N Basics Transmit without waiting for ACK
Each frame received correctly is acknowledged with a sequence number corresponding to the NEXT frame expected The receiver discards all out of sequence packets If number of outstanding frames is WS: Go-Back-N

What does 𝑾 𝒔 Represent? WS is called the sending window size
Each frame must be buffered (stored) until a valid ACK arrives (in case retransmission is needed) WS represents the maximum number of frames that may be outstanding (i.e., stored) simultaneously

Sliding Window Concept
Go Back N is an example of what is called a sliding window protocol An ACK received means that all previous frames have been received correctly (previous ACKs were lost or receiver chose not to send them) Example Go Back 4 Sending Window Sender Receiver Fr_0 Fr_0 Fr_1 ACK_1 Fr_1 Fr_2 ACK_2 Fr_2 Fr_3 ACK_3 Fr_3 Fr_4 Frame Sequences that are allowed to be sent Fr_5 Fr_6 Outstanding Frame Sequences

Go Back N is an example of what is called a sliding window protocol An ACK received means that all previous frames have been received correctly (previous ACKs were lost or receiver chose not to send them) Example Go Back 4 Sending Window Sender Receiver Fr_4 Fr_0 Fr_1 ACK_1 Fr_5 Fr_2 ACK_2 Fr_6 Fr_3 ACK_3 Fr_7 Fr_4 ACK_4 Frame Sequences that are allowed to be sent Fr_5 Fr_6 Outstanding Frame Sequences Transmitter has nothing to send

Go-Back-N ARQ Modeling
Assumptions 𝑳 𝒂 ≪ 𝑳 𝒑𝒌 , 𝒕 𝒂 ≪ 𝒕 𝒑𝒌 ⇒ 𝒕 𝒂 ≈𝟎 𝒕 𝒑𝒓𝒐𝒄 ↓↓ ⇒ 𝒕 𝒑𝒓𝒐𝒄 ≈𝟎 Forward Channel BER 𝝐 Backward Channel (i.e., ACK/NAK) is Error Free Infinite number of retransmissions Machine A Machine B Physical Channel

Go-Back-N ARQ Markov Model
Model Details The time step is equal to 𝒕 𝒑𝒌 Probability of packet error 𝒑 𝒆 𝒑 𝒆 =𝟏− 𝟏−𝝐 𝑳 𝒑𝒌 For 𝝐 𝑳 𝒑𝒌 ≪𝟏⇒ 𝒑 𝒆 ≈𝝐 𝑳 𝒑𝒌 Define 𝝅 𝒌 as the probability of 𝒔 𝒌 Define 𝜹 𝒌𝒋 as the transition probability from 𝒔 𝒌 to 𝒔 𝒋 𝟏− 𝒑 𝒆 𝟏 𝟏 𝒔 𝟏 𝒔 𝟐 𝟏 𝒔 𝟑 𝒔 𝑵−𝟏 𝒔 𝑵 𝒑 𝒆 𝚷= 𝝅 𝟏 𝝅 𝟐 𝝅 𝟑 … 𝝅 𝑵 𝚫= 𝜹 𝟏,𝟏 𝜹 𝟏,𝟐 … 𝜹 𝟏,𝑵−𝟏 𝜹 𝟏,𝑵 𝜹 𝟐,𝟏 𝜹 𝟐,𝟐 … 𝜹 𝟐,𝑵−𝟏 𝜹 𝟐, 𝑵 ⋮ ⋮ ⋮ ⋮ ⋮ 𝜹 𝑵−𝟏,𝟏 𝜹 𝑵−𝟏,𝟐 … 𝜹 𝑵−𝟏,𝑵−𝟏 𝜹 𝑵−𝟏,𝑵 𝜹 𝑵,𝟏 𝜹 𝑵,𝟐 … 𝜹 𝑵,𝑵−𝟏 𝜹 𝑵,𝑵

Go-Back-N ARQ Markov Model
𝚷= 𝝅 𝟏 𝝅 𝟐 𝝅 𝟑 … 𝝅 𝑵 𝟏− 𝒑 𝒆 𝚫= 𝟎 𝟏 𝟎 … 𝟎 𝟎 𝟎 𝟎 𝟏 … 𝟎 𝟎 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 𝟎 𝟎 𝟎 … 𝟎 𝟏 𝒑 𝒆 𝟎 𝟎 … 𝟎 𝟏− 𝒑 𝒆 𝟏 𝟏 𝟏 𝒔 𝟏 𝒔 𝟐 𝒔 𝟑 𝒔 𝑵−𝟏 𝒔 𝑵 𝒑 𝒆 At steady State 𝚷×𝚫=𝚷 With boundary condition 𝒌=𝟏 𝑵 𝝅 𝒌 =𝟏 Solving: 𝝅 𝑵 = 𝟏 𝟏+ 𝒑 𝒆 𝑵−𝟏 𝝅 𝟏 = 𝝅 𝟐 =…= 𝝅 𝑵−𝟏 = 𝒑 𝒆 𝟏+ 𝒑 𝒆 𝑵−𝟏 𝝅 𝟐 = 𝝅 𝟏 𝝅 𝟑 = 𝝅 𝟐 ⋮ 𝝅 𝑵−𝟏 = 𝝅 𝑵−𝟐 𝝅 𝑵 = 𝝅 𝑵−𝟏 + 𝟏− 𝒑 𝒆 𝝅 𝑵 𝝅 𝟏 = 𝒑 𝒆 𝝅 𝑵 𝒌=𝟏 𝑵 𝝅 𝒌 =𝟏 𝑵−𝟏 𝝅 𝟏 + 𝝅 𝑵 =𝟏 𝑵−𝟏 𝝅 𝟏 + 𝝅 𝟏 𝒑 𝒆 =𝟏 𝒑 𝒆 𝑵−𝟏 𝝅 𝟏 + 𝝅 𝟏 = 𝒑 𝒆 𝝅 𝟏 = 𝒑 𝒆 𝟏+ 𝒑 𝒆 𝑵−𝟏 𝝅 𝑵 = 𝟏 𝟏+ 𝒑 𝒆 𝑵−𝟏

Go-Back-N ARQ Throughput
Throughput measures the percentage of time slots that are utilized for successful transmissions 𝑻𝒉= 𝒌=𝟎 ∞ 𝝅 𝒌 ×𝑷𝒓 𝑺𝒖𝒄𝒄. 𝑻𝒙 𝒂𝒕 𝒔 𝒌 𝑻𝒉= 𝟏− 𝒑 𝒆 𝝅 𝑵 𝑻𝒉= 𝟏− 𝒑 𝒆 𝟏+ 𝒑 𝒆 𝑵−𝟏 𝟏− 𝒑 𝒆 𝟏 𝟏 𝒔 𝟏 𝒔 𝟐 𝟏 𝒔 𝟑 𝒔 𝑵−𝟏 𝒔 𝑵 𝒑 𝒆

S/W and GBN Comparison

References NETW 707 Lectures slides by A. Prof. Tallal El-Shabrawy, 2016 & 2017 “Simulation Modeling and Analysis” by Averill M. Law, 5th Edition, 2015

Modeling and Simulation NETW 707

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