[컴퓨터 통신] L4 Transport Layer_2

No jam

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Transport Layer Protocols

• Simple Protocol
• Stop-and-Wait Protocol
• Go-Back-N Protocol
• Selective-Repeat Protpcol
  -Bidirectional Protocols: Piggybacking
  
  

Simple Protocol

• Simple and Connectionless protocol
• Flow, Error control 없음!!
  
• Receiver can immediately handle any frame it receives.

FSM for the simple protocol

• Sender sends packets without thinking about the receiver
  
  

Stop-and-Wait Protocol

• Connection-oriented protocol
  

• Sliding window size: 1

• Flow control is achieved

  • By forcing the sender to wait for an acknowledgment
  • 일반적인 Flow control 방법: Buffer
  • Stop-and-Wait의 buffer size: 1

• Error control is achieved

  • By discarding corrupted packets and letting the sender resend unacknowledged packets when the timer expires
  • Error control을 위해 사용하는 번호: Sequence Number
  • Stop-and-Wait에서는 1 bit field를 sequence num 저장하기 위해 사용
    • m=1 → sequence numbers: modulo 2 (0 or 1)
    • the ack num always announces in modulo-2 the seq num of the next packet expected
      

• All calculation in the Stop-and Wait protocol is in modulo 2
  FSM for the stop-and-wait protocol
  

• Example
  

• Inefficient

  • Lost Data
    

  • Lost Ack
    

  • Delayed Ack/Data

• Bandwidth-delay product

  • Channel: pipe
  • Bandwidth-delay product: pipe capacity
  • e.g.) bandwidth: 1Mbps, delay: 20msec 일때 Bandwidth-delay product: (110^6)(20*10^-3) = 20,000 bit

• Example

  • In a Stop-and-Wait system, the bandwidth of the line is 1Mbps, and 1 bit takes 20msec to make a round trip. What is the bandwidth-delay product? If the system sends one packet (1,000 bits in length), what is the utilization percentage of the link?
    1,000 bits / 20,000 bits = 5% (Too low!)
  • What is the maximal utilization percentage of the above link if we have a protocol that can send up to 15 packets before stopping and worrying about the acknowledgments?
    15,000 bits / 20,000 bits = 75%

• So, sneding up to multiple packets are more efficient!

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