**Flow Control Protocols-**

In computer networking, there are various flow control protocols-

In this article, we will discuss about sliding window protocol.

**Sliding Window Protocol-**

- Sliding window protocol is a flow control protocol.
- It allows the sender to send multiple frames before needing the acknowledgements.
- Sender slides its window on receiving the acknowledgements for the sent frames.
- This allows the sender to send more frames.
- It is called so because it involves sliding of sender’s window.

Maximum number of frames that sender can send without acknowledgement = Sender window size |

**Optimal Window Size-**

In a sliding window protocol, optimal sender window size = 1 + 2a |

**Derivation-**

We know,

To get 100% efficiency, we must have-

η = 1

T_{t} / (T_{t} + 2T_{p}) = 1

T_{t} = T_{t} + 2T_{p}

Thus,

- To get 100% efficiency, transmission time must be T
_{t}+ 2T_{p }instead of T_{t}. - This means sender must send the frames in waiting time too.
- Now, let us find the maximum number of frames that can be sent in time T
_{t}+ 2T_{p}.

We have-

- In time T
_{t}, sender sends one frame. - Thus, In time T
_{t}+ 2T_{p}, sender can send (T_{t}+ 2T_{p}) / T_{t}frames i.e. 1+2a frames.

Thus, to achieve 100% efficiency, window size of the sender must be 1+2a.

**Required Sequence Numbers-**

- Each sending frame has to be given a unique sequence number.
- Maximum number of frames that can be sent in a window = 1+2a.
- So, minimum number of sequence numbers required = 1+2a.

To have 1+2a sequence numbers, Minimum number of bits required in sequence number field = ⌈log |

**NOTE-**

- When minimum number of bits is asked, we take the ceil.
- When maximum number of bits is asked, we take the floor.

**Choosing a Window Size-**

The size of the sender’s window is bounded by-

**1. Receiver’s Ability-**

- Receiver’s ability to process the data bounds the sender window size.
- If receiver can not process the data fast, sender has to slow down and not transmit the frames too fast.

**2. Sequence Number Field-**

- Number of bits available in the sequence number field also bounds the sender window size.
- If sequence number field contains n bits, then 2
^{n}sequence numbers are possible. - Thus, maximum number of frames that can be sent in one window = 2
^{n}.

For n bits in sequence number field, Sender Window Size = min (1+2a , 2^{n}) |

**Implementations of Sliding Window Protocol-**

The two well known implementations of sliding window protocol are-

**Efficiency-**

Efficiency of any flow control protocol may be expressed as-

**Example-**

In **Stop and Wait ARQ**, sender window size = 1.

Thus,

Efficiency of Stop and Wait ARQ = 1 / 1+2a

**PRACTICE PROBLEMS BASED ON SLIDING WINDOW PROTOCOL-**

**Problem-01:**

If transmission delay and propagation delay in a sliding window protocol are 1 msec and 49.5 msec respectively, then-

- What should be the sender window size to get the maximum efficiency?
- What is the minimum number of bits required in the sequence number field?
- If only 6 bits are reserved for sequence numbers, then what will be the efficiency?

**Solution-**

Given-

- Transmission delay = 1 msec
- Propagation delay = 49.5 msec

**Part-01:**

To get the maximum efficiency, sender window size

= 1 + 2a

= 1 + 2 x (T_{p} / T_{t})

= 1 + 2 x (49.5 msec / 1 msec)

= 1 + 2 x 49.5

= 100

Thus,

For maximum efficiency, sender window size = 100

**Part-02:**

Minimum number of bits required in the sequence number field

= ⌈log_{2}(1+2a)⌉

= ⌈log_{2}(100)⌉

= ⌈6.8⌉

= 7

Thus,

Minimum number of bits required in the sequence number field = 7

**Part-03:**

If only 6 bits are reserved in the sequence number field, then-

Maximum sequence numbers possible = 2^{6} = 64

Now,

Efficiency

= Sender window size in the protocol / Optimal sender window size

= 64 / 100

= 0.64

= 64%

**Problem-02:**

If transmission delay and propagation delay in a sliding window protocol are 1 msec and 99.5 msec respectively, then-

- What should be the sender window size to get the maximum efficiency?
- What is the minimum number of bits required in the sequence number field?
- If only 7 bits are reserved for sequence numbers, then what will be the efficiency?

**Solution-**

Given-

- Transmission delay = 1 msec
- Propagation delay = 99.5 msec

**Part-01:**

To get the maximum efficiency, sender window size

= 1 + 2a

= 1 + 2 x (T_{p} / T_{t})

= 1 + 2 x (99.5 msec / 1 msec)

= 1 + 2 x 99.5

= 200

Thus,

For maximum efficiency, sender window size = 200

**Part-02:**

Minimum number of bits required in the sequence number field

= ⌈log_{2}(1+2a)⌉

= ⌈log_{2}(200)⌉

= ⌈7.64⌉

= 8

Thus,

Minimum number of bits required in the sequence number field = 8

**Part-03:**

If only 6 bits are reserved in the sequence number field, then-

Maximum sequence numbers possible = 2^{7} = 128

Now,

Efficiency

= Sender window size in the protocol / Optimal sender window size

= 128 / 200

= 0.64

= 64%

To gain better understanding about sliding window protocol,

**Next Article-****Practice Problems On Sliding Window Protocol**

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