GATE Overflow | Mock GATE | Test 1 | Question: 38

Question

Consider a half-duplex link with a one-way propagation delay of $P$ seconds and a bandwidth of $B$ bits/sec. The sender and the receiver decide to share the link using Time Division Multiplexing i.e. the sender sends for $T_s$ time and then receiver sends for $T_r$ time and so on. Refer to the figure below. Assume that the sender sends data in packets of size $F$ bits and receiver sends $ACK$ in packets of size $A$ bits.
Suppose you are implementing Stop and Wait protocol. What is the natural value for sender timeout for transmission?

• A. $2* (F/B)+P+(A/B)$
• B. $(2/B)*(F+A)+P$
• C. $(1/B)*(F+A)+2*P$
• D. $(2/B)*(F+A)+2*P$

Comments

Just more readability and understandability purpose.

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@Ruturaj Mohanty

Here Ts will be equal to Tt Transmission time of frame + Tp?

Otherwise Sender can't send complete frame in Tt+Tp because next slot Tr is given to reciever

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The value in option C is the minimum time required for transmission and acknowledgement. Why is this value taken as a timeout value?