1. Suppose that Bob sends 90,000 bits to Alice over a 2.25 Kbps link (note, 1 Kb
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Question
1. Suppose that Bob sends 90,000 bits to Alice over a 2.25 Kbps link (note, 1 Kbps 103 bps) that uses circuit switching. Suppose that circuit switching on the link is implemented using TDM, that a frame is 1 sec, and that a frame consists of 6 slots. Bob is assigned one slot in every frame. How many minutes does it take Bob to transmit all bits to Alice? (7 points) 2. Suppose that Router A and Router B are connected via a 10 Kbps link, and that the two routers are 50 meters apart. Suppose that Router A is transmitting a packet, and that there are 6 additional packets in the queue of the router, when a new packet arrives at Router A. Suppose that 700 bits of the currently-being-transmitted packet has been transmitted by Router A. Also, assume the following a. Every packet consists of 1000 bits. b. Propagation speed of a bit over the link connecting the two routers is 2 x 10 meters/sec. c. Processing delay of a (whole) packet at a router takes 10 milliseconds (i.e., 0.01 seconds). What is the total delay of the new packet, i.e., how long does it take the new packet to arrive at Router B from the time it arrived at Router A? In your solution, clearly identify all four delay components the new packet (10 points) experiencesExplanation / Answer
1. Speed of link is 2.25 kbps i.e. it can send 2250 bits per second.
There are 6 slots in a frame (of 1 second), so 2250 bits can be transmitted in these 6 slots i.e each slot contains at most 375 bits (2250/6).
Bob is assigned 1 slot out of 6, so Bob can send 375 bits per second. So for sending 90000 bits it needs 90000/375 seconds i.e. (90,000/375)*(1/60) minutes = 4.61 minutes.
2. The total delay consist of Processing Delay + Transmission delay+ Queueing delay + Propogation delay. The packet is the 7th packet in the queue (as there are 6 packets in queue already).
Processing Delay = 0.01 * 6 = 0.06seconds
Transmission Delay = 1000 bits/10,000 bits (i.e. Total bits need to be transmitted(Size of packets)/Bandwith (10Kbps = 10000 bits per second)) = 0.1 seconds
Propogation delay = 50 / 2*10,000 (i.e. distance/speed) = 0.0025 seconds
Queueing delay = transmission delay*Num of packets = 6*0.1 = 0.6 seconds
Total Delay = (0.06 +0.1+0.0025+0.6) = 0.7625 seconds i.e. 762.5 milliseconds.
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