Question

2. Consider the figure below, which follows the analogy between store-and-forward link transmission and propagation of bits in a packet along a link, and cars in a caravan being serviced at a toll booth and then driving along a road to the next tollbooth, as discussed in the lecture (see slides 1-39 and 1-40). 200 km 200 km toll booth toll booth 20 car caravan toll booth (aka 20-bit packet) (aka link) Suppose the caravan has 20 cars, and that the tollbooth services (that is, “transmits”) a car at a rate of one car per 2 seconds. Once a car has been serviced, it proceeds to the next tool both, which is 200 kilometers away, at a rate of 10 kilometers per second. Also, assume that whenever the first car of the caravan arrives at a tollbooth, it must wait at the entrance to the tollbooth until all of the other cars in its caravan have arrived, and lined up behind it before being serviced at the toll booth. That is, the entire caravan must be “stored” at the tollbooth (like a packet at a router) before the first car in the caravan can pay its toll and begin driving towards the next tollbooth. a. How long does it take for the entire caravan to receive service at the tollbooth (that is the time from when the first car enters service until the last car leaves the tollbooth)? b. Once the first car leaves the tollbooth, how long does it take until it is serviced at the next tollbooth?

Answer 1

Here is the solution for the above problem,

Hope it helps!!

20 Cars x 2 sec/ Car > 40 seconds / It takes 40 seconds to receive service. (6) Time Taken for the first can to receive service at the next toll booth is; (C20 20- 1 Cars x a sec/ car) + Icm) + (doskum / okul 38 + 20 -> 58 seconds,