I just need help with number 2 please. I am confused as to why 2 Mbps is 2 * 10^
ID: 3882587 • Letter: I
Question
I just need help with number 2 please.
I am confused as to why 2 Mbps is 2 * 10^3 and not 2 * 10^6??
Show your work please. Suppose that all connections are between switches A and C. What is the maximum number of simultaneous connections that can be in progress? c. Suppose we want to make four connections between switches A and C, and another four connections between switches B and D. Can we route these calls through the four links to accommodate all eight connections? If yes/no, why? 2. We consider sending real-time voice from Host A to Host B over a Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A 56-byte packets. There is one link between Hosts A and B: its transmission rate is 2 Mbps and its propagation delay is 10 msec. As soon as Host A gathers a packet. it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet's bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)? packet-switched network (VolP). then groups the bits into 3 pts] 3. Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queued. Each packet is of length L and the link has transmission rate R. What is the average queuing delay for the N packets? 3 pts
Explanation / Answer
given: transmission rate = 2 Mbps, packet length = 56 bytes;
transmission time = (56*8)/(2*106) = 224 micro seconds = 0.224 msec
given: propogation delay = 10 msec
time to convert analog to digital and vice-versa = 1/(64 * 103) = 0.015625 msec
So, overall time taken = time taken to convert analog to digital + transmission time + propogation delay + time taken to convert digital to analog = 0.015625 + 0.224 + 10+ 0.015625 = 10.25525 msec
Hope it helps, feel free to comment in case of any query.