A router receives a packet and determines the outbound link for the packet. When the packet arrives, 2/3 of one other packet is done being transmitted on this outbound link and five other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 3,000 bytes, and the link rate is 4 Mbps. What is the queuing delay for the packet

Queuing Delay is 0.08 seconds

Explanation:

The answer follows a formula that is relatively easy to use and is detailed below.

Queuing Delay = [(L - x) + (nL)] / R

where,

L is packet length given as 3,000 bytes

x is the currently transmitted packet given as 2/3 * 3,000 = 2,000

n is the number of packets waiting in the Que given as 5

R is the rate of transmission given as 4 Mbps (4 * 10^6 Bps)

We can simply plug in the above information in the equation for computing Queuing Delay.

Lets take the numerator first which would be [(3000 - 2000) + (5 * 3000)]

The numerator would be 16000 bytes. These are being transmitted at 4Mbps. So, 16000*4*5 = 320,000 "bits"

Queuing Delay= 320,000/4000000

Queuing Delay = 0.08 seconds.

As we can see, the formula is quite intuitive to use. You are simply taking the number of packets to be transmitting, incorporating the partially transmitted packet, multiplying by the number of packets and the rate of transmission and then dividing the product by the rate of transmission to compute what the delay in the Que is.

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given the options, i would say your best bet is c. the modem supports up to 56 kbps, which is too low for video streaming. dial up modems use actual phone lines, which only can go up to 56 kbps as that's more than sufficient for traditional phone calls. this is also the reason why when a dial-up modem connects, you hear all those sounds. it's literally using the phone line to establish a connection.