CS 330 Homework: Chapter 1
Due: Tuesday, Sep 15, 2026 by 11:59 PM
Network Transmission Scenario (15 pts)
Consider a single router transmitting packets, each of size L bits, over a single link to another router. The link has a transmission rate of R Mbps.
Given:
- Packet size: L = 16,000 bits
- Link transmission rate: R = 400 Mbps
Questions:
- Compute the one-hop transmission delay.
- Express your answer in seconds, rounded to two decimal places after leading zeros.
Transmission delay = L / R = 16,000 / (400 × 10^6) = 0.00004 sec
- Express your answer in seconds, rounded to two decimal places after leading zeros.
- Determine the maximum number of packets per second that can be transmitted over this link.
Packets/sec = R / L = (400 × 10^6) / 16,000 = 25,000 packets/sec
Multi-Link Network Delay Calculation (25 pts)
Consider a network with three links, each with the specified transmission rate and link length:
- Link 1: Transmission rate = 10 Mbps, Length = 3 Km
- Link 2: Transmission rate = 100 Mbps, Length = 1000 Km
- Link 3: Transmission rate = 1000 Mbps, Length = 1 Km
The packet being transmitted is 4,000 bits in size.
Assume the speed of light is 3 × 10⁸ m/sec.
Questions:
- Calculate the transmission and propagation delays for each of the three links.
- Express all answers in seconds, rounded to two decimal places after leading zeros.
- Compute the total end-to-end delay for transmitting the packet from the source to the destination across all three links.
Link 1 transmission delay = L/R = 4000 bits / (10 Mbps *10^6) = 0.0004 seconds
Link 1 propagation delay = d/s = (3 Km * 1000) / 3*10^8 m/sec = 0.00001 seconds
Link 1 total delay = 0.0004 seconds + 0.00001 seconds = 0.00041 seconds
Link 2 transmission delay = L/R = 4000 bits / (100 Mbps *10^6) = 0.00004 seconds
Link 2 propagation delay = d/s = (1000 Km * 1000) / 3*10^8 m/sec = 0.00333 seconds
Link 2 total delay = 0.00004 seconds + 0.00333 seconds = 0.0034 seconds
Link 3 transmission delay = L/R = 4000 bits / (1000 Mbps * 10^6) = 0.000004 seconds
Link 3 propagation delay = d/s = (1 Km * 1000) / 3*10^8 m/sec = 0.00000333 seconds
Link 3 total delay = 0.000004 seconds + 0.00000333 seconds = 0.00000733 seconds
The total delay = 0.00041 seconds + 0.0034 seconds + 0.00000733 seconds = 0.0038 seconds
End-to-End Packet Delay Analysis (10 pts)
Consider sending a packet from a source host to a destination host over a fixed network route.
Questions:
-
List and briefly describe the major components of the end-to-end delay experienced by the packet.
The delay components are processing delays, transmission delays, propagation delays, and queuing delays. -
Classify each delay component as either constant or variable, and explain why.
All of these delays are fixed, except for the queuing delays, which are variable.
Circuit vs. Packet Switching Scenario(45 pts)
Consider the two scenarios below:
- A circuit-switching scenario in which Ncs users, each requiring a bandwidth of 10 Mbps, must share a link of capacity 150 Mbps.
- A packet-switching scenario with Nps users sharing the same 150 Mbps link, where each user again requires 10 Mbps when transmitting, but only needs to transmit 20% of the time.
Round all your answers to two decimal places after leading zeros.
Questions:
-
When circuit switching is used, what is the maximum number of users that can be supported?
Max Users: 150 Mbps / 10 Mbps = 15 Users -
Suppose packet switching is used, if there are 29 users, can this many users be supported under circuit-switching? Why?
No. 29 Users * 10 Mbps = 290 Mbps, which is greater than 150 Mbps of link capacity available -
If there are 29 packet-switching users, what is the probability that a given (specific) user is transmitting, and the remaining users are not transmitting?
p = 0.20
𝑝 ∗ (1 − 𝑝)(29 − 1) = (0.20) * (0.80)28 ~ 0.00038685 ~ 0.00039 -
What is the probability that one user (any one among the 29) is transmitting, and the remaining users are not transmitting?
(Assume packet switching is used.)
29 ∗ 𝑝 ∗ (1 − 𝑝)(29 − 1) = 29 * 0.20 * (0.80)28 ~ 0.01121 ~ 0.011 -
When one user is transmitting, what fraction of the link capacity is used by this user?
Write your answer as a decimal number. (Assume packet switching is used.)
10 Mbps over the 150 Mbps link, or 10 Mbps / 150 Mbps = 6.7% of the link's capacity when busy. -
When packet switching is used, what is the probability that any 16 users (of the total 29 users) are transmitting and the remaining users are not transmitting?
(29 choose 16) * 𝑝16 ∗ (1 − 𝑝)(29-16) = (29 choose 16) * 0.2016 * 0.80(29-16) ~ 0.000024450 ~ 0.0000245
Wolfram Alpha -
When packet switching is used, what is the probability that more than 15 users are transmitting?
Sum{(29 choose n) * p n * (1 - p)(29 - n)}, for n = 16 to 29 => sum{(29 choose n) * 0.20n * 0.80(29-n)}, for n = 16 to 29 => 0.000030032 ~ 0.00003
Wolfram Alpha
TCP/IP Stack Concept Check (5 pts)
Which layer of the TCP/IP protocol stack is responsible for handling messages from various network applications?
Application layer
Submit
Post your solutions in Marmoset by the scheduled due date in the syllabus.
