[HW] HW#01Q 2019

[Prof. Hsiao]

Q1. What are the pros and cons of ADSL and cable access network?

A1.

ADSL pros:

 a) build on the top of old technology -- phone line which most of the household has.

 b) asymmetric upstream and downstream design which is quite suitable for general users (more download than upload)

 c) dedicated line. Guaranteed for your use only; good for real-time communications, such as gaming, VoIP.

 d) can be extended to VDSL easily; higher speed without changing underlying technique.

ADSL cons:

 a) vanilla ADSL is expensive (in Taiwan) and out-of-dated; most of the time, Taiwan users use VSDL or fiber.

 b) vanilla ADSL is slower (comparing to VDSL, fiber or even Cable)

 c) the phone line (between your house to the central office) might be too old to provide high-speed ADSL.

Cable pros:

 a) Bigger bandwidth (than vanilla ADSL)

 b) Can provide TV programs at the same time

 c) Cheaper, because multiple households share the cable infrastructure.

 d) coax can provide better transmission rate and longer range transmission since it is thicker than a phone line.

Cable cons:

 a) need cable available near your house; otherwise, a new cable is needed from the headend to your house.

 b) shared cable; the throughput is not guaranteed from the modem to the headend. 

 c) throughput highly depends on the number of users in the shared cable.

 d) In Taiwan, most of the Cable ISP are not tier-one ISP.

Q2. What are the pros and cons of the wired and wireless network?

A2.

Wired pros:

 a) delicate line; no interfering

 b) high speed (up to 10Gbps), such as fiber

 c) less noise; physical link is usually protected.

Wired cons:

 a) connected through a physical line; not convenient

 b) not easy to upgrade physical links

Wireless pros:

 a) no physical link, high mobility; can transmit through the wall

Wireless cons:

 a) high interference; high noise

 b) may need multiple access points to cover a wide area

 c) the transmission rate is lower. (comparing with a physical link)

 d) may have a security problem, e.g., a malicious user may connect to your home WiFi.

 e) a shared media; low throughput.

Q3. What is the reason why application messages must be fragmented?

A3. As we use packet-switching network, the fragmented messages (i.e., packets) are sent one by one. Thus, we do not need to retransmit the whole message when it is lost on the Internet. For the intermediated routers, they can set up a queue to perform store-and-forward style transmission for the incoming packets from multiple hosts. Therefore, a physical link would not be occupied by one host; rather is shared among multiple hosts. 

Q4. Assume an ISP deploys routers along with Taiwan’s High-Speed Rail train stations (total 12 stations, i.e., 11 hops), and the total length of the railroad is 349.5 KM. How long (in seconds) does it take for a packet of length 1,000 bytes to propagate over these fiber links (the fiber propagation speed is 2.5 x 10^8 m/s and the transmission rate is 10Gbps)?

A4.

Total propagation delay = (the end-to-end distance) / (propagation speed) =  (349.5 KM) / (2.5 x 10^8 m/s) = (3.495 *10^5m) / (2.5 * 10^8 m/s) = 1.398 * 10^-3 s = 1.398 ms

One-hop transmission delay = (packet length)/(transmission rate) = (1000 bytes) / (10Gbps) = ( 8* 10^3 bits) / (10 * 10^9 bps) = 0.8 * 10^-6 s = 0.8 mu second

Total transmission delay = 11 hops delay = 11* 0.8 mu sec = 8.8 mu sec

=> Total delay of propagation and transmission is 1.398 ms + 8.8 mu sec = 1.4068 ms

Q5. We plan to develop an in-game VoIP service over a packet-switched network. Assume the analog voice signals are converted to a digital 64 kbps bitstream on the fly and voice sender then groups the bits into 56-byte packets. Assume Internet bandwidth is 100 Mbps on the average and its average propagation delay is 100 msec. How long (in seconds) does it take for you to wait for a voice response from your friend? Is there any other possible delay? Please list them as more as possible.

A5.

Note that there is no "standard answer" for this problem.

a) Think of the first 56-byte packet sent from you to your friend.

It takes x seconds to collect the first 56-byte voice data with a 64 kbps encoder.

64 kbps = 64 * 10^3 bps = 8 *10^3 Byte/sec = 8000 bytes/sec 

So, 56 bytes needs 0.007 second (i.e., x) = 7 msec.

The propagation delay is 100 msec.

The one-hop transmission delay is (56 bytes) / (transmission rate) = (56 *8 bit) / 100Mbps) = 4.48 *10^-6 s

However, we do not know how many hops are there between you and your friends.

But notice that the one-hop transmission delay (several mu seconds) is much smaller than the propagation delay (100 msec) and encoder's processing delay (7 msec).

So the transmission delay is negligible.

Thus, the (two-way) response time is

=> 7 msec (encoding) + 100 msec (propagation) + n hops * 4.48 * 10^-6 s (negligible transmission delay ) + 7 msec (decoding) 

  + 7 msec (encoding response) + 100 msec (response propagation) + n hops * 4.48 * 10^-6 s (negligible response transmission delay) + 7 msec (decoding response) 

b) queuing delay in the router, processing delay by routers and hosts, retransmission delay (if packets are lost), encoding/decoding delay

Q6. Suppose users share a 3 Mbps link. Also, suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time.

a. When circuit switching is used, how many users can be supported?

b. For the remainder of this problem, suppose packet switching is used. Find the probability that a given user is transmitting.

c. Suppose there are 120 users. Find the probability that at any given time, exactly n users are transmitting simultaneously. (Hint: binomial distribution.)

d. Find the probability that there are 21 or more users transmitting simultaneously.

A6.

a) number of users = (total capacity) / (one user demand) = 3Mbps / 150 kbps = 3 *10^6 / 150 * 10^3 = 20 users.

b) each user transmits only 10 percent of the time => that is, given any time and any arbitrary user, the probability of tranmission on this user is 0.1.

c) P{k} = P{exactly k users are transmitting out of m users} = C(m, k) * (0.1)^k * (0.9)^(m-k). Now let m be 120 and k be n.

 The equation means that at any given time, selecting k users out of 120 users, and only k users are transmitting with probability = 0.1 and the rest of the users (i.e., 120-k) are not transmitting with probability = 0.9.

d) P{ >= 21 users} = 1.0 - P{<= 20 users} = 1.0 - Sigma{k = 0 to 20}(P{k}) 

Note that you can find the Binomial Distribution Table at the end of your statistics textbook to find this value.

However, this table usually only provides the results when m <= 20.

So, you can use normal distribution and Z score to approximate this value,

i.e.,

the average of this distribution = 120 users * activate user prob 0.1 = 12

the standard deviation of this distribution = (120*0.1*0.9)^(1/2) = 3.286335345030997

then Z score is => (Z-12)/3.286335345030997

then P(<=20) = P{Z < (20-12)/(3.286335345030997)} = P{Z< 2.434322477800738} ~= 0.9925

so 1.0 - Sigma{k = 0 to 20}(P{k}) = 1.0 - P(<=20) ~= 1.0 - P{Z< 2.434322477800738} ~= 0.0075