SYSC4602 Computer Communications | Carleton卡尔顿大学 | CS代写 | midterm exam代考

FALL TERM MIDTERM
EXAMINATION
October 2021
DURATION: 1.5 HOURS No. Of Students: 171
Department Name & Course Number: Systems and Computer Engineering
SYSC 4602
Course Instructor (s): Changcheng Huang
AUTHORIZED MEMORANDA
Simple calculator
Students MUST count the number of pages in this examination question paper before beginning to
write, and report any discrepancy to a proctor. This question paper has 1 pages + cover
page = 2_ pages in all.
This examination question paper may not be taken from the examination room.
In addition to this question paper, students require: an examination booklet yes
Scantron Sheet no
Name:
Student Number:

1. (25%) Consider sending a large file of F bits from Host A to Host B, traversing three uncongested links with
   lengths D1, D2, D3 km respectively (and two switches) between A and B. Signal propagation speed in the three
   links is p km/sec. Host A segments the file into segments of S bits each and adds H bits of header to each segment,
   forming packets of L = H + S bits. Each link has a transmission rate of R bps. Find the value of S that minimizes
   the delay of moving the file from Host A to Host B. Ignore processing delay.
   Solution:
   Because each link has the same bandwidth R bps and processing delay is negligible, the queuing delay will be
   0.
   There are F/S packets (when F is large, the fractional part can be ignored). Each packet is S + H bits. Time at
   which the first packet is received at Host B is 3 ×
   !”#
   $ + %!”%””%#
   & sec. Thus delay in sending the whole file is
   (
   ‘
   !

• 2) ×
  !”#
  $ + %!”%””%#
  & .
  To calculate the value of S which leads to the minimum delay,
  �
  �� ����� = 0 ⇒ � = 0��
  2

1. (25%) A CDMA receiver gets the following chips: (0 -2 0 +2 +2 0 0 2). Assuming the following chip sequences
   used by four sending stations, which stations transmitted, and which bits did each one send?
   Solution: Let X = (0 -2 0 +2 +2 0 0 2). Then � ∙ � = 8, � ∙ � = 0, � ∙ � = 0, � ∙ � = −8. Station A sent 1,
   Stations B and C did not send, Station D sent 0.
2. (25%) Consider a CRC scheme with the 5-bit generator, G = 10011, and suppose that D has the value
3. What is the value for R?
   Solution:
   If we divide 10011 into 1010101010 0000, we get 1011011100, with a remainder of R=0100. Note that,
   G=10011 is CRC-4-ITU standard.
4. (25%) Consider two stations, A and B, that use the slotted ALOHA protocol to contend for a channel. Suppose
   node A has more data to transmit than node B, and node A’s transmission/retransmission probability in a slot pA
   is greater than node B’s transmission/retransmission probability in a slot, pB.
   a. Provide a formula for node A’s average throughput measured in frames/slot. What is the total
   efficiency of the protocol with these two nodes?
   b. If pA= 2pB, is node A’s average throughput twice as large as that of node B? Why or why not? If not,
   how can we choose pA and pB to make that happen?
   Solution:
   a. A’s average throughput is given by pA(1-pB).
   Total efficiency is pA(1-pB) + pB(1-pA).
   b. A’s throughput is pA(1-pB)=2pB(1-pB)= 2pB- 2(pB)
   2
   .
   B’s throughput is pB(1-pA)=pB(1-2pB)= pB- 2(pB)
   2
   .
   Clearly, A’s throughput is not twice as large as B’s.
   In order to make pA(1-pB)= 2 pB(1-pA), we need that pA= 2pB/(1+pB)