Problem Set #1 RTT

CSC458 
Filip Tomin
1001329984
1. (a) The minimum RTT would be the RTT for a single bit, which would just be
(3.85 × 108
m)/(3 × 108
m/s) = 1.283s × 2 = 2.57s.
(b) Bandwidth-delay product is (1.0 × 109
) × 2.57 = 2.57 × 109 or 2.57Gb.
(c) The bandwidth-delay product is extremely large. Assuming the wire is being
used fully, there is a lot of data that the wire can hold unacknowledged by the
destination. Slow response time is also a problem for protocols such as TCP.
(d) Assuming no errors and the image is sent as one packet, the delivery time will be
the transmission time + propagation delay.
Transmission time: (2.5 × 107
B) × 8/(1.0 × 109
b/s) = 0.2s
Propagation delay is just 1.28s, half of the RTT.
If we assume the request to be extremely small, such as 1 bit, then the total time
it would take would be 0.2 + 1.28 + 1.28 = 2.76 seconds.
2. (a) Latency = P
i
(P ropi + M/Ri)
Latency = 2((1.0 × 10−5
) + (1.2 × 104
)/(1.0 × 108
)) = 0.00026s or 260µs.
(b) Latency = 4((1.0 × 10−5
) + (1.2 × 104
)/(1.0 × 108
)) = 0.00052s or 520µs.
(c) Latency = M/Rmin +
P
i P ropi
Latency = (1.2 × 104
)/(1.0 × 108
) + 2(1.0 × 10−5
) = 0.00014s or 140µs.
3. (a) RTT is the one-way delay times 2, so RTT = 2 × 10µs = 20µs
Bandwidth-delay product is (1.0 × 108
) × (2.0 × 10−5
) = 2000b or 2kb.
(b) RTT = 520µs from #16.
Bandwidth-delay product = (1.0 × 108
) × (5.2 × 10−4
) = 52000b or 52kb.
(c) RTT = 2 × 50ms= 100ms. Bandwidth-delay product = (1.5 × 106
) × (0.1) =
1.5 × 105 or 150kb.
(d) RTT = 4 × ((3.59 × 107
)/(3 × 108
)) = 479ms. Bandwidth-delay product = (1.5 ×
106
) × (0.479) = 7.185 × 105 or 718kb.
4. (a) Total bytes per second = 640 × 480 × 3 × 30 = 27.7MB.
Bandwidth must be ≥ 27.7MB/s or 221.6Mbps.
1
(b) Total bytes = 160 × 120 × 1 × 5 = 96kB.
Bandwidth must be ≥ 96kB/s or 768kbps.
(c) Total bytes = (6.5 × 108
)/4500 = 144kB.
Bandwidth must be ≥ 144kB/s or 1.16Mbps.
(d) Since it’s black and white, assume each pixel requires only one bit.
Total bits = 8 × 72 × 10 × 72 = 414720 bits.
414720/14400 = 28.8s.
5. The Internet checksum is usually calculated by taking the ones complement sum in 16-
bit units. This method is equivalent since the 32-bit sum is the same as two side-by-side
16-bit sums, even maintaining the overflow rules. The 32-bit sum is converted into the
16-bit sum next. The only problem is this 16-bit value has been ones complemented 2
times, so it is inverted from what it should be. Thus, we take the ones complement of
the result again, to produce the checksum value.
32-bit:
0001101000010101 0000000100110000
+ 0000000000011101 0000001001010010
0001101000110010 0000001110000010
1’s comp - 1110010111001101 1111110001111101
1110010111001101
+ 1111110001111101
1110001001001011
1’s comp - 0001110110110100
1’s comp - 1110001001001011
16-bit:
0001101000010101
0000000100110000
0000000000011101
+ 0000001001010010
0001110110110100
1’s comp - 1110001001001011
As shown above, both methods ended up with the same 16-bit checksum value.
2
6. (a) M(x) = 11100011, G(x) = 1001, x
rM(x) = 11100011000
_____________
1001 | 11100011000
1001
1100
1001
1011
1001
1000
1001
10 = Remainder
T(x) = 11100011010, this is the message that should be transmitted.
(b) The received bits would be 01100011010. Then:
_____________
1001 | 01100011010
1001
1110
1001
1101
1001
1010
1001
110 = Remainder
Since the remainder does not equal 0, the receiver knows there has been an error.
7. In this scenario, the successful transmissions occur in the order of C, B, A, the attempted transmissions occur in the order of A, B, C, and there are at least 4 collisions.
D----------| C----------| B----------| A----------|
^ ^ ^ ^ ^ ^ ^
A B C A B A A
| | | |
\____\___________\____________\___-- Collisions.
3
8. Based on the graph:
A:
Destination Next
B C
C C
D C
E C
F C
B:
Destination Next
A E
C E
D E
E E
F E
C:
Destination Next
A A
B E
D E
E E
F F
D:
Destination Next
A E
B E
C E
E E
F E
E:
Destination Next
A C
B B
C C
D D
F C
F:
Destination Next
A C
B C
C C
D C
E C
9. For this answer, A-B means that the bridge has knowledge that to get to A, it must
send the packet to B.
A sends to C
B1 B2 B3 B4
A-A A-B1 A-B2 A-B2
C sends to A
B1 B2 B3 B4
A-A A-B1 A-B2 A-B2
C-B2 C-B3 C-C
D sends to C
B1 B2 B3 B4
A-A A-B1 A-B2 A-B2
C-B2 C-B3 C-C
D-B4 D-B2 D-D
10. (a) The packet will go to both B1 and B2, which will record M-M. Then they will
send the packet to L and the other bridge. Since neither of the bridges has an
entry for L in their table, they will continue to send the packet to each other in
a loop, while also giving L the packet multiple times. This will also replace the
M-M records with M-B1 or B2.
(b) Let’s assume that the packet from (a) is already circling clockwise around the
bridges. When L sends its packet, B1 has just obtained the packet from M. B1
believes M is at the bottom port and B2 believes M is at the top port. The packet
from L which hits B1 is sent to B2 from the bottom and then B1 again from the
top, a counter-clockwise loop. Now, a packet from M is moving clockwise around
the loop and a packet from L is moving counter-clockwise around the loop.
4