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Title 1 (14 minutes)
Network as shown in FIG. A time at t = 0 starts transmitting the file to a 2Mbits C; B at t = 0.1 + e s (e infinitely small positive real number close to 0) is sent to the file 1Mbits D. Ignoring propagation delay and processing delay node.
Please answer the following questions:
(1) if the figure using the network storage - packet switching and forwarding method, then A file will be delivered to the C 2Mbits how long? B 1Mbits files will be delivered to the D How long?
(2) if the network using FIG Storage - forwarding a packet switched mode, packet length, etc. of the 1 kbits long, and ignores the packet header overhead and overhead packet disassembly, then A will deliver documents to 2Mbits C takes approximately how long time? B 1Mbits files will be delivered to D takes approximately how long?
(3) Compared with the packet switching packet switching, which exchange more equitable? (I.e., a small amount of data transmission with less transmission data with greater duration)
Note: K = 10 . 3, M = 10 . 6.
Knowledge Point 2 Review
- Packet switching: refers to the complete information sent from a source host to the target host.
- Packet switching: the packet a split into small packets for transmission, then reassembled into a complete information on the target host.
- Statistical multiplexing, when sharing routing, packet sequence when two or more hosts uncertain demand shared link.
- Packet switching and packet switching are used in storage - forward switching mode. The difference is in a complete packet switching packets "store - forward." Packet switching performed in small groups "store - forward"
- Packet switching: transmission delay = L / R
- Message exchange: for each transmission packet for an M / R s
- Packet-switched packet delivery time: T = M / R + ( h-1) L / R = M / R + nL / R.
Note: L refers to the length of the message packet, R link bandwidth, M refers to the packet length, h refers to the number of hops, n refers to the number of routes.
3 My answer analysis
Title Analysis: Suppose A, B sharing route for E, C, D share routing is F. t = 0 (s), A -> 2Mbits -> C; t = (0.1 + e) (s) (e infinitely small positive real number close to 0), B-> 1Mbits -> D , and ignoring propagation delay and processing delay node.
Solution: (1) Analysis: The store - and-forward message switching. A need therefore to wait until the entire message to the route E by E Forward all the time to F, B began to send messages to the routing E.
A transmitting 2Mbits file F is the time required to route 2/10 + 2/20 = 0.3 (s)
Thus, A document will be delivered to 2Mbits C is the time required for 0.3 + 2/10 = 0.5 ( s)
time B will be delivered to the desired file 1Mbits D-E is +. 1 0.3-0.1 / 10 + (2 / 10-0.1). 1 + / + 20 is. 1 / 0.55 = 10-E (S)
(2) analysis: using store - and-forward packet switching and ignore the packet header overhead and to the demolition of the overhead message. Accordingly packet length of 1 kbits, the packets may be split into 2000 2Mbits groups, packet 1Mbits 1000 may split into groups, on a packet-switched transmission 10Mb / s link delay of 0.001 / 10 = 0.1 (ms ), transmission delay on the 20Mb / s link to 0.05 (ms). When 0.1 seconds before, and only sends packets in the packet A, at 0.1 + e second, B is also added thereto, and share the shared link shared routing resources. Case it is assumed A, B ideal state of cross transmission.
During 0-0.1 seconds, the route A to E transmit a packet transmission amount is 100 / 0.1 = 1000 (a), the number of packets transmitted to the route E A route F is 999, the number of packets transmitted to the route F is C 998, in this time period, since the route a to E restriction link transmission rate of the link between the routing route E to F only 50% utilization.
At this time, start adding B, B, and A and A each first transmission packet 1000, the transmission B, and both routes are alternately occupied shared link shared resources, but due to the route between the route E to F link is 20Mb / s, and at this time, utilization rate is 100%, it can be seen as a to C, B to D is sent over the data exchange link two separate 10Mb / s of.
The resulting sum, A document will be delivered to 2Mbits C is the time required to 2/10 2 + 0.0001 = .2001 (S)
B of documents to be delivered 1Mbits D is the time required for 1/2 10 + 0.0001 = 0.1001 ( S)
(3) in the context of this question, compared with packet switching packet switching, packet switching exchange fairer.
4 standard answer comparative analysis
- Reflections on the first question in the B error has occurred, and there is no dynamic to analyze the link between the two, resulting in an error. After correction, B A transmission process will be transmitted to the packet transmission after the start of the message E to E. At this time should be 0.2- (0.1 + e) + 1/10 + 1/20 + 1/10 = 0.35-e≈0.35s = 350ms, this time coincides with the answer.
- As for the cause of the error, because I think that A transmits the packet to E, then E to F packet transmission process in, B should be unable to transmit the message to the E, E because there are A packet transmission buffer the same token, the route F I too consider, which led to errors.
- A second question in C is delivered to the 0.20015s was my second mistake, after careful analysis found that there is a link bandwidth is 20Mbps, propagation delay at this time should be half of 10Mbps, 10Mbps and should not be considered in accordance with general, leading to calculate the details wrong thinking, there is also the details of the calculation error, careless computing, resulting in less 0.001s calculations, thinking ill less 0.0005s. Correct time A to C 2/10 + 1000/20000000 + 1000/10000000 = 0.20015s≈0.2s = 200ms, the correct time from B to D is 1/10 + 1000/1000 + 20000000/20000000 + 1000 / 10000000 = 0.1002s≈0.1s = 100ms.
5 standard answer
Score guidance:
[scoring]:
1. 14 points out of the question, as long as serious homework (rightly or wrongly), you can get basic points: 5 points;
2. Each small ask the right answer, the answer in reference to the sub-score mark for 14 minutes;
3. However, an incorrect analysis result is incorrect or partially correct, as appropriate to the sub.
Answer []:
1) A first transmitter since the packet, so the packet A is ahead of B in the message queue of the router output link. Thus, A to C deliver desired packets 2Mbits time is:
2/2 + 10 / + 20 is 2/10 = 0.5s = 500ms; (. 3 minutes)
B D deliver documents to 1Mbits time required is:
1/10 + 2/20 (queuing delay) + 1/20 + 1/ 10 = 0.35s = 350ms. (3 minutes)
2) A time required for file delivery 2Mbits to about C:
2/10 1000 + / 1000 + 20000000/10000000 = 0.20015s≈0.2s = 200ms; (3 minutes)
B D required for delivery of documents to 1Mbits time is about:
1/1000 + 10/20000000 (queue) 1000 + / 1000 + 20000000/10000000
= 0.1002s≈0.1s = 100ms. (3 points)
3) packet switching is more than fair exchange of messages. (2 minutes)
Summary 6
- Exercise should be considered in the next round, do not know the details should grasp the knowledge points clear, it can not be taken for granted.
- Calculation should be more careful attention to detail.
