Chapter 8 - Problems

Figure 8.27 shows a switch (router) in a datagram network. Find the output port for packets with the following destination addresses: a. Packet 1: 7176 b. Packet 2: 1233 c. Packet 3: 8766 d. Packet 4: 9144

Packet 1: 2 Packet 2: 3 Packet 3: 3 Packet 4: 2

We need a three-stage space-division switch with N = 100. We use 10 crossbars at the first and third stages and 4 crossbars at the middle stage. a. Draw the configuration diagram. b. Calculate the total number of crosspoints. c. Find the possible number of simultaneous connections. d. Find the possible number of simultaneous connections if we use a single crossbar (100 × 100). e. Find the blocking factor, the ratio of the number of connections in part c and in part d.

a. See Figure 8.1. b. The total number of crosspoints are Number of crosspoints = 10 (10 × 4) + 4 (10 × 10) + 10 (4 × 10) = 1200 c. Only four simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is 40. d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100 simultaneous connections. e. The blocking factor is 40/100 or 40 percent.

Repeat Problem 8-12 if we use 6 crossbars at the middle stage.

a. See Figure 8.2. b. The total number of crosspoints are Number of crosspoints = 10 (10 × 6) + 6 (10 × 10) + 10 (6 × 10) = 1800 c. Only six simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is 60. d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100 simultaneous connections. e. The blocking factor is 60/100 or 60 percent.

We mentioned that two types of networks, datagram and virtual-circuit, need a routing or switching table to find the output port from which the information belonging to a destination should be sent out, but a circuit-switched network has no need for such a table. Give the reason for this difference.

A datagram or virtual-circuit network handles packetized data. For each packet, the switch needs to consult its table to find the output port in the case of a datagram network, and to find the combination of the output port and the virtual circuit identifier in the case of a virtual-circuit network. In a circuit-switched network, data are not packetized; no routing information is carried with the data. The whole path is established during the setup phase

Redesign the configuration of Problem 8-12 using the Clos criteria.

According to Clos, n = (N/2)^1/2 = 7.07. We can choose n = 8. The number of crossbars in the first stage can be 13 (to have similar crossbars). Some of the input lines can be left unused. We then have k = 2n − 1 = 15. Figure 8.3 shows the configuration. 3 (8 × 15) + 15 (13 × 13) + 13 (15 × 8) = 5655 crosspoints The number of crosspoints is still much less than the case with one crossbar (10,000). We can see that there is no blocking involved because each 8 input line has 15 intermediate crossbars. The total number of crosspoints here is a little greater than the minimum number of crosspoints according to Clos using the formula 4N[(2N)^1/2 − 1], which is 5257.

Transmission of information in any network involves end-to-end addressing and sometimes local addressing (such as VCI). Table 8.2 shows the types of networks and the addressing mechanism used in each of them. Answer the following questions: a. Why does a circuit-switched network need end-to-end addressing during the setup and teardown phases? Why are no addresses needed during the data transfer phase for this type of network? b. Why does a datagram network need only end-to-end addressing during the data transfer phase, but no addressing during the setup and teardown phases? c. Why does a virtual-circuit network need addresses during all three phases?

In a circuit-switched network, end-to-end addressing is needed during the setup and teardown phase to create a connection for the whole data transfer phase. After the connection is made, the data flow travels through the already-reserved resources. The switches remain connected for the entire duration of the data transfer; there is no need for further addressing. In a datagram network, each packet is independent. The routing of a packet is done for each individual packet. Each packet, therefore, needs to carry an endto-end address. There is no setup and teardown phases in a datagram network (connectionless transmission). The entries in the routing table are somehow permanent and made by other processes such as routing protocols. . In a virtual-circuit network, there is a need for end-to-end addressing during the setup and teardown phases to make the corresponding entry in the switching table. The entry is made for each request for connection. During the data transfer phase, each packet needs to carry a virtual-circuit identifier to show which virtual-circuit that particular packet follows.

An entry in the switching table of a virtual-circuit network is normally created during the setup phase and deleted during the teardown phase. In other words, the entries in this type of network reflect the current connections, the activity in the network. In contrast, the entries in a routing table of a datagram network do not depend on the current connections; they show the configuration of the network and how any packet should be routed to a final destination. The entries may remain the same even if there is no activity in the network. The routing tables, however, are updated if there are changes in the network. Can you explain the reason for these two different characteristics? Can we say that a virtual-circuit is a connection-oriented network and a datagram network is a connectionless network because of the above characteristics?

In circuit-switched and virtual-circuit networks, we are dealing with connections. A connection needs to be made before the data transfer can take place. In the case of a circuit-switched network, a physical connection is established during the setup phase and the is broken during the teardown phase. In the case of a virtual-circuit network, a virtual connection is made during setup and is broken during the teardown phase; the connection is virtual, because it is an entry in the table. These two types of networks are considered connection-oriented. In the case of a datagram network no connection is made. Any time a switch in this type of network receives a packet, it consults its table for routing information. This type of network is considered a connectionless network.

Figure 8.28 shows a switch in a virtual-circuit network. Find the output port and the output VCI for packets with the following input port and input VCI addresses: a. Packet 1: 3, 78 b. Packet 2: 2, 92 c. Packet 3: 4, 56 d. Packet 4: 2, 71

Packet 1: 2, 70 Packet 2: 1, 45 Packet 3: 3, 11 Packet 4: 4, 41

The minimum number of columns in a datagram network is two; the minimum number of columns in a virtual-circuit network is four. Can you explain the reason? Is the difference related to the type of addresses carried in the packets of each network?

The switching or routing in a datagram network is based on the final destination address, which is global. The minimum number of entries is two; one for the final destination and one for the output port. Here the input port, from which the packet has arrived is irrelevant. The switching or routing in a virtual-circuit network is based on the virtual circuit identifier, which has a local jurisdiction. This means that two different input or output ports may use the same virtual circuit number. Therefore, four pieces of information are required: input port, input virtual circuit number, output port, and output virtual circuit number.

A path in a digital circuit-switched network has a data rate of 1 Mbps. The exchange of 1000 bits is required for the setup and teardown phases. The distance between two parties is 5000 km. Answer the following questions if the propagation speed is 2 × 10^8 m: a. What is the total delay if 1000 bits of data are exchanged during the datatransfer phase? b. What is the total delay if 100,000 bits of data are exchanged during the data-transfer phase? c. What is the total delay if 1,000,000 bits of data are exchanged during the data-transfer phase? d. Find the delay per 1000 bits of data for each of the above cases and compare them. What can you infer?

We assume that the setup phase is a two-way communication and the teardown phase is a one-way communication. These two phases are common for all three cases. The delay for these two phases can be calculated as three propagation delays and three transmission delays or 3 [(5000 km)/ (2 ×10^8 m/s)]+ 3 [(1000 bits/1 Mbps)] = 75 ms + 3 ms = 78 ms We assume that the data transfer is in one direction; the total delay is then delay for setup and teardown + propagation delay + transmission delay a. 78 + 25 + 1 = 104 ms b. 78 + 25 + 100 = 203 ms c. 78 + 25 + 1000 = 1103 ms d. In case a, we have 104 ms. In case b we have 203/100 = 2.03 ms. In case c, we have 1103/1000 = 1.101 ms. The ratio for case c is the smallest because we use one setup and teardown phase to send more data.

Five equal-size datagrams belonging to the same message leave for the destination one after another. However, they travel through different paths as shown in Table 8.1. We assume that the delay for each switch (including waiting and processing) is 3, 10, 20, 7, and 20 ms respectively. Assuming that the propagation speed is 2 × 108 m, find the order the datagrams arrive at the destination and the delay for each. Ignore any other delays in transmission.

We assume that the transmission time is negligible in this case. This means that we suppose all datagrams start at time 0. The arrival timed are calculated as: First: (3200 Km) / (2 × 10^8 m/s) + (3 + 20 + 20) = 59.0 ms Second: (11700 Km) / (2 × 10^8 m/s) + (3 + 10 + 20) = 91.5 ms Third: (12200 Km) / (2 × 10^8 m/s) + (3 + 10+ 20 + 20) = 114.0 ms Fourth: (10200 Km) / (2 × 10^8 m/s) + (3 + 7 + 20) = 81.0 ms Fifth: (10700 Km) / (2 × 10^8 m/s) + (3 + 7 + 20 + 20) = 103.5 ms The order of arrival is: 3 → 5 → 2 → 4 → 1

It is obvious that a router or a switch needs to search to find information in the corresponding table. The searching in a routing table for a datagram network is based on the destination address; the searching in a switching table in a virtual-circuit network is based on the combination of incoming port and incoming VCI. Explain the reason and define how these tables must be ordered (sorted) based on these values.

When a packet arrives at a router in a datagram network, the only information in the packet that can help the router in its routing is the destination address of the packet. The table then is sorted to make the searching faster. Today's routers use some sophisticated searching techniques. When a packet arrives at a switch in a virtual-circuit network, the pair (input port, input VCI) can uniquely determined how the packet is to be routed; the pair is the only two pieces of information in the packet that is used for routing. The table in the virtual-circuit switch is sorted based on the this pair. However, since the number of port numbers is normally much smaller than the number of virtual circuits assigned to each port, sorting is done in two steps: first according to the input port number and second according to the input VCI.

Consider an n × k crossbar switch with n inputs and k outputs. a. Can we say that the switch acts as a multiplexer if n > k? b. Can we say that the switch acts as a demultiplexer if n < k?

a. If n > k, an n × k crossbar is like a multiplexer that combines n inputs into k outputs. However, we need to know that a regular multiplexer discussed in Chapter 6 is n × 1. b. If n < k, an n × k crossbar is like a demultiplexer that divides n inputs into k outputs. However, we need to know that a regular demultiplexer discussed in Chapter 6 is 1 × n.

Answer the following questions: a. Can a routing table in a datagram network have two entries with the same destination address? Explain. b. Can a switching table in a virtual-circuit network have two entries with the same input port number? With the same output port number? With the same incoming VCIs? With the same outgoing VCIs? With the same incoming values (port, VCI)? With the same outgoing values (port, VCI)?

a. In a datagram network, the destination addresses are unique. They cannot be duplicated in the routing table. b. In a virtual-circuit network, the VCIs are local. A VCI is unique only in relationship to a port. In other words, the (port, VCI) combination is unique. This means that we can have two entries with the same input or output ports. We can have two entries with the same VCIs. However, we cannot have two entries with the same (port, VCI) pair

We need to have a space-division switch with 1000 inputs and outputs. What is the total number of crosspoints in each of the following cases? a. Using a single crossbar. b. Using a multi-stage switch based on the Clos criteria.

a. Total crosspoints = N^2 = 1000^2 = 1,000,000 b. Total crosspoints ≥ 4Ν[(2Ν)^1/2 −1] ≥ 174,886. With less than 200,000 crosspoints we can design a three-stage switch. We can use n = (N/2)^1/2 =23 and choose k = 45. The total number of crosspoints is 178,200.

We need a three-stage time-space-time switch with N = 100. We use 10 TSIs at the first and third stages and 4 crossbars at the middle stage. a. Draw the configuration diagram. b. Calculate the total number of crosspoints. c. Calculate the total number of memory locations we need for the TSIs.

a. We first solve the problem using only crossbars and then we replace the crossbars at the first and the last stage with TSIs. Figure 8.1 shows the solution using only crossbars. We can replace the crossbar at the first and third stages with TSIs as shown in Figure 8.4. The total number of crosspoints is 400 and the total number of memory locations is 200. Each TSI at the first stage needs one TDM multiplexer and one TDM demultiplexer. The multiplexer is 10 × 1, but the demultiplexer is 1 × 4. In other words, the input frame has 10 slots and the output frame has only 4 slots. The data in the first slot of all input TSIs are directed to the first switch, the output in the second slot are directed to the second switch, and so on. b. We can see the inefficiency in the first solution. Since the slots are separated in time, only one of the switches at the middle stage is active at each moment. This means that, instead of 4 crossbars, we could have used only one with the same result. Figure 8.5 shows the new design. In this case we still need 200 memory locations but only 100 crosspoints.