计网一些题

2. (单选题, 3分)Consider a router with the switching fabric based on sharing memory. The memory access speed (read and write) is B packets per second. The overall forwarding throughput is always _______

• A. greater than B packets per second
• B. greater than B/2 packets per second
• C. less than B/2 packets per second
• D. less than (√¯B¯)/2 packets per second

Consider a subnet with prefix 101.101.101.64/26. Give an example of one IP address (of from xxx.xxx.xxx.xxx) that can be assigned to this network. Suppose an ISP owns the block of addresses of the form 101.101.128/19. Suppose it wants to create four subnets from this block, with each block having the same number of IP addresses. What are the prefixes (of from a.b.c.d/x) for the four subnets?

101.101101.64/26

.65~126 （展开之）

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.00……

.01……

.10……

.11……

22. (简答题, 10分)

Consider a datagram network using 32-bit host addresses. Suppose a router has four links, numbered 0 through 3, and packets are to be forwarded to the link interfaces as follows:

1)       Provide a forwarding table that has five entries, uses longest prefix matching, and forwards packets to the correct link interfaces. (6 points)

110000：2

11000011：3

相同部分之后不是全一 单拎出来一个

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1. (简答题) You are sending a message with 10 segments, each having a sequence number from 1 to 10. The flow control algorithm used between the sender and receiver is Go-Back-N using cumulative ACKs where the sender window size is 3. Under the set of assumptions below, how long does it take before the sender deems the transmission complete if:

   ·       The first transmission of segment 5 is not received by the receiver.

   ·       The second transmission of segment 5 is received by the receiver but the ACK is slow, taking 60 ms (instead of 20ms as mentioned below).  It still takes 20 ms for the segment to be sent from the sender to the receiver.

   ·       The first ACK sent by the receiver after receiving segment 8 is not received.

   ·       接收器未接收段5的第一次传输。

   ·       接收器接收段 5 的第二次传输，但 ACK 很慢，需要 60 毫秒（而不是下面提到的 20 毫秒）。段从发送方发送到接收方仍需要 20 毫秒。

   ·       接收端在接收段8后发送的第一个ACK未收到。

Several assumptions:

• Sequence numbers are assigned to segments rather than bytes.

• Unless indicated otherwise, all segments were received properly and received in the same order as they were sent.

• It takes the sender  10 ms to“push” the segment onto the network. This means that if segment 1 pushed onto the network starting at time 0, then segment 2 can started  to be pushed onto the network at 10 ms.

• Unless indicated  otherwise, each successfully transmitted segment has a round trip time of  exactly 40ms (20 ms each way).  This  time includes transmission time.

• The sender only  retransmits on timeouts. The timeout timer is set at 60 ms for all  segments.

• No congestion  control is used.

• There is no      processing delay. As soon as an ACK is received by the sender, it can      immediately (in the same time instant) start pushing a segment onto the  network if appropriate.

• 发送方需要 10 毫秒才能将分段“推送”到网络上。这意味着，如果分段 1 从时间 0 开始推送到网络上，则分段 2 可以在 10 毫秒时开始推送到网络上。

• 除非另有说明，否则每个成功传输的段的往返时间正好为 40 毫秒（单程 20 毫秒）。此时间包括传输时间。

• 发送方仅在超时时重新传输。所有分段的超时计时器设置为 60 毫秒。

• 不使用拥塞控制。

• 没有处理延迟。一旦发送方收到 ACK，它就可以立即（同时立即）开始将网段推送到网络上（如果适用）。

Show your work by filling in the table on the following page. For each time interval, write the sequence number of any segment that is sent (i.e., starting to be pushed onto the network) in the SEND column.  Similarly, write the sequence number included in any ACK received in the RECV column.  To help you get started, segments 1 is done for you.  Not all of the rows will be necessary.（10 points）

GBN

SR