内容发布更新时间 : 2024/12/23 13:20:46星期一 下面是文章的全部内容请认真阅读。

第 1 列，流 2 的第 1 列，流 3 的第 2 列，流 2 的第 2 列， 以此类推，最后形成 270 列宽 9

的第 1 列，随后是流 1

行高的帧。

为了使分组交换比电路交换快，必须：

OC-3c 流中的用户实际数据传输速率比 OC-3 流的速率略高（149.760Mbps所以：

和 43. Suppose that x bits of user data are to be transmitted over a k-hop path in a 148.608Mbps），因为路径开销仅在 SPE 中出现一次，而不是当使用 3 条单独 packet-switched network as a series of packets, each containing p data bits and h OC-1 header bits, with x >>p + h. The bit rate of the lines is b bps and the propagation 流时出现的 3 次。换句话说，OC-3c 中 270 列中的 delay is negligible. What value of p minimizes the total delay?（M） 260 列可用于用户数据，而在 所需要的分组总数是 x/p，因此加上头信息的总数据量为(p+h)x/p 位。 OC-3 中仅能使用 258 列。更高层次的串联帧（如 OC-12c）也存在这样的情况。]] 源端发送这些位需要时间为(p+h)x/pb， OC-12c 帧有 12*90=1080 列和 9 行。其中段开销和线路开销占 12*3=36中间的路由器重传最后一个分组所花的总时间为(k-1)(p+h)/b，

列，这

样同步载荷信封就有 1080-36=1044 列。SPE 中仅 1 列用于路径开销，结果就是 1043 列用于用户数据。

由于每列 9 个字节，因此一个 OC-12c 帧中用户数据比特数是 8 ×\慇2X9×1043＝75096。 每秒 8000 帧，得到用户数据速率 75096×8000 =600768000bps，即 600.768Mbps。 所以，在一条 OC-12c

连接中可提供的用户带宽是 600.768Mbps。 因此我们得到的总的延迟为 41. Three packet-switching networks each contain n nodes. The first network has a star topology with a central switch, the second is a (bidirectional) ring, and the third is fully interconnected, with a wire from every node to every other node. What are the best-, average-, and-worst case transmission paths in hops?（E） 三种网络的特性如下：

星型：最好为 2，最差为 2，平均为 2； 环型：最好为 1，最差为 n/2，平均为 n/4 如果考虑 n 为奇偶数，

则 n 为奇数时，最坏为（n-1）/2，平均为（n+1）/4 n 为偶数时，最坏为 n/2，平均为 n2/4(n1) 全连接：最好为 1，最差为 1，平均为 1。 ，

对该函数求 p 的导数，得到 ，令 得到

因为p＞0，所以 故 时能使总的延迟最小。 对于电路交换， t= s 时电路建立起来；t＝s+x/d 时报文的最后一位发送完毕；t= s+x/d+kb 时报文到达目的地。而对于分组交换，最后一位在 t=x/b时发送完毕。

为到达最终目的地，最后一个分组必须被中间的路由器重发 k-1 次，每次重发花

42. Compare the delay in sending an x-bit message over a k-hop path in a 时间 p/b，所以总的延迟为 circuit-switched network and in a (lightly loaded) packet-switched network. The circuit setup time is s sec, the propagation delay is d sec per hop, the packet size is p

bits, and the data rate is b bps. Under what conditions does the packet network have

a lower delay?(M)

44. In a typical mobile phone system with hexagonal cells, it is forbidden to reuse

a frequency band in an adjacent cell. If 840 frequencies are available, how many can

be used in a given cell?（B）

每个蜂窝单元都有 6 个邻居。假定中间的单元使用频率 A，它的邻居则可以依次

使用 B，C，B，C，B，C，也就是说，只需用三个相互独立的单元就足够 another,

the current call is abruptly terminated, even though all transmitters and receivers are functioning perfectly. Why?（E）

在邻近的蜂窝单元中频率不能复用。所以当一名用户从一个单元移动到另一个单 元时，必须给他分配一个新的频率。如果当用户移到一个新的单元，但是当前的新 单元中的所有频率都在使用中，则用户的呼叫必须被终止。

了。所以，

每个单元可以使用 280 种频率。

47. Sometimes when a mobile user crosses the boundary from one cell to

- 7 -

48. D-AMPS has appreciably worse speech quality than GSM. Is this due to the requirement that D-AMPS be backward compatible with AMPS, whereas GSM had

no such constraint? If not, what is the cause?（M） It is not caused directly by the need for backward compatibility. The 30 kHz channel

was indeed a requirement, but the designers of D-AMPS did not have to stuff three users

into it. They could have put two users in each channel, increasing the payload before error correction from 260 ×50=13 kbps to 260×\慇2X75 =19.5 kbps. Thus, the quality loss was

an intentional trade-off to put more users per cell and thus get away with bigger cells. 49. Calculate the maximum number of users that D-AMPS can support simultaneously within a single cell. Do the same calculation for GSM. Explain the difference.（M）

D-AMPS uses 832 channels (in each direction) with three users sharing a single channel. This allows D-AMPS to support up to 2496 users simultaneously per cell. GSM

uses 124 channels with eight users sharing a single channel. This allows GSM to support

up to 992 users simultaneously. Both systems use about the same amount of spectrum (25

MHz in each direction).

D-AMPS uses 30 KHz×\慇2X832 = 24.96 MHz. GSM uses 200 KHz ×\慇2X124 =24.80 MHz.

The difference can be mainly attributed to the better speech quality provided by GSM (13

Kbps per user) over D-AMPS (8 Kbps per user). 50. Suppose that A, B, and C are simultaneously transmitting 0 bits, using a CDMA system with the chip sequences of Fig. 2-45(b). What is the resulting chip sequence?（E）

传输 0，则时间片序列取其补码，传输 1，则时间序列取其本身。 将 A，B，C 相加后，取补码得结果为：(+3 +1 +1 -1 -3 -1 -1 +1).

51. In the discussion about orthogonality of CDMA chip sequences, it was stated

that if S?T = 0 then S??T is also 0. Prove this.(E)

percent

chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through? （E）

由于每一帧有 0.8 p=0.810=0.107。

为使信息完整的到达接收方，发送一次成功的概率是 p 的概率正确到达，整个信息正确到达的概率为

，二次成功的概率

By definition

If T sends a 0 bit instead of 1 bit, its chip sequence is negated, with the i-th element

是

becoming Ti . Thus,

(1-p)p，三次成功的概率为(1-p)2 p，i 次

52. Consider a different way of looking at the orthogonality property of - 8 -

CDMA

chip sequences. Each bit in a pair of sequences can match or not match. Express the

orthogonality property in terms of matches and mismatches.

When two elements match, their product is +1. When they do not match, their product

is -1. To make the sum 0, there must be as many matches as mismatches. Thus, two chip

sequences are orthogonal if exactly half of the corresponding elements match and exactly

half do not match.

53. A CDMA receiver gets the following chips: (-1 +1 -3 +1 -1 -3 +1 +1). Assuming

the chip sequences defined in Fig. 2-45(b), which stations transmitted, and which

bits did each one send?（E）

分别与（-1 +1 -3 +1 -1 -3 +1 +1）做内积：

A：（-1 +1 -3 +1 -1 -3 +1 +1）d (-1 -1 -1 +1 +1 -1 +1 +1) /8 = 1 B：（-1 +1 -3 +1 -1 -3 +1 +1）d (-1 -1 +1 -1 +1 +1 +1 -1) /8 = -1 C：（-1 +1 -3 +1 -1 -3 +1 +1）d (-1 +1 -1 +1 +1 +1 -1 -1) /8 = 0 D：（-1 +1 -3 +1 -1 -3 +1 +1）d (-1 +1 -1 -1 -1 -1 +1 -1) /8 = 1

所以 A，D 发送的都是 1 bits，B 发送的是 0 bit,，C 没有发送 Chapter 3 The Data Link Layer Problems

1. An upper-layer packet is split into 10 frames, each of which has an 80

次成功的概率为(1-p)i-1p，因此平均的发送