计算机网络-自顶向下方法（7th) 第五章 Problems 英文题目1-15+中文答案

P1.

Looking at Figure 5.3 , enumerate the paths from y to u that do not contain any loops.

(y,x,u)
(y,w,x,u)
(y,z,w,x,u)
(y,x,v,u)
(y,w,v,u)
(y,x,w,v,u)
(y,z,w,v,u)
(y,w,u)
(y,z,w,u)
(y,x,w,u)
(y,x,v,w,u)

(不太知道怎么列举，我是根据后缀列举的，比如前面的分别是以x,u/v,u/w,u…)

P2.

Repeat Problem P1 for paths from x to z, z to u, and z to w.
略

P3.

Consider the following network. With the indicated link costs, use Dijkstra’s shortest-path algorithm to compute the shortest path from x to all network nodes. Show how the algorithm works by computing a table similar to Table 5.1.

P4.

Consider the network shown in Problem P3. Using Dijkstra’s algorithm, and showing your work using a table similar to Table 5.1 , do the following:
a. Compute the shortest path from t to all network nodes.
b. Compute the shortest path from u to all network nodes.
c. Compute the shortest path from v to all network nodes.
d. Compute the shortest path from w to all network nodes.
e. Compute the shortest path from y to all network nodes.
f. Compute the shortest path from z to all network nodes.

离谱…

P5.

Consider the network shown below, and assume that each node initially knows the costs to each of its neighbors. Consider the distance-vector algorithm and show the distance table entries at node z.

最后一步我不能理解，明明没有更新任何内容

P6. Consider a general topology (that is, not the specific network shown above) and a synchronous version of the distance-vector algorithm. Suppose that at each iteration, a node exchanges its distance vectors with its neighbors and receives their distance vectors. Assuming that the algorithm begins with each node knowing only the costs to its immediate neighbors, what is the maximum number of iterations required before the distributed algorithm converges? Justify your answer.

设d为最长无环路径长度。
则最多迭代d-1次。

原因：对于任意点x，第i次迭代可以确认长度为i+1之内的代价最小的路径。

P7.

Consider the network fragment shown below. x has only two attached neighbors, w and y. w has a minimum-cost path to destination u (not shown) of 5, and y has a minimum-cost path to u of 6. The complete paths from w and y to u (and between w and y) are not shown. All link costs in the network have strictly positive integer values.

a. Give x’s distance vector for destinations w, y, and u.
b. Give a link-cost change for either c(x, w) or c(x, y) such that x will inform its neighbors of a new minimum-cost path to u as a result of executing the distance-vector algorithm.
c. Give a link-cost change for either c(x, w) or c(x, y) such that x will not inform its neighbors of a new minimum-cost path to u as a result of executing the distance-vector algorithm

a, Dx(w) = 2, Dx(y) = 4, Dx(u) = 7
b, c(x,w) = 8（大于6就可以）
c, c(x,y) = 6 (大于等于1就可以)

P8.

Consider the three-node topology shown in Figure 5.6 . Rather than having the link costs shown in Figure 5.6 , the link costs are c(x,y)=3, c(y,z)=6, c(z,x)=4。Compute the distance tables after the initialization step and after each iteration of a synchronous version of the distance-vector algorithm (as we did in our earlier discussion of Figure 5.6 ).

P9.

Consider the count-to-infinity problem in the distance vector routing. Will the count-to-infinity problem occur if we decrease the cost of a link? Why? How about if we connect two nodes which do not have a link?

不会。这样的结构不会形成环路。环路通常由链路代价突然增高导致。

P10.

Argue that for the distance-vector algorithm in Figure 5.6 , each value in the distance vector D(x) is non-increasing and will eventually stabilize in a finite number of steps.

每一次更新，至少有一个距离向量的某维度的值减小。(假如没有任何距离向量发生改变，更新就停止。)
由于这些值都是整数、不可能增加、也不可能小于0、所以减小的次数必然是有限的，所以更新也必然是有限的。

P11.

Consider Figure 5.7. Suppose there is another router w, connected to router y and z. The costs of all links are given as follows: c(x,y)=4, c(x,z)=50, c(y,w)=1, c(z,w)=1, c(y,z)=3. Suppose that poisoned reverse is used in the distance-vector routing algorithm.

a. When the distance vector routing is stabilized, router w, y, and z inform their distances to x to each other. What distance values do they tell each other?

这时官方给的答案。我这部分理解的不是太好，貌似：最短线路只要不是末端直接连接着y，就不必把距离标记为正无穷。

b. Now suppose that the link cost between x and y increases to 60. Will there be a count-to-infinity problem even if poisoned reverse is used? Why or why not? If there is a count-to-infinity problem, then how many iterations are needed for the distance-vector routing to reach a stable state again? Justify your answer.

会的，问题出在z上，因为z和x的最短路径经过了y，z却告知y的Dz(x)=6（这是毒性逆转作用不到的地方），y就会把Dy(x)更新成9（实际上并不是）,然后陷入了无穷计数。

c. How do you modify c(y, z) such that there is no count-to-infinity problem at all if c(y,x) changes from 4 to 60?

将c(y,z)设置为无穷大。

P12.

Describe how loops in paths can be detected in BGP.
在寻路时假如发现路径上有己方路由器，那么就是出现了环。

P13.

Will a BGP router always choose the loop-free route with the shortest ASpath length? Justify your answer.
不是。在BGP中，路径上允许并存不同的代价，而最终的结果可能是权衡各种代价来得出的，未必是最短路。

P14.

Consider the network shown below. Suppose AS3 and AS2 are running OSPF for their intra-AS routing protocol. Suppose AS1 and AS4 are running RIP for their intra-AS routing protocol. Suppose eBGP and iBGP are used for the inter-AS routing protocol. Initially suppose there is no physical link between AS2 and AS4.

a. Router 3c learns about prefix x from which routing protocol: OSPF, RIP, eBGP, or iBGP?
b. Router 3a learns about x from which routing protocol?
c. Router 1c learns about x from which routing protocol?
d. Router 1d learns about x from which routing protocol?

a, eBGP(从4c得知)
b, iBGP
c, eBGP
d, iBGP

其实不太理解。猜测：前缀x是BGP协议的一部分，只能通过iBGP或者eBGP传播

P15.

Referring to the previous problem, once router 1d learns about x it will put an entry (x, I) in its forwarding table.
a. Will I be equal to l1l_1l1 or l2l_2l2 for this entry? Explain why in one sentence.
b. Now suppose that there is a physical link between AS2 and AS4, shown by the dotted line. Suppose router 1d learns that x is accessible via AS2 as well as via AS3. Will I be set tol1l_1l1 or l2l_2l2 ? Explain why in one sentence.
c. Now suppose there is another AS, called AS5, which lies on the path between AS2 and AS4 (not shown in diagram). Suppose router 1d learns that x is accessible via AS2 AS5 AS4 as well as via AS3 AS4. Will I be set to l1l_1l1 or l2l_2l2? Explain why in one sentence.

a, l1，因为l1的距离最短（跳数成本最小）
b, l2, 因为它距离NEXT-HOP最近
c, l1,它的AS-PATH最小。