计算机网络第四章作业.doc
【精品文档】如有侵权，请联系网站删除，仅供学习与交流计算机网络第四章作业.精品文档.第四章P450-22: Consider the following network. With the indicated link costs, use Dijkstras shortest-path algorithm to compute the shortest path from x to all network nodes. Show how the algorithm works by computing a table.zxyVwtus126681483425914AnswerProblem 22StepND(s),p(s) D(t),p(t)D(u),p(u)D(v),p(v)D(w),p(w)D(y),p(y)D(z),p(z)0x8,x6,x6,x1xw14,w8,x6,x2xwy15,y14,w7,y18,y3xwyv11,v10,v18,y4xwyvu14,u11,v18,y5xwyvut12,t16,t6xwyvuts16,t7xwyvutsz y t zx v s w uRouting Table:DESCOSTNEST HOPx0y6yw6wv7yt11yu10yz16ys12yP451-25: Consider the network fragment shown below. x has only two attached neighbors, w and y. w has a minimum-cost path to destination u 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 xs 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.wxy25AnswerProblem 25a. Dx(y) = 5, Dx(w) = 2, Dx(u) = 7b. First consider what happens if c(x,y) changes. If c(x,y) becomes larger or smaller (as long as c(x,y) > 0), the least cost path from x to u will still have cost at least 7. Thus a change in c(x,y) will not cause x to inform its neighbors of any changes. Now consider if c(x,w) changes. If c(x,w) = e £ 1, then the least-cost path to u continues to pass through w and its cost changes to 5 + e; x will inform its neighbors of this new cost. If c(x,w) = d > 6, then the least cost path now passes through y and has cost 11; again x will inform its neighbors of this new cost. c. Any change in link cost c(x,y) will not cause x to inform its neighbors of a new minimum-cost path to u .
