maximum speed last?（E） 令最大突发时间长度为 Δ t
秒。在极端情况下，漏桶在突发期间的开
始是充满的
（1MB），这期间数据流入桶内 10Δ t MB，流出包含 50Δ t MB，由等式
1+10Δ t=50Δ
t，得到 Δ t=1/40s，即 25ms。因此，以最大速率突发传送可维持 25ms 的时- 23 - 间。
29. The network of Fig. 5-37 uses RSVP with multicast trees for hosts 1 and 2 as shown. Suppose that host 3 requests a channel of bandwidth 2 MB/sec for a flow from host 1 and another channel of bandwidth 1 MB/sec for a flow from host 2. At the same time, host 4 requests a channel of bandwidth 2 MB/sec for a flow from host
1 and host 5 requests a channel of bandwidth 1 MB/sec for a flow from host 2. How 34. Suppose that host A is connected to a router R 1, R 1 is connected to another much total bandwidth will be reserved for these requests at routers A, B, C, E, H, J, router, R 2, and R 2 is connected to host B. Suppose that a TCP message that K, and L?(E) contains 900 bytes of data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B. Show the Total length, Identification, DF, MF, and
Fragment offset fields of the IP header in each packet transmitted over the three
links. Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of
512 bytes, including an 8-byte frame header, and link R2-B can support a maximum
frame size of 512 bytes including a 12-byte frame header.（M）
在 I1 最初的 IP 数据报会被分割成两个 IP 数据报，以后不会再分割了。 链路 A-R1：Length = 940; ID = x; DF = 0; MF = 0; Offset = 0 链路 R1-R2：
(1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0 (2) Length = 460; ID = x; DF = 0; MF = 0; Offset = 60 链路 R2-B:
(1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0 (2) Length = 46