Take the afternoon
off turn it in on Thursday
Resubmissions must be handwritten
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1. For this diagram as modified from the text, identify for each node tell what characteristics guide a host to forward data to that node. For instance, Ethernet is characterized by two addresses in the E’net header (identify them for a.). Getting to other nodes may simply depend on an additional bit of information in the E’net header (E’type). Identify it. Other nodes are located by additional headers. What are they? What additional data must be supplied in these additional headers. Use the diagrams on the web site page “TCP/IP Headers.” As an example I might describe access to a POP3 mail process as:
E’net (DMAC, SMAC, E’type = 0x800); IP (ver. = 0x04, length = 20/0x14, Protocol = 0x06 –TCP, Src IP, Dest IP); TCP (Src port, Dest port = 110, Seq, Ack)
The headers required for this service are E’net, IP, TCP. For parameters that are dynamically assigned the name is sufficient, unless they will be assigned a standard dynamic value such as 0.0.0.0 or ff-ff-ff-ff-ff-ff. Note for instance, that E’net (a) will only have an E’net header. Remember also that ARP and RARP are nearly the same protocol so only ARP is included. You don’t need to include the minor fields. Remember that the E’net fields and for the most part the IP fields will be repetitive. If you have a question about which field to use, please e-mail me. Do not do IGMP or RARP.
The j. node will be, for instance:
E’net (DMAC, SMAC, E’type = 0x806); IP (ver. = 0x04, length = 20/0x14, Protocol = 17/0x11 – UDP). To reach UDP alone is an incomplete transaction so there could be no Src/Dest IP addresses or Src/Dest ports. Likewise, E’net, IP, and TCP alone have to be incomplete.
(Please excuse the graphics, I couldn’t get the XML graphics to work right.)
2. a. As a network administrator your company has assigned you 8 class C-equivalent subnetworks that include the following address: 172.18.10.5. What subnet mask will be required to select these 8 subnetworks from the 172,18.0.0 Class B network? Remember the technique of using the maximum value for a byte and subtracting the aggregation number to obtain the last octet of the subnetmask.
b. What is the network address that this subnetwork mask defines in the Class B network 172.18.0.0?
c. What is the broadcast address for this subnetwork?
a. 109.145.15.38
b. 131.109.33.18
c. 12.209.254.18
d. 195.209.254.48
e. 192.192.208.201
f. 191.192.129.13
g. 225.239.245.22
a. 109.145.16.38 and 109.145.17.39 – classful network masks
b. 109.145.16.38/255.255.0.0 and 109.145.17.39/255.255.0.0
c. 109.145.16.38/255.255.240.0 and 109.145.17.39/255.255.240.0
d. 109.145.16.38/21 and 109.145.17.39/255.255.240.0
e. 109.145.16.38/20 and 109.145.17.39/255.255.240.0
f. 109.145.16.38/255.255.255.0 and 109.145.17.39/255.255.255.0
g. 109.145.16.38/255.255.255.0 and 109.145.17.39/24
h. 109.145.31.38/20 and 109.145.33.39/20