The correct physical address in mark 1 in the following network isA/10To another | X/44networkSenderUpper layers DataF/20T/99APDia Network layerAP DRouter 1AP Deta20 10 AP Da20 10 AP Data 12JAPDa T2Data link layerPhysicaladdressechangedLAN 1LAN 2LAN 3Data link layer95 66 AP Deta T295 66Data 12AP Data 12AP D Network layerUpper layers DataRouter 2AP DZ66N33PhysicaladdressesReceiverTo anotherchangednetworkI YSSP/95Select one:a. T/Nb. 10/20C. 99/33d. A/P

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The correct physical address in mark 1 in the following network is A/10 To another | X/44 network Sender Upper layers Data F/20 T/99 APDia Network layer AP D Router 1 AP Data 20 10 AP D Data link layer 20 10 AP Dao T2 JAPDat T2 Physical addresse changed LAN 1 LAN 2 LAN 3 Data link layer 95 66 AP Deta T2 95 66 Data 12 AP Data T2 AP Di Network layer Router 2 AP D Upper layers Data N/33 Z66 Physical addresses changed Receiver To another networkI Y/SS P/95 Select one: a. T/N b. 10/20 C 99/33 d. A/P

The correct physical address in mark 1 in the following network is A/10 To another | X/44 network Sender Upper layers Data F/20 T/99 APDia Network layer AP D Router 1 AP Data 20 10 AP D Data link layer 20 10 AP Dao T2 JAPDat T2 Physical addresse changed LAN 1 LAN 2 LAN 3 Data link layer 95 66 AP Deta T2 95 66 Data 12 AP Data T2 AP Di Network layer Router 2 AP D Upper layers Data N/33 Z66 Physical addresses changed Receiver To another networkI Y/SS P/95 Select one: a. T/N b. 10/20 C 99/33 d. A/P

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter13: Internet And Distributed Application Services

Section: Chapter Questions

Problem 15VE

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Frame 42: 1033 bytes on wire (8264 bits), 1033 bytes captured (8264 bits) Ethernet II, Src: Pegatron_b2:31:e9 (4c:72: b9:b2:31:e9), Dst: IPv4mcast_7f:ff:fa (01:00:5e:7f:ff:fa) Internet Protocol Version 4, Src: 192.168.0.116, Dst: 239.255.255.250 Src Port: 50178, Dst Port: 3702 Source Port: 50178 Destination Port: 3702 Length: 999 Checksum: Oxb50f [unverified] [Checksum Status: Unverified] [Stream index: 3] [Timestamps] [Time since first frame: 0.000000000 seconds] [Time since previous frame: 0.000000000 seconds] ▸ Data (991 bytes) Observe the captured packet, answer the following questions: 1) The packet is sent from server to client or from client to server? Briefly justify your answer. 2) Which protocol is used in this communication at the transport layer? Briefly justify your answer 3) The protocol at transport layer is suitable for streaming applications or not? Briefly justify your answer
Design a classfull network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 25% growth in all areas. IP Address 172.16.0.0 FO/0 S/0/1 FO/O Sales Administrative Router A FO/1 SO/0/0 185 Hosts 30 Hosts Marketing 50 Hosts
2. Subnetting a subnet or using variable-length subnet mask (VLSM) was designed to maximize addressing efficiency. You are required to subnet, based on the number of hosts, including router interface and WAN interface and WAN connections for the following scenario: •Toronto: 60 host addresses •Montreal: 28 host addresses •Ottawa: 12 host addresses •Vancouver: 12 host addresses •WAN1: 2 host addresses •WAN2: 2 host addresses •WAN3: 2 host addresses Using the given address block of 192.168.15.0/24, you can design an address scheme for the network to meet its requirements and save potential addresses. The data for Toronto is given to help you to fill the remaining all other columns. Name Toronto Montreal Ottawa Vancouver WAN1 WAN2 WAN3 Required no of addresses 60 28 12 12 ~~ 2 2 2 Subnet Address 192.168.15.0 Address Range .1 - .62 Broadcast Address .63 Network/ Prefix 192.168.15.0 /26 19
R6Fragmentation of an IP datagram takes place if its size is larger than the MTU of the subnet over which the datagram will be routed. Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose a datagram is fragmented into four fragments. The first three fragments arrive, but the last one is delayed. Eventually the timer goes off and the three fragments in the receiver's memory are discarded. A little later, the last fragment stumbles in. What should be done with it?
Block Address: 192.168.16.0/24 Port to Point Link WANI Usable Hof Hosts: 2 Users LAN: Usable #of Hosts: 25 Network Address: LANI Usable #of Hosts: 100 Users Network Address: 7 Point to Point Link WAN2 Usable 4of Hosts: 2 Users LAN2: Usable #of Hosts: 50 Network Address: The network given in the above figure consists of three local area networks and two wide area networks are connected with two serial links. With an ID range 192.168.15.0/24, design an IP plan for this network. Find the network address and broadcast address, and subnet mask for each network.
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Design a classfull network addressing scheme that will supply the minimum number of hosts per subnet, and allow enough extra subnets and hosts for 25% growth in all areas. IP Address 172.16.0.0 FO/O S//0/1 FO/O Sales Administrative FO/1 185 Hosts 30 Hosts SO/0/0 Marketing 50 Hosts,
Consider the figure below. The IP and MAC addresses are shown for nodes A, B, C and D, as well as for the router's interfaces. Consider an IP datagram being sent from node D to node A. Give the source and destination Ethernet addresses, as well as the source and destination addresses of the IP datagram encapsulated within the Ethernet frame at points (5), (4), (2), and (1) in the figure above.
19. ARP is used for: O packet path determination O finding of a mapping IPV6 → MAC address O finding of a mapping IPV4 → MAC address O finding of a mapping IPV4 → TCP address
Explain the protocol's route across each OSI layer.
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how can iterated dns querries improve the overall perfomance?