Complete the following table by identifying the source IP address and destination IP address of each packet (P₁–P₅, shown as Process 1–5 in the…

Complete the following table by identifying the source IP address and destination IP address of each packet (P₁–P₅, shown as Process 1–5 in the network diagram).

PacketSource IPDestination IP
P1
P2
P3
P4
P5

From the BPDB Assistant Engineer (CSE) October 2026 – Written (BUET) exam, question 30 of 30.

Answers (1)

The idea

Every IP packet carries two addresses: the source (who sent it) and the destination (who it is for). In a diagram like this, each arrow is one packet:

  • Tail of the arrow = the device that sends it = source IP.
  • Head (the point) of the arrow = the device that receives it = destination IP.

Method, for any question of this type

  1. Write down each device and its IP from the figure.
  2. For each packet, find its arrow and see which way it points.
  3. Source = the IP of the device at the tail; destination = the IP of the device at the head.
  4. Fill in one row per packet, then check: no packet should have the same source and destination.

Step 1 — the devices

  • Router: 192.168.0.15
  • Computer (bottom left): 192.168.0.10
  • Cloud server: 192.168.0.17
  • Computer (top right): 192.168.0.18

Steps 2–3 — each packet

  • P1: arrow from the bottom-left computer up to the router → 192.168.0.10 sends to 192.168.0.15.
  • P2: arrow from the router down to the cloud → 192.168.0.15 sends to 192.168.0.17.
  • P3: arrow from the cloud up to the top-right computer → 192.168.0.17 sends to 192.168.0.18.
  • P4: arrow from the top-right computer to the router → 192.168.0.18 sends to 192.168.0.15.
  • P5: arrow from the router down to the bottom-left computer → 192.168.0.15 sends to 192.168.0.10. (P1 and P5 run between the same two devices, in opposite directions — so their addresses simply swap.)

Step 4 — the answer

PacketSource IPDestination IP
P1192.168.0.10192.168.0.15
P2192.168.0.15192.168.0.17
P3192.168.0.17192.168.0.18
P4192.168.0.18192.168.0.15
P5192.168.0.15192.168.0.10

About 192.168.0.19: the figure also prints 192.168.0.19 beside P3’s arrowhead. If the examiner meant that label as the address P3 arrives at, P3’s destination is 192.168.0.19. Since 192.168.0.18 is the label on the computer itself, .18 is the safer reading; mention both in an exam.

The trap to watch for: routers

Here every device is on the same network (all addresses are 192.168.0.x), so each arrow is a separate packet between its two ends. If instead a question shows one packet travelling from a PC through a router to a different network, the IP addresses do not change along the way — source and destination stay the original sender and the final receiver. What changes at each hop is the MAC address (the frame is re-addressed to the next device). So: arrow by arrow when each arrow is its own packet; end to end when one packet crosses a router.

Quick check

If a sixth arrow went from the cloud server to the bottom-left computer, what would it be? Tail = cloud, head = computer → source 192.168.0.17, destination 192.168.0.10.

— rhs

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