Consider sending an IP datagram of size 1420 bytes (including 20 bytes of IP header) from a sender to a receiver over a path of two links with a router between them. The first link (sender to router) has an MTU (Maximum Transmission Unit) size of 542 bytes, while the second link (router to receiver) has an MTU size of 360 bytes. The number of fragments that would be delivered at the receiver is _______

### 1. Initial IP Datagram

• Total size = $1420\text{ bytes}$

• IP header = $20\text{ bytes}$

• Data payload = $1420 - 20 = 1400\text{ bytes}$

### 2. Fragmentation on Link 1 (Sender to Router)

• $\text{MTU} = 542\text{ bytes}$

• Maximum data per fragment must be a multiple of $8$:

  $$\text{Max data payload} = \left\lfloor \frac{542 - 20}{8} \right\rfloor \times 8 = \lfloor 65.25 \rfloor \times 8 = 65 \times 8 = 520\text{ bytes}$$

• The original $1400\text{ bytes}$ payload is divided into:

  • Fragment 1: $520\text{ bytes}$ data (Total size $= 540\text{ bytes}$)

  • Fragment 2: $520\text{ bytes}$ data (Total size $= 540\text{ bytes}$)

  • Fragment 3: $1400 - (520 + 520) = 360\text{ bytes}$ data (Total size $= 380\text{ bytes}$)

### 3. Fragmentation on Link 2 (Router to Receiver)

• $\text{MTU} = 360\text{ bytes}$

• Maximum data per fragment on link 2:

  $$\text{Max data payload} = \left\lfloor \frac{360 - 20}{8} \right\rfloor \times 8 = \lfloor 42.5 \rfloor \times 8 = 42 \times 8 = 336\text{ bytes}$$

• Now the router fragments each of the 3 fragments:

  • Fragment 1 ($520\text{ bytes}$ data):

    • Sub-fragment 1a: $336\text{ bytes}$

    • Sub-fragment 1b: $184\text{ bytes}$

    *(2 fragments)*

  • Fragment 2 ($520\text{ bytes}$ data):

    • Sub-fragment 2a: $336\text{ bytes}$

    • Sub-fragment 2b: $184\text{ bytes}$

    *(2 fragments)*

  • Fragment 3 ($360\text{ bytes}$ data):

    • Sub-fragment 3a: $336\text{ bytes}$

    • Sub-fragment 3b: $24\text{ bytes}$

    *(2 fragments)*

### 4. Total Fragments Delivered at Receiver