A disk of size 512M bytes is divided into blocks of 64K bytes. A file is stored in the disk using linked allocation. In linked allocation, each data block reserves 4 bytes to store the pointer to the next data block. The link part of the last data block contains a $NULL$ pointer (also of 4 bytes). Suppose a file of 1M bytes needs to be stored in the disk. Assume, 1K = $2^{10}$ and 1M = $2^{20}$. The amount of space in bytes that will be wasted due to internal fragmentation is _______. (Answer in integer)

To determine the internal fragmentation, we calculate the number of blocks required and the unused space in the last block:

1. Block Capacity:

   • Total block size $= 64\text{ KB} = 64 \times 2^{10} = 65536\text{ bytes}$.

   • Each block reserves $4\text{ bytes}$ for the next pointer (or NULL pointer).

   • Data capacity per block $= 65536 - 4 = 65532\text{ bytes}$.

2. Number of Blocks Needed:

   • Total file size $= 1\text{ MB} = 2^{20} = 1048576\text{ bytes}$.

   • Number of blocks required $n = \left\lceil \frac{1048576}{65532} \right\rceil = \lceil 16.001038 \rceil = 17\text{ blocks}$.

3. Space Used in the Last ($17^{\text{th}}$) Block:

   • Data stored in the first 16 blocks $= 16 \times 65532 = 1048512\text{ bytes}$.

   • Data stored in the $17^{\text{th}}$ block $= 1048576 - 1048512 = 64\text{ bytes}$.

   • Pointer space used in the $17^{\text{th}}$ block $= 4\text{ bytes}$.

   • Total space used in the $17^{\text{th}}$ block $= 64 + 4 = 68\text{ bytes}$.

4. Internal Fragmentation:

   • Internal fragmentation is the unused space in the allocated $17^{\text{th}}$ block:

   $$\text{Internal Fragmentation} = 65536 - 68 = 65468\text{ bytes}$$

Hence, the amount of wasted space is 65468 bytes.