ISRO CSE 2014 PYQ

The IPv4 fragment offset is measured in 8-byte units. The fragment begins at byte offset 100$\times 8 = 800$ in the original payload. HLEN = 5 means a 20-byte IP header, so the fragment payload length is 200-20 = 180 bytes. Numbering the first payload byte at offset 800, the last byte is 800+180-1 = 979. This assumes the question’s “byte number” refers to the original datagram payload offset numbering.

The leading zero makes 0570 an octal integer. In decimal, 0570₈ = 376₁₀. Right-shifting by 4 gives 23₁₀ (027₈); left-shifting that result by 6 gives 1472₁₀ = 02700₈. The %o format specifier prints the value in octal, so the output ends in 2700. The code uses void main, which is not a strictly conforming hosted C main signature; the arithmetic result is nevertheless the intended textbook result.

The Calculation:

\(\text{Total\ Memory}=2\times (\text{Width}\times \text{Height}\times \text{Bit\ Depth})\)
\(\text{Total\ Bits}=2\times 512\times 512\times 24=12,582,912\text{\ bits}\)
Convert to Bytes:
\(\frac{12,582,912\text{\ bits}}{8\text{\ bits/byte}}=1,572,864\text{\ bytes}\)
Convert to Megabytes (using binary \(1 \text{ MB} = 1,048,576 \text{ bytes}\)):
\(\frac{1,572,864\text{\ bytes}}{1,048,576\text{\ bytes/MB}}=\mathbf{1.5}\text{\ MB}\)

The physical address 25 lies in frame 0, with offset 25 because the page/frame size is 2048 words. The page table maps virtual page 2 to physical frame 0. Therefore the corresponding virtual address is page number$\times page size + offset = 2\times 2048+25 = 4121$. The page table mapping is explicitly provided in the question text.