NIMCET Previous Year Questions (PYQs) – Operating System
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The first instruction of bootstrap loader program of an operating system is stored in
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Solution
When a computer is powered on, the CPU looks for the first instruction in the BIOS (stored in ROM). The BIOS runs the Power-On Self Test (POST) and loads the bootstrap loader, which then loads the OS from a bootable device.
A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation look-aside buffer (TLB) which can hold a total of 128 page table entries and is 4 -way set associative. The minimum size of the TLB tag is:
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Solution
✅ Given Information:
The CPU generates 32-bit virtual addresses with a 4 KB page size. The TLB holds 128 page table entries and is 4-way set associative.
✅ Step 1: Virtual Address Breakdown:
We split the 32-bit virtual address into:
- Page Offset: 12 bits (because 4 KB = 212)
- Page Number: 20 bits (32 - 12)
✅ Step 2: TLB Breakdown:
The TLB has 128 entries, and with 4-way set associativity, there are:
128 / 4 = 32 sets, so we need 5 bits for the index.
✅ Step 3: Tag Calculation:
The total number of bits for the virtual address is 32:
- Page Offset: 12 bits
- Index: 5 bits (because we have 32 sets)
- Tag: Remaining bits = 32 - 12 - 5 = 15 bits
✅ Final Answer:
The minimum size of the TLB tag is: 15 bits
A CPU generates 32 bits virtual
addresses. The page size is 4 KB. The processor has a translation look-aside
buffer (TLB) which can hold a total of 128-page table entries and is 4- way set
associate. The minimum size of the TLB tag is
1
2
3
4
Solution
TLB Tag Size Calculation
Given:
- Virtual Address = 32 bits
- Page Size = 4 KB =
- VPN = 32 − 12 = 20 bits
- TLB entries = 128, 4-way set associative → 32 sets
Set index bits:
TLB Tag = VPN − Set Index = 20 − 5 = 15 bits
✅ Final Answer: 15 bits
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