# CPU Scheduling Numerical Questions

**1. Find the size of the memory if its address consists of 22 bits. Assume the memory is 2-byte addressable.****Solution – **

If the given address consists of ‘k’ bits, then 2k locations are possible.

Size of memory = 2k x Size of one location.

According to the question, number of locations with 22 bits = 222 locations

Given that the size of the memory is 2-bytes addressable, which means that the size of one location is 2 bytes.

Hence, the size of memory = 222 x 2 bytes = 223 bytes = 8.3 MB ≈ 8 MB.

**2. Calculate the number of address bits, if the memory has a size of 16 GB. Assume the memory is 4-byte addressable.****Solution –**

If the given address consists of ‘k’ bits, then 2k locations are possible.

Size of memory = 2k x Size of one location.

Given that the size of memory is 16 GB. (16 GB = 234 B)

2k x 4 = 234

2k = 234 / 22 (∵ 22 = 4)

2k = 232

∴ k = 32 bits

**3. Consider a machine with 32 bit logical addresses, 4 KB page size and page table entries of 4 bytes each. Find the size of the page table in bytes. Assume the memory is byte addressable.****Solution –**

Given – No. of bits = 32 bits

Page Size = 4 KB = 4 x 103 bytes

Page table entry size = 4 bytes

Size of page table = Number of entries in page table x Page table entry size

Number of entries in page table = Process size / Page size

Process Size = Number of address bits

Thus, Process size = 232 bytes = 230 x 22 bytes ≈ 109 x 4 bytes (∵ 230 ≈ 109)

Number of entries in page table = 4 x 109 / 4 x 103 = 106 pages

Size of page table = 106 x 4 = 4 x 106 bytes.

**4. Consider a system with page table entries of 8 bytes each. If the size of the page table is 256 bytes, what is the number of entries in the page table? ****Solution –**

Size of page table = Number of entries in the page table x page table entry size

Given – Size of page table = 256 bytes, page table entry size = 8 bytes.

Thus, 256 = number of entries in page table x 8

Number of entries in the page table = 256 / 8 = 32 = 25

**5. Find the total number of frames. If a system, size of the main memory is 230 bytes, the page size is 4 KB and the size of each page table entry is 32-bit. ****Solution –**

Given – Size of main memory = 230 B, page size = 4 KB = 22 KB = 212 B (∵ 1 KB = 210 B)

Size of main memory = Total number of frames x page size.

230 = Total number of frames x 212

Total number of frames = 230 / 212 = 218