A computer has 64-KB virtual address space and 32-KB physical address space. The

Question

A computer has 64-KB virtual address space and 32-KB physical address space. The page size is 4 KB. At an instant ti, the mapping between virtual address space and physical address space is shown in Fig.2. Answer the following questions.

1- Compute virtual page number and offset for the virtual addresses 4090, 8300, 16500, and 48400.

2- Compute the physical address for the virtual addresses 4090, 8300, 16500, and 48400.

Virtual address space 60K-64K X 56K-60KX 52K-56KX 48K-52KX 44K-48K 7 40K-44K X Virtual page Physical memory address 36K-40K.5 32K-36KX 28K-32KX 24K-28K X 28K-32K 24K-28K 20K-24K 20K-24K 3 16K-20K | 4 12K-16K O 16K-20K 12K-16K 8K-12K 4K-8K 8K-12K 6 4K-8K 1 OK-4K2 OK-4K Page frame Fig.2

Explanation / Answer

Dear Student,

4090, 8300, 16500, and 48400

1)
4 KB = 4 kibibyte = 2^12 bytes = 4096

page number = 4090/4096=0
offset = (4090 mod 4096) = 4090

pagenumber = 8300/4096 = 2
offset = (3904 mod 4096) = 3904

pagenumber = 16500/4096 = 4
offset = (16500 mod 4096) = 116

pagenumber = 48400/4096 = 11.8~12(approx 12)
offset = (48400 mod 4096) = 3344

2)
4090: physical address: 8k+4090=12090
8300: physical address: 24k+(8300-8k) =24300
16500: physical address: 0k + 16k=16k
48400: physical address: 28k+(48400-16k)=60400

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