Math

A computer provides each process with 65536 bytes of address space divided into pages of 4096 bytes. A particular program has a text size of 32768 bytes, a data size of 16386 bytes and a stack size of 15870 bytes.
1. Will this program fit in the address space?

2. If the page size were 512 bytes, would it fit?

Give the reason for all answer.

Answer:

Q1.

 Address space for each process: 65536  bytes
 Page Size: 4096 bytes
 Available address space =65536 / 4096=16 pages

 For the text size 32768 bytes address space is =(32768/4096) = 8 pages
 For the text size 16386 bytes address space is =(16386/4096) = 4.0034 = 5 pages
 For the stack size 15870 bytes address space is =(15870/4096) = 3.874 = 4 pages

 The available space is 16 pages but the required space is (8+5+4)=17 pages which is greater than 
 the available space. Hence this program does not fit in the available address space.




Q.2

 If the page size were of 512 bytes then

 Address space for each process: 65536 bytes
 Page Size: 512 bytes
 Available address space = (65536/512) = 128 pages

 then,
 For the text size 32768 bytes address space is =(32768/512) = 64 pages
 For the text size 16386 bytes address space is =(16386/512) = 32.0039 = 33 pages
 For the stack size 15870 bytes address space is =(15870/512) = 30.996 = 31 pages

 The available space is 128 pages but the required space is  (64+33+31)=128 pages which is equal to
 the available space.  Hence this program does not fit in the available address space.