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In a system which supports overlay two programs want to run.

Size of program1 = 90kb

Size of program 2 = 100kb

Size of common code = 10kb

Size of overlay driver = 20kb

Size of error routine = 50kb

If no error is occured then what will be the minimum memory requirement to execute program

 

a) 130kb.                                       b)180kb

c) 270kb.                                       d) 220kb

Please give explanation also thank you.

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Overlay can share same memory space for different programs

So, if we take large program , then both program could run

So, 100kB memory is taken

Common code size =10 kB

Overlay driver size 20kB

Error has not occured . So, no need to take space for error routing

So, total space required 100+10+20=130kB
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