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GAte1996_2.18

. A 1000 Kbyte memory is managed using variable partitions but to compaction. It currently has two partitions of sizes 200 Kbytes and 260 Kbytes respectively. The smallest allocation request in Kbytes that could be denied is for
(a) 151
(b) 181
(c) 231
(d) 541

Answer

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3Comments
Amit Pal @amitpal101
25 Sep 2016 11:27 pm

total memory space = 1000 KB

Currently given partitions = 200 KB and 260 KB 
 so maximum partition we can create except these (200 KB and 260 KB) = 3 
so rest space (1000 -200-260)KB divide in equal 3 parts = 540/3 =180
 so if we try insert who size is less than 181 . we can successfully insert. 

so ans = 181 .

pankaj soni @pspankajsoni43
18 Dec 2016 04:49 pm

Sir Please Explain How you calculate Maximum partition except these=3 ?

Sumit Verma @sumitverma
19 Dec 2016 12:56 pm

@pspankajsoni43 The partitions for which the smallest allocation request can be denied if we have the partition sizes to be equal. Examine a little bit by taking any example.