UGC NET CSE | December 2012 | Part 2 | Question: 14

Question

Given a memory partitions of 100 K, 500 K, 200 K, 300 K and 600 K (in order) and processes of 212 K, 417 K, 112 K and 426 K (in order), using the first fit algorithm, in which partition would the process requiring 426 K be placed?

• A. 500 K
• B. 200 K
• C. 300 K
• D. 600 K

Comments

dewritt i think the word  " (in order)" is ambiguous ...if we consider it as " in order "  like pre- order/post -order

then..partitions will be 100k,200k,300k,500,600k as..in order is increasing order..

and processsse will be 112,212,417, 426...as..in order is increasing order.. now we can fulfill the request

 of 426 with the slot 600k....i guess the question paper writer may have thought that/or else question is total wrong..

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No tauhin, according to me question is wrong.
Because inorder and preorder concept is totally different
and by the way for BST we arrange elements in increasing order(for inorder) not for anyone.
this is my expalanation

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yes u r correct....i 4get that

Answer 1

426 cannot be placed. This question was removed from final answer keys published officially by ugc net

Answer 2

process requiring 426 Kb should wait

Comments

you are using variable size partitioning, why not fixed???