Unformatted text preview: Split to IE 0K E‘I—I “ions CHAPTER 4 Memory Management 4.13 On a system with mm of memory using the buddy system, what is the ﬁrst request
that will fail in the following string of requests due to lack of available memory? Requests: 50K. 150K. 90K. 130K, 70K, 80K, 120K, 180K, 60K. Answer:
Memory blocks will be adjusted for each request as follows: 4.14 In the previous problem,
wasted due to internal fragmentaﬁon ﬁ'agmentaﬁon'? Answer:
There is one 64K block that is unallocated but unusable. Thus, external fragmentation is 64K. 01' the remaining 1024K—64K=960K of allocated space, only
0K+70K+80K=570K has been requested. Thus, 960K , 570K = 390K is wasted due to internal fragmentation. [email protected] Why are page sizes always a power of 2? Answer:
All addresses are binary and ate divi§¢Q_iFL°,Eﬁﬂ gage frame number and offset.
2, the page n , the page ﬂame number, and By making the page sine aphiisif
the oﬁ‘set can be determined by looking at palﬁculﬂl' bits in the address; no
mathematical calculations are mixed. The physical address can be generated by 'Ilii 1n paging (assuming the page size is a power of 2), why is it not messaty to add the
_7_,/ page oﬁ‘set to the staining address of the page frame to generate a physical address? —-——® ...
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- Fall '08