Ace_OperatingSystem_Subject_test

Question

I think the answer given here is wrong ,

Comments

No Jason I think  you done that in a wrong way

---

@Magma, The solution provided by them should be wrong...

 

Note that given Process Size = 2³². and Page Size = 1 Kb = 2¹⁰ b = 2⁷ Bytes

no.of pages in the process ( no.of entries in PT₁ ) = $\frac{2^{32}}{2^{7}}$ =  ${2^{25}}$

===> Size of 1^st level Page Table = ${2^{25}}$ * 2² B = ${2^{27}}\;B$

 

no.of pages in the PT₁ ( no.of entries in PT₂ )= $\frac{2^{27}}{2^{7}}$ =  ${2^{20}}$

===> Size of 2^nd level Page Table = ${2^{25}}$ * 2² B = ${2^{22}}\;B$

 

no.of pages in the PT₂ ( no.of entries in PT₃ ) = $\frac{2^{22}}{2^{7}}$ =  ${2^{15}}$

===> Size of 3^rd level Page Table = ${2^{25}}$ * 2² B = ${2^{17}}\;B$

 

no.of pages in the PT₃ ( no.of entries in PT₄ ) = $\frac{2^{17}}{2^{7}}$ =  ${2^{10}}$

===> Size of 4^th level Page Table = ${2^{10}}$ * 2² B = ${2^{12}}\;B$

 

no.of pages in the PT₄ ( no.of entries in PT₅ ) = $\frac{2^{12}}{2^{7}}$ =  ${2^{5}}$

===> Size of 5^th level Page Table = ${2^{5}}$ * 2² B = ${2^{7}}\;B$ ===> Fit in 1 Page

 

Over Head By page tables =  $ ({2^{27}} + {2^{22}} + {2^{17}} + {2^{12}} + {2^{7}} ) \;B$

---

thanks Shaik Masthan

you explanations are just awsome and easy to understand