Operating system subjective question(paging and TLB) for competition exam(8)

Q1.Consider a paging system it takes 10ns to search TLB and 80ns to access main memory. If the TLB hit ratio is 80% the effective memory access time is.

Hit ratio(h)=0.8
memory access time(m)=80ns
TLB access time(t)=10ns
EAT=h(t+m)+(1-h)(t+2m)
EAT=0.8(80+10)+(1-0.8)(10+2*80)
EAT=72+34=106ns

Q2.A 3 level paging scheme uses a TLB.A TLB access takes 20ns and the main memory access takes 70ns.What is the effective access time if the TLB hit ratio is 70%.

Hit ratio(h)=0.7
memory access time(m)=70ns
TLB access time(t)=20ns
page level(k)=3
EAT=h(t+m)+(1-h)(t+k*m+m)
EAT=0.7(20+70)+(1-0.7)(20+3*70+70)
EAT=63+90=153ns

Q3.Consider a paging system it takes 10ns to search TLB and 80ns to access main memory.If the effective memory access time(EMAT) is 106ns the find the TLB hit ratio.

memory access time(m)=80ns
TLB access time(t)=10ns
EMAT=106ns
EAT=h(10+80)+(1-h)(10+2*180)
106=90h+170-170h
106=-80h+170
80h=170-106
h=64/80=0.8

Q4.A single-level paging system uses a TLB where memory access takes 100ns and hit ratio of TLB 80%.If EMAT is 140ns then find TLB access time.

m=10ns
h=0.8
EMAT=140ns
EAT=h(t+m)+(1-h)(t+2m)
140=0.8(t+100)+(1-0.8)(t+2*100)
140=0.8t+80+0.2t+40
140=t+120
t=20ns

Q5.A single level paging system uses a TLB.A TLB access take 20ns as well as a TLB hit ratio is 80%.What is the main memory access takes is EMAT is 140ns access time.

t=20
h=0.8
EMAT=140ns
EAT=h(t+m)+(1-h)(t+2m)
140=0.8(20+m)+(1-0.8)(20+2*m)
140=16+0.8m+4+0.4m
140=20+1.2m
140-20=1.2m
120=1.2m
m=120/1.2=100ns