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Triss [41]
3 years ago
15

Question 1 : Replacement [42 Pts] Consider the following page reference string:

Engineering
2 answers:
mariarad [96]3 years ago
7 0

Answer:

2,970

Explanation:

The string are arranged as follow: a,a,b,b,c,c,d,e,e,g,g,g

There are 12 combinations and with four frames

Then, it can be computed thus: nCr, where n = 12 and r = 4 = n!/r(n - r)!

∴ 12C4 = 12!/4(8)! = 2970

MakcuM [25]3 years ago
4 0

Answer:

Optimal

Time 123456789101112

RS. ecbeagdcegda

F0. eeeeeeeee e e a

F1 c c c c c c c c c c c

F2 b b b g g g g g g g

F3 a a d d d d d d

Page fault? * * * * * * *

Total page fault:7

2. LRU

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 e e e e e e e c c c c a

F1 c c c c g g g g g g g

F2 b b b b d d d d d d

F3 a a a a e e e e

Page fault? Y Y Y N Y Y Y Y Y N N Y Total page fault:9

3. LRU approximation algorithm: Second chance

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 0,e 0,e 0,e 1,e 1,e 0,e 0,e 0,e 1,e 1,e 1,e 0,e

F1 0,c 0,c 0,c 0,c 0,g 0,g 0,g 0,g 1,g 1,g 0,g

F20,b0,b0,b0,b0,d0,d0,d0,d1,d0,dF30,a0,a0,a0,c0,c0,c0,c0,a

Page fault? YYYNYYYYNNNY

Total page fault: 8

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Read 2 more answers
For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.
rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

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Stiffness of the members = 2.6 MN/mm per bolt

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Pmax =60 kN,  Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

n_{p} = \frac{S_{p}A_{t}  }{CP_{max}+ F_{i}  }  ------ ( 1 )

Sp = 380 MPa,   At = 20.1 mm^2,   C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = \frac{380*20.1}{0.277*7500*5728.5} = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

n_{L} =  \frac{S_{p}A_{t}-F_{i}   }{C(P_{max} )}  ------- ( 2 )

Sp =  380 MPa,   At =  20.1 mm^2, C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

input values into equation 2 above

hence : n_{L} = 0.9191n_{L}  = 0.9191

<u>C)  Determine the factor of safety based on joint separation</u>

n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }

Fi =  5728.5 N,  Pmax = 7500 N,  C = 0.277,

input values into equation above

Hence n_{0} = 1.056

<u>D)  Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

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3 years ago
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