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deff fn [24]
3 years ago
14

For the system in problem 4, suppose a main memory access requires 30ns, the page fault rate is .01%, it costs 12ms to access a

page not in memory (this time includes the time necessary to transfer the page into memory, update the page table, and access the data). Also suppose a TLB hit requires 7ns, the cache miss rate is 3%, the TLB hit rate is 95%, a cache hit requires 15 ns. On a TLB or cache miss, the time required for access includes a TLB and/or cache update, but the access is not restarted. On a page fault, the page is fetched from disk, all updates are performed but the access is restarted . All references are sequential (no overlap, nothing done in parallel)
a.) Calculate the time for a TLB hit and a cache hit.

b.) Calculate the EAT (effective access time) for a TLB hit.
Engineering
1 answer:
raketka [301]3 years ago
8 0

Answer:

a. 7.75

b. 24.4

Explanation:

The Operating system uses virtual memory and page tables maps these virtual address to physical address. TLB works as a cache for such mapping.

program >>> TLB >>> cache >>> Ram

A program search for a page in TLB, if it doesn't find that page it's a TLB miss and then further looks for the page in cache.

If the page is not in cache then it's a cache miss and further looks for the page in RAM.

If the page is not in RAM, then it's a page fault and program look for the data in secondary storage.

So, typical flow would be

Page Requested >> TLB miss >> cache miss >>main memory>> page fault >> looks in secondary memory.

Here,

Main memory access time= 30 ns

Page fault rate=.01%

page fault service time= 12ns

TLB access time=7 ns

TLB hit rate= .95%

TLB miss rate =1-.95=.05%

cache access time = 15 ns

cache miss rate= .3%

cache hit rate = 1-.3=.97%

So,

a) TLB hit time= TLB access time = 7 ns

cache hit time = TLB hit rate * TLB access time + TLB miss rate * ( TLB access time + cache hit time)

= .95 * 7 + .05 * (7+15)

= 7.75 ns

b) EAT for TLB hit= 7ns

Total EAT = TLB hit rate *( TLB access time + Cache hit rate * cache access time + cache miss rate * (cache + main memory access time))+ TLB miss rate ( TLB access time + main memory access time + cache hit rate * cache access time + cache miss rate ( cache + main memory access time))

= .95 *( 7 + (.97*15) + .03(15+30))+ .05*(7+30+(.97*15) + .03 ( 15 + 30))=24.4 ns

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melinda is using a rectangular brass bar in a sculpture she is creating. the brass bar has a length that is 4 more than 3 times
lawyer [7]

Answer:

180 x 60 inches

Width = 60 inches

Length = 180 inches

Explanation:

Given

Let L = Length

W = Width

P = Perimeter

Length = 3 * Width

L = 3W

Perimeter of Brass = 480 inches

P = 480

Perimeter is given as 2(L + W);

So, 2 (L + W) = 480

L + W = 480/2

L + W = 240

Substitute 3W for L; so,

3W + W = 240

4W = 240

W = 240/4

W = 60 inches

L = 3W

L = 3 * 60

L = 180 inches

6 0
3 years ago
Read 2 more answers
La probabilidad de que un nuevo producto tenga éxito es de 0.85. Si se eligen 10 personas al azar y se les pregunta si compraría
liq [111]

Answer:

La probabilidad pedida es 0.820196

Explanation:

Sabemos que la probabilidad de que un nuevo producto tenga éxito es de 0.85. Sabemos también que se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto. Para responder a la pregunta, primero definiremos la siguiente variable aleatoria :

X: '' Número de personas que adquirirán el nuevo producto de 10 personas a las que se les preguntó ''

Ahora bien, si suponemos que la probabilidad de que el nuevo producto tenga éxito se mantiene constante (p=0.85) y además suponemos que hay independencia entre cada una de las personas al azar a las que se les preguntó ⇒ Podemos modelar a X como una variable aleatoria Binomial. Esto se escribe :

X ~ Bi(n,p) en donde ''n'' es el número de personas entrevistadas y ''p'' es la probabilidad de éxito (una persona adquiriendo el producto) en cada caso.

Utilizando los datos ⇒ X ~ Bi(10,0.85)

La función de probabilidad de la variable aleatoria binomial es :

p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}    con x=0,1,2,...,n

Si reemplazamos los datos de la pregunta en la función de probabilidad obtenemos :

P(X=x)=\left(\begin{array}{c}10&x\end{array}\right)(0.85)^{x}(0.15)^{10-x} con x=0,1,2,...,10

Nos piden la probabilidad de que por lo menos 8 personas adquieran el nuevo producto, esto es :

P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)

Calculando P(X=8), P(X=9) y P(X=10) por separado y sumando, obtenemos que P(X\geq 8)=0.820196

7 0
3 years ago
The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0
3 years ago
4. The instant the ignition switch is turned to the start position,
geniusboy [140]

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

4 0
3 years ago
Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)
IceJOKER [234]

Answer:

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

\mathbf{M_n = 49163.56431  \ g/mol }

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa)            Number- Average Molecular Weight  (g/mol)

82                                                  12,700

156                                                 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

T_S = T_{S \infty} - \dfrac{A}{M_n}

where;

T_S = Tensile Strength

T_{S \infty} = Tensile Strength (Infinity)

M_n = Number- Average Molecular Weight  (g/mol)

SO;

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

From equation (1) ; collecting the like terms; we have :

T_{S \infty} =82+ \dfrac{A}{12700}

From equation (2) ; we have:

T_{S \infty} =156+ \dfrac{A}{28500}

So; T_{S \infty} = T_{S \infty}

Then;

T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}

Solving by L.C.M

\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}

\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}

By cross multiplying ; we have:

({4446000 + A})*  {12700} ={28500} *({1041400 + A})

(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)

Collecting like terms ; we have

(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)

2.67843*10^{10}  = 15800 \ A

Dividing both sides by 15800:

\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}

A = 1695208.861

From equation (1);

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

Replacing A = 1695208.861 in the above equation; we have:

82= T_{S \infty} - \dfrac{1695208.861}{12700}

T_{S \infty}= 82 + \dfrac{1695208.861}{12700}

T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}

T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}

T_{S \infty}= \dfrac{2736608.861 }{12700}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

From equation(2);

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

Replacing A = 1695208.861 in the above equation; we have:

156= T_{S \infty} - \dfrac{1695208.861}{28500}

T_{S \infty}= 156 + \dfrac{1695208.861}{28500}

T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}

T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}

T_{S \infty}= \dfrac{6141208.861}{28500}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

T_S = T_{S \infty} - \dfrac{A}{M_n}

Then:

181 = 215.481- \dfrac{1695208.861 }{M_n}

Collecting like terms ; we have:

\dfrac{1695208.861 }{M_n} = 215.481-  181

\dfrac{1695208.861 }{M_n} =34.481

1695208.861= 34.481 M_n

Dividing both sides by 34.481; we have:

M_n = \dfrac{1695208.861}{34.481}

\mathbf{M_n = 49163.56431  \ g/mol }

5 0
3 years ago
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