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prohojiy [21]
3 years ago
11

Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 second

s. In Eager Mode, the app utilizes a faster timer resolution for its computations, so the execution time in Eager Mode is 2 seconds (i.e., Eager Mode execution time is 60% of Lazy Mode execution time).
After finishing computation, the app sends some data to the cloud, regardless of the mode it’s in. The data size sent to the cloud is 600 MB. The bandwidth of communication is 15 MBps for WiFi and 5 MBps for 4G. Assume that the communication radio is idle during the computation time. Assume that the communication radio for WiFi has a power consumption of 75 mW when active and 15 mW when idle. Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization. Assume that the power consumption of the CPU is a linear function of its utilization. In other words: P = (Idle Power) + (Utilization)*(Power per unit Utilization). A configuration of the mobile app involves choosing a timer resolution (Lazy or Eager) and choosing a type of radio (WiFi or 4G). For example, faster timer resolution (Eager) and 4G network is a configuration, while slower resolution (Lazy) and WiFi is another. There are four possible configurations in all.

Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?
Engineering
1 answer:
lara [203]3 years ago
3 0
N didn’t do it for you toroeriot everyone wwas wowowowoww
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A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft
grandymaker [24]

Answer:

peak-hour volume = 1890 veh/h

Explanation:

<u>Determine the peak-hour Volume </u>

Applying the equation below

Vp =  v / ( PHF * N * Fg * Fdp )  -------------- ( 1 )

where :

Vp = 1250

v ( peak - hour volume ) =  ?

PHF ( peak hour factor ) = 0.84

N  = 2 lanes per direction

Fg ( grade adjustment for rolling terrain ) = 0.99 ≈ 1

Fdp = 0.90

<u>Back to equation 1 </u>

v = Vp (  PHF * N * Fg * Fdp )  

  = 1250 ( 0.84 * 2 * 1 * 0.90 )

  = 1890 veh/h

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3 years ago
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Hitman42 [59]

Answer:

woah

Explanation:

7 0
4 years ago
A water tank is completely filled with liquid waterat 20°C.The tank material is such that it can withstand tensioncaused by a vo
Xelga [282]

Answer:

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Explanation:

From literature, the coefficient of volume expansion of water between 20°C and 50°C = β = (0.377 × 10⁻³) K⁻¹

Volume expansivity is given by

ΔV = V β ΔT

ΔV = Change in volume

V = initial volume

β = Coefficient of volume expansion = (0.377 × 10⁻³) K⁻¹ = 0.000377 K⁻¹

ΔT = Change in temperature = ?

It is given in the question that maximum volume increase the tank can withstand is

(ΔV/V) × 100% = 0.8%

(ΔV/V) = 0.008

V β ΔT = ΔV

β ΔT = (ΔV/V)

β ΔT = 0.008

ΔT = (0.008/β)

ΔT = (0.008/0.000377)

ΔT = 21.22°C

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Hope this Helps

5 0
4 years ago
When handling chemicals and solvents, technicians are recommended to
Luda [366]

Answer:

  1. To wear PPE
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7 0
3 years ago
Read 2 more answers
Two very long concentric cylinders of diameters D1 = 0.42 m and D2 = 0.5 m are maintained at uniform temperatures of T1 = 950 K
MrRa [10]

Answer:

Q=33.34 KW/m

Explanation:

Given that

D₁=0.42 m

A₁= π D₁ L

For unit length

A₁= π D₁ = 0.42 π m²

D₂=0.5 m

A₂= 0.5 π m²

ε₁= 1 ,ε₂= 0.55

T₁=950 K  ,T₂ = 500 K

Q=\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{A_1\epsilon _1}+\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+\dfrac{1}{A_1F_{12}}}

F₁₁+F₁₂= 1

F₁₁= 0

So, F₁₂= 1

Q=A_1\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{\epsilon _1}+A_1\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+1}

Q=0.42\pi \dfrac{5.67\times 10^{-8}(950^4-500^4)}{\dfrac{1-1}{1}+0.42\pi\times \dfrac{1-0.55}{0.5\pi\times 0.55}+1}

Q=33.34 KW/m

3 0
4 years ago
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