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3 years ago

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Suppose you are designing a sliding window protocol for a 1 Mbps point-to-point link to the moon, which has a one-way latency of

1.25 seconds. Assume that each frame carries 1 KB of data. (a) How many frames need to be in transit to keep this pipe full? (b) What is the maximum reasonable size for SWS and RWS needed for such a system? (c) For the SWS and RWS sizes found in part b, what is the number of bits needed for the sequence number?

1 answer:

mylen [45]3 years ago

7 0

Answer:

approx 306 frames need to be in transit to keep this pipe full

maximum reasonable size is 306 frames

sequence number = 10 bit

Explanation:

given data

transmission rate = 1 Mbps

size of frame Tt = 1 KB

one way latency Tp = 1.25 s

to find out

How many frames need and What is the maximum reasonable size for SWS and RWS and what is the number of bits needed for the sequence number

solution

we know useful time is = $\frac{1 KB}{1 Mbps} = \frac{1024*8}{1000*1000}$

so total time is

total time = Tt + 2Tp

total time = $\frac{1024*8}{1000*1000}$ + 2 × 1.25

so

possible frames are

$\frac{n*Tt}{Tt+2Tp}$ ≤ 1

n ≤ 1 + $\frac{2.5*1000000}{1024*8}$

n ≤ 306.17

so approx 306 frames need to be in transit to keep this pipe full

and

we know here

size of sending window is equal to size of receiving window

so maximum reasonable size is 306 frames

and

bit need for sequence number is

sequence number = $log^{306+306}_2$

sequence number = 10 bit

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Answer:

See Answer below- Explanation is the entire answer

Explanation:

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Properties: Ductile, good heat conductivity, good electrical conductivity, high strength;

Drawbacks: Relatively high weight, reactive with oxygen to create oxides- corrosion is presented;

Examples: steel, aluminum alloys, brass, copper, titanium

Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

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Drawbacks: Deforms by fracturing, shock resistance is low, no conductivity of electricity;

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Properties: amorphous, transparent, high weight

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Applications: windows, protection screens;

Polymers:

Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

Drawbacks: low strength, low operating temperatures;

Examples: polyethylene, nylon, ABS-plastic, rubber;

Applications: toys, tires, insulation covers for the wires.

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Answer:

q = 1.73 W

Explanation:

given data

small end = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

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solution

first we will get here side of cross section that is express as

$S = S1 + \frac{S2-S1}{L} x$ ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + $\frac{10-5}{15} x$

S = $\frac{0.15 + x}{3}$ m

and

now we get here Area of section at distance x is

area A = S² ...............2

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and

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put here value and we get

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it will be express as

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Answer:

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3 years ago