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nataly862011 [7]
3 years ago
11

A buffer is filled over a single input channel and emptied by a single channel with a capacity of 64 kbps. Measurements are take

n in the steady state for this system with the following results:
Average packet waiting time in the buffer = 0.05 seconds
Average number of packets in residence = 1 packet
Average packet length = 1000 bits

The distributions of the arrival and service processes are unknown and cannot be assumed to be exponential.

Required:
What are the average arrival rate λ in units of packets/second and the average number of packets w waiting to be serviced in the buffer?
Computers and Technology
1 answer:
ELEN [110]3 years ago
3 0

Answer:

a) 15.24 kbps

b) 762 bits

Explanation:

Using little law

a) Determine the average arrival rate (  λ  ) in units of packets/s

λ  = r / Tr  --- 1

where ; r = 1000 bits , Tr = Tw + Ts = 0.05 + (( 1000 / (64 * 1000 ))  = 0.0656

back to equation 1

λ = 1000 / 0.0656 = 15243.9 = 15.24 kbps

b) Determine average number of packets <em>w</em> to be served

w =  λ * Tw =  15243.9 * 0.05 = 762.195 ≈ 762 bits

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What will be displayed after code corresponding to the following pseudocode is run? Main Set OldPrice = 100 Set SalePrice = 70 C
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Answer:

A jacket that originally costs $ 100 is on sale today for $ 80                                    

Explanation:

Main : From here the execution of the program begins

Set OldPrice = 100  -> This line assigns 100 to the OldPrice variable

Set SalePrice = 70   -> This line assigns 70to the SalePrice variable

Call BigSale(OldPrice, SalePrice)  -> This line calls BigSale method by passing OldPrice and SalePrice to that method

Write "A jacket that originally costs $ ", OldPrice  -> This line prints/displays the line: "A jacket that originally costs $ " with the resultant value in OldPrice variable that is 100

Write "is on sale today for $ ", SalePrice  -> This line prints/displays the line: "is on sale today for $ " with the resultant value in SalePrice variable that is 80

End Program -> the main program ends

Subprogram BigSale(Cost, Sale As Ref)  -> this is a definition of BigSale method which has two parameters i.e. Cost and Sale. Note that the Sale is declared as reference type

Set Sale = Cost * .80  -> This line multiplies the value of Cost with 0.80 and assigns the result to Sale variable

Set Cost = Cost + 20  -> This line adds 20 to the value of Cost  and assigns the result to Cost variable

End Subprogram  -> the method ends

This is the example of call by reference. So when the method BigSale is called in Main by reference by passing argument SalePrice to it, then this call copies the reference of SalePrice argument into formal parameter Sale. Inside BigSale method the reference &Sale is used to access actual argument i.e. SalePrice which is used in BigSale(OldPrice, SalePrice) call. So any changes made to value of Sale will affect the value of SalePrice

So when the method BigSale is called two arguments are passed to it OldPrice argument and SalePrice is passed by reference.

The value of OldPrice is 100 and SalePrice is 70

Now when method BigSale is called, the reference &Sale is used to access actual argument SalePrice = 70

In the body of this method there are two statements:

Sale = Cost * .80;

Cost = Cost + 20;

So when these statement execute:

Sale = 100 * 0.80 = 80

Cost = 100 + 20 = 120

Any changes made to value of Sale will affect the value of SalePrice as it is passed by reference. So when the Write "A jacket that originally costs $ " + OldPrice Write "is on sale today for $ " + SalePrice statement executes, the value of OldPrice remains 100 same as it does not affect this passed argument, but SalePrice was passed by reference so the changes made to &Sale by statement in method BigSale i.e.  Sale = Cost * .80; has changed the value of SalePrice from 70 to 80 because Sale = 100 * 0.80 = 80. So the output produced is:

A jacket that originally costs $ 100 is on sale today for $ 80                              

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