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HACTEHA [7]
2 years ago
8

Consider a DASH system for which there are N video versions (at N different rates and qualities) and N audio versions (at N diff

erent rates and qualities). Suppose we want to allow the player to choose at any time any of the N video versions and any of the N audio versions. a. If we create files so that the audio is mixed in with the video, so server sends only one media stream at given time, how many files will the server need to store (each a different URL)? b. If the server instead sends the audio and video streams separately and has the client synchronize the streams, how many files will the server need to store?
Computers and Technology
1 answer:
Ksju [112]2 years ago
5 0

Answer: a) N² files. b) 2N files.

Explanation:

If an user can choose freely both the video and the audio quality, if they are stored separately, but the user can download any mix of them as a single file, the server must store one audio version for each video version, so it will need to store NxN = N² files.

If, instead, the server can send the audio and video streams separately, so the user can choose one of the N videos available, and one of the N audio versions, the server will need to store only N video files + N audio audio files, i.e. , 2 N files.

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We Deliver trains its truck loaders how to set the packages in the delivery vehicles, so that when delivery drivers are pulling
damaskus [11]

Answer:

We Deliver trains its truck loaders how to set the packages in the delivery vehicles, so that when delivery drivers are pulling packages off their trucks, they are organized in a specific order and with the label facing forward to reduce errors and save time. A <u>Procedure </u>is being implemented when directing that the trucks be loaded in this specific manner.

Explanation:

Procedure is the method or set of instructions to perform some operation in specific order to complete the task in an organized manner. It is the type of directions to perform some operations.

In above mentioned case, if we want that the packages should set in some organized manner such as labeling face of package in forward direction so that, chance of error should reduce and time save. We should provide some procedure to the truck loaders.

5 0
2 years ago
You are consulting for a trucking company that does a large amount ofbusiness shipping packages between New York and Boston. The
likoan [24]

Answer:

Answer explained with detail below

Explanation:

Consider the solution given by the greedy algorithm as a sequence of packages, here represented by indexes: 1, 2, 3, ... n. Each package i has a weight, w_i, and an assigned truck t_i. { t_i } is a non-decreasing sequence (as the k'th truck is sent out before anything is placed on the k+1'th truck). If t_n = m, that means our solution takes m trucks to send out the n packages.

If the greedy solution is non-optimal, then there exists another solution { t'_i }, with the same constraints, s.t. t'_n = m' < t_n = m.

Consider the optimal solution that matches the greedy solution as long as possible, so \for all i < k, t_i = t'_i, and t_k != t'_k.

t_k != t'_k => Either

1) t_k = 1 + t'_k

    i.e. the greedy solution switched trucks before the optimal solution.

    But the greedy solution only switches trucks when the current truck is full. Since t_i = t'_i i < k, the contents of the current truck after adding the k - 1'th package are identical for the greedy and the optimal solutions.

    So, if the greedy solution switched trucks, that means that the truck couldn't fit the k'th package, so the optimal solution must switch trucks as well.

    So this situation cannot arise.

  2) t'_k = 1 + t_k

     i.e. the optimal solution switches trucks before the greedy solution.

     Construct the sequence { t"_i } s.t.

       t"_i = t_i, i <= k

       t"_k = t'_i, i > k

     This is the same as the optimal solution, except package k has been moved from truck t'_k to truck (t'_k - 1). Truck t'_k cannot be overpacked, since it has one less packages than it did in the optimal solution, and truck (t'_k - 1)

     cannot be overpacked, since it has no more packages than it did in the greedy solution.

     So { t"_i } must be a valid solution. If k = n, then we may have decreased the number of trucks required, which is a contradiction of the optimality of { t'_i }. Otherwise, we did not increase the number of trucks, so we created an optimal solution that matches { t_i } longer than { t'_i } does, which is a contradiction of the definition of { t'_i }.

   So the greedy solution must be optimal.

6 0
2 years ago
All the employees of Delta Corporation are unable to access the files stored in the server. What do you think is the reason behi
CaHeK987 [17]
C serve crash hope this is correct
5 0
2 years ago
Nancy would like to configure an automatic response for all emails received while she is out of the office tomorrow, during busi
Stella [2.4K]

she can appoints someone she trust to act on her behalf

4 0
2 years ago
Read 2 more answers
1. Create a view named customer_addresses that shows the shipping and billing addresses for each customer.
Verdich [7]

Answer:

Answer given below

Explanation:

1.

CREATE VIEW CustomerAddresses AS

SELECT custo. CustomerID, EmailAddress , LastName ,FirstName,

bill.Line1 AS BillLine1, bill.Line2 AS BillLine2, bill.City AS BillCity, bill.State AS BillState, bill.ZipCode AS BillZip,

ship.Line1 AS ShipLine1, ship.Line2 AS ShipLine2, ship.City AS ShipCity, ship.State AS ShipState, ship.ZipCode AS ShipZip

FROM Customers custo , Addresses ship , Addresses bill

WHERE custo. BillingAddressID= bill.AddressID AND custo.ShippingAddressID= ship. AddressID;

2.

SELECT CustomerID, LastName, FirstName, BillLine1 FROM CustomerAddresses;

3.

CREATE VIEW OrderItemProducts

AS

SELECT Orders.OrderID, OrderDate, TaxAmount, ShipDate,

ItemPrice, DiscountAmount, (ItemPrice- DiscountAmount) AS FinalPrice,

Quantity, and (Quantity * (ItemPrice-DiscountAmount)) AS ItemTotal,

ProductName FROM

Orders, OrderItems, Products

WHERE

Orders.OrderID = OrderItems.OrderID AND

OrderItems.ProductID = Products. ProductID;

4.

CREATE VIEW ProductSummary

AS

SELECT distinct

ProductName, COUNT(OrderID) AS OrderCount, SUM(ItemTotal) AS OrderTotal

FROM

OrderItemProducts

GROUP BY ProductName;

5.

SELECT ProductName, OrderTotal

FROM ProductSummary P

WHERE 5> (select count(*) FROM ProductSummary S

WHERE P.OrderTotal<S.OrderTotal)

ORDER BY OrderTotal;

5 0
2 years ago
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