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Ternary A server has customers waiting to be served. The service time required by eachcustomer is known in advance: it is ciminutes for customer i. So if, for example, the customers are served in order of increasing i , then the ithcustomer has to wait Pij=1tjminutes. We wish to minimize the total waiting time.

T=Xni=1(time spent waiting by customer ).

Give an efficient algorithm for computing the optimal order in which to process the customers.

Short Answer

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Step by step solution

01

Introduction

Algorithms providing optimum customer processing ordering:

Its greedy algorithm is utilised to assist the client who serves first in the shortest period of time.

• That is, a large number of consumers are sorted according to their time "t" values, and then served in that order.

It has been discovered that it is no issue what, the total time that deals inside the consumers serve does not alter.

• The total time taken is always equal to the sum of the service times for all of the clients.

02

Total time

The total time can be calculated as:

T=i=1n(numberofcustomersstillwaitingattimet)

• Reduce consequently, this number of customers who should wait is increasing, implying that the optimal course of action is for consumers to be served with greater service time.

Algorithm:

Explanation: Presume that there is indeed a better option than the greedy one.

• When sort data number of customers by time, the Greedy approach is employed. Customers are served once they have been sorted.

• A minimum of one pair of consecutive consumers is required for the best solution. Serving the second consumer takes less time than serving the first.

Take, for example, the requirement that "out of order" must be met by a pair of customers.

• Pair of customers denotes as ciand ci+1.

• Service time for the pair of customers denotes as tiand ti+1.

• By assumption, the service time of first customer is greater than the service time of second customer. That is,ti>ti+1.

o Swap the order of the pair of customers will produce the better ordering. So, the second customer ci+1is served before the first customer ci.

• It does not change any waiting time of other customers.

o The waiting time for the customer ciwill increase by ti+1and customer ci+1will increase by ti. So, the assumption of ti>ti+1will reduces the overall waiting time.

• As a result, changing the order of the consumers results in the shortest overall waiting time.

Sorting the number of clients "n" will take a total of n minutes O(nlogn).

Each and every swap will reduce the total waiting time. So, after all the swaps are performed, the customers will be sorted in ascending order. The optimal solution becomes the greedy solution.

Therefore, the greedy solution must be the optimal solution.

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Most popular questions from this chapter

Consider the following graph.

(a) What is the cost of its minimum spanning tree?

(b) How many minimum spanning trees does it have?

(c) Suppose Kruskal’s algorithm is run on this graph. In what order are the edges added to the MST? For each edge in this sequence, give a cut that justifies its addition.

Suppose we want to find the minimum spanning tree of the following graph.

(a) Run Prim’s algorithm; whenever there is a choice of nodes, always use alphabetic ordering (e.g., start from node A). Draw a table showing the intermediate values of the cost array.

(b) Run Kruskal’s algorithm on the same graph. Show how the disjoint-sets data structure looks at every intermediate stage (including the structure of the directed trees), assuming path compression is used.

A binary counter of unspecified length supports two operations: increment (which increases its value by one) and reset (which sets its value back to zero). Show that, starting from an initially zero counter, any sequence of n increment and reset operations takes time O(n); that is, the amortized time per operation is O(1) .

Suppose you implement the disjoint-sets data structure usingunion-by-rank but not path compression. Give a sequence ofm union and find operations onnelements that take Ω(mlogn)time.

Show that for any integer n that is a power of 2 , there is an instance of the set cover problem (Section 5.4) with the following properties:

  1. There are n elements in the base set.
  2. The optimal cover uses just two sets.
  3. The greedy algorithm picks at least log n sets.

Thus the approximation ratio we derived in the chapter is tight.

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