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To determine the optimal schedule for talks, such that total number of attendees is maximized.

Short Answer

Expert verified

Therefore, the optimal schedule for talks \( = \) \(talks1,3,7\).

Step by step solution

01

Given data

The number of attendees of the talks \({w_i}\); where \(i = 1,2, \ldots .7\)

\(20,10,50,30,15,25,40\).

02

Concept used of recurrence relation

A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms (Expressing\({F_n}\)as some combination of\({F_i}\)with\(i < n)\).

03

Solve for possible configurations

We have total seven number of talks

from the given data, Let us assume,

\({\rm{ talk }}1 = 20,{\rm{ talk }}2 = 10,{\rm{ tal}}k3 = 50,{\rm{ talk }}4 = 30,{\rm{ talk }}5 = 15,{\rm{ tal}}k6 = 25,{\rm{ talk }}7 = 40\)

There is no talks during first talks begin

\(P(1) = 0 \Rightarrow P(2) = 0\)

Talk 3 and talk 1 are compatible, where as Talk 3 and talk 2 are not

\(P(3) = 1\)

\({\rm{talk }}4\)and \({\rm{talk }}5\) are not compatible with any other talks

\(P(4) = 0 \Rightarrow P(5) = 0\)

Similarly,

\(P(6) = 2 \Rightarrow P(7) = 4\)

We have

\(T(i) = \max \left( {{w_i} + T(P(i)),T(i - 1)} \right)\)

With the use of above formula

\(T(7) = 110\)

\(S(i)\)is the optimal schedule for talks

Therefore, the optimal schedule fot talks = \(talks1,3,7\).

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Most popular questions from this chapter

Suppose that f(n)=f(n/3)+1when is a positive integer divisible by 3, and f(1)=1. Find:

a)f(3).

b)f(27).

c)localid="1668607414775" f(729).

A sequence \({a_1},{a_2},.....,{a_n}\) is unimodal if and only if there is an index \(m,1 \le m \le n,\) such that \({a_i} < {a_i} + 1\) when \(1{1 < i < m}\) and \({a_i} > {a_{i + 1}}\) when \(m \le i < n\). That is, the terms of the sequence strictly increase until the \(m\)th term and they strictly decrease after it, which implies that \({a_m}\) is the largest term. In this exercise, \({a_m}\) will always denote the largest term of the unimodal sequence \({a_1},{a_2},.....,{a_n}\).

a) Show that \({a_m}\) is the unique term of the sequence that is greater than both the term immediately preceding it and the term immediately following it.

b) Show that if \({a_i} < {a_i} + 1\) where \(1 \le i < n\), then \(i + 1 \le m \le n\).

c) Show that if \({a_i} > {a_{i + 1}}\) where \(1 \le i < n\), then \(1 \le m \le i\).

d) Develop a divide-and-conquer algorithm for locating the index \(m\). (Hint: Suppose that \(i < m < j\). Use parts (a), (b), and (c) to determine whether \(((i + j)/2) + 1 \le m \le n,\) \(1 \le m \le ((i + j)/2) - 1,\) or \(m = ((i + j)/2)\)

Find a closed form for the generating function for the sequence\(\left\{ {{a_n}} \right\}\), where,

a) \({a_n} = 5\) for all\(n = 0,1,2, \ldots \).

b) \({a_n} = {3^n}\)for all\(n = 0,1,2, \ldots \)

c) \({a_n} = 2\)for\(n = 3,4,5, \ldots \)and\({a_0} = {a_1} = {a_2} = 0\).

d) \({a_n} = 2n + 3\)for all\(n = 0,1,2, \ldots \)

e) \({a_n} = \left( {\begin{array}{*{20}{l}}8\\n\end{array}} \right)\)for all\(n = 0,1,2, \ldots \)

f) \({a_n} = \left( {\begin{array}{*{20}{c}}{n + 4}\\n\end{array}} \right)\)for all\(n = 0,1,2, \ldots \)

Use generating functions to find the number of ways to select 14 balls from a jar containing 100 red balls, 100 blue balls, and 100 green balls so that no fewer than 3 and no more than 10 blue balls are selected. Assume that the order in which the balls are drawn does not matter.

Find the sequence with each of these functions as its exponential generating functionf(x)=e3x-3e2x.

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