Question

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

Short answer

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

.

Step-by-step solution

Given data

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

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

.

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\(i < n)\).

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 = \(talks1,3,7\).