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Check my work Suppose a firm has 16.3 million shares of common stock outstanding and eight candidates are up for election to

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Answer #1

a. If the firm uses cumulative voting to elect its board, what is the minimum number of votes needed to ensure election of one member to the board?

Total number of shares of common stock outstanding = 16.3 million shares

For election to four seats on the board of directors, if the firm uses cumulative voting to elect its board,

Total available votes =16.3 million shares *4 = 65.20 million

The four candidates with the highest number of votes will be elected to the board, therefore the minimum number of votes needed to ensure election of one member to the board = [Total available votes / (number of seats on the board of directors+1)] +1   (one is added to avoid any tie)

= [65.20 million/ (4+1)] +1

=13,040,000 + 1

=13,040,001

Now remaining number of votes are = 65,200,000 - 13,040,001 = 52,159,999 if it is divided between four other candidates, one will definitely get less than 13,040,001 votes and candidate with minimum 13,040,001 votes will definitely win.

Therefore Minimum number of votes: 13,040,001

b. If the firm uses straight voting to elect its board, what is the minimum number of votes needed to ensure election of one member to the board?

If the firm uses straight voting to elect its board, total votes available per candidate is number of shares of common stock outstanding = 16.3 million

The minimum number of votes needed to ensure election of one member to the board is half of the total votes available per candidate plus one

= 16.3 million/2 +1    (one is added to avoid any tie)

= 8,150,000 +1 = 8,150,001

Minimum number of votes: 8,150,001

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