Suppose a firm has 16.8 million shares of common stock outstanding and six candidates are up for election to five seats on the board of directors.
If the firm uses cumulative voting to elect its board, what is the minimum number of votes needed to ensure the election of one member to the board?

The vote on the board of directors occurs one director at a time. So, the number of votes that are eligible for each director is 16,800,000, equal to the number of shares available. The minimum number of votes that will be required to make an election is one half (1/2) of 17 million votes available, or we can say 8,400,000. To avoid candidates having equal votes, one would actually need one more vote than 8,350,000 to win the election.

batolisis batolisis · Answer 2 · 2020-03-14T12:58:16+01:00

Answer:

14000001

Explanation:

using the cumulative voting system the total number of votes that could be made would be

= shares of common stock outstanding * Number of available board of directors

= 16800000 * 5 = 84000000

hence the minimum number of votes from this total votes that is required to win a seat will be

= (1/6 * 84000000 ) + 1 ( since there are six contestants and 5 available seats)

= 14000000 + 1 = 14000001

note : the addition of 1 to the result is to break up an even position since no candidate can score more than 14000001 votes at a time

Suppose a firm has 16.8 million shares of common stock outstanding and six candidates are up for election to five seats on the board of directors. If the firm uses cumulative voting to elect its board, what is the minimum number of votes needed to ensure the election of one member to the board?

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Suppose a firm has 16.8 million shares of common stock outstanding and six candidates are up for election to five seats on the board of directors.
If the firm uses cumulative voting to elect its board, what is the minimum number of votes needed to ensure the election of one member to the board?