Question

Suppose a firm has 55 million shares of common stock outstanding and eight candidates are up for election to six seats on the board of directors.

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

(For all requirements, enter your answers in dollars, not millions of dollars.)

Answer 1

a.) Under cumulative voting scenario,

Total number of votes available = Common Shares Outstanding × No of directors

= 55 x 6 million

= 330 million

As there are eight candidates for the six board positions, the six candidates with highest number of votes will be elected to the board and the candidate with the least total votes will not be elected.

Minimum votes needed to ensure election =1/8 x 330 million + 1 vote to break any ties

= 41,250,001 votes

If one candidate receives 41,250,001 votes, the leftover is total 2,88,749,999 votes.

No matter how these votes are spread over the remaining 6 director candidates, it is impossible for each of the 6 to receive more than 41,250,001. This would require more than 5 × 41,250,001 votes, or more than the remaining 2,88,749,999votes.

b.) Now, in case of straight voting,

Vote on board of directors occurs one director at a time.

=> Number of votes eligible for each director = Number of Shares Outstanding = 55,000,000

Minimum number of votes needed to ensure election is through simple majority i.e. = 55,000,000/6+ 1 = 9.16 million votes

This solution is provided with detailed explanation. Please discuss in case of Doubt.

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