Honeywell to buy EMS Technologies for $491 million

Share price of $33 a one-third premium for maker of mobile computing products

By  

In a move to bolster its mobile computing business, Honeywell (NYSE: HON) on Monday announced an agreement to purchase EMS Technologies (NASDAQ: ELMQ) for $491 million.

The $33-per-share price tag is, coincidentally, exactly 33 percent above EMS's Friday's closing price of 24.80.

EMS shares were up as high as 32.88 in Monday morning trading, while Honeywell shares were up 36 cents, or 0.7 percent, to 55.89. Clearly Wall Street isn't punishing Honeywell for paying a premium to acquire EMS.

The $33 share price is 124 percent above EMS's 14.75 closing price a year ago. Shares have risen steadily over the past year, surging in April after the company announced it was "exploring strategic alternatives."

EMS makes mobile computing products (and offers services) for field environments, including transportation, warehousing and search and rescue. Products include terminals, antennas, data storage and surveillance applications.

(Wait a minute. Surveillance applications, Honeywell, government contractor...someone call Alex Jones!)

A month ago EMS reported record adjusted first-quarter earnings of $9.5 million.

In a statement announcing the acquisition, Honeywell Automation and Control Solutions President and CEO Roger Fradin said, "EMS strengthens our core mobile computing business and expands our addressable market with complementary new products, channel partners, and entry into the warehousing and port segments that we believe will be growth drivers for the business."

Honeywell said it expects the transaction, which needs shareholder and regulatory approval, to close in the third quarter.

Join us:
Facebook

Twitter

Pinterest

Tumblr

LinkedIn

Google+

Answers - Powered by ITworld

ITworld Answers helps you solve problems and share expertise. Ask a question or take a crack at answering the new questions below.

Join us:
Facebook

Twitter

Pinterest

Tumblr

LinkedIn

Google+

Ask a Question
randomness