The San Francisco 49ers agreed to terms on a four-year deal worth up to $54 million with Tampa Bay Buccaneers linebacker and impending free agent Kwon Alexander.

We now have some details on the four-year deal courtesy of Aaron Wilson of the Houston Chronicle. The contract cannot be signed until the start of the new league year on Wednesday.

Potential worth: $54 million
Guaranteed: $27.5 million
Signing bonus: $4 million
2019 roster bonus: $8.5 million

More details, including yearly salaries, can be found within Wilson's tweet below.


Alexander's $13.5 million annual salary will make him the second-highest paid player on the 49ers roster behind quarterback Jimmy Garoppolo. If things don't work out in San Francisco, the team can move on after the 2019 season with just a $3 million hit in dead money.

Here is a year-by-year breakdown of Alexander's contract courtesy of Spotrac.com.

Year Base Salary Cap Hit
2019 $1,750,000 $11,531,250
2020 $11,250,000 $13,100,000
2021 $12,550,000 $14,400,000
2022 $12,650,000 $14,500,000