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.
Kwon Alexander (49ers), $54M, $27.5M gtd, $4M signing bonus, $8.5M fourth day 2019 roster bonus, salaries $1.75M (gtd), $11.25M (gtd inj at signing, skill-capp if on roster April 1, 2020; $12.55M, $12.65M, $46,875 per game active annual
— Aaron Wilson (@AaronWilson_NFL) March 12, 2019
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 |