Kwon Alexander leads Buccaneers in performance-based pay distributions

TAMPA, Fla. -- It's bonus time in the NFL, and the league has released its annual performance-based pay distributions report, with linebacker Kwon Alexander coming in first for the Tampa Bay Buccaneers.

Performance-based pay compensates players based on playing time for their salary levels. It's designed to reward those with smaller contracts who contribute significantly to their teams.

Each team is given $3,995,000 to allocate to players and where that money goes sheds light on those whose play hasn't yet been rewarded with a big paycheck but recognition is still warranted. It's particularly good for showcasing players on rookie contracts, especially undrafted free agents and those earning the league minimum.

Alexander, a former fourth-round draft pick, is in the second year of his four-year rookie contract. He had a play-time percentage of 72.77 percent and he'll take home an extra $291,500.78, which bumps his total compensation to $650,625.

Offensive lineman Kevin Pamphile came up second on the list, as he played 64.23 percent of snaps, filling in all season for starting left guard J.R. Sweezy. He takes home an extra $255,317.95, which makes his adjusted compensation $655,670.

Coming in third was right guard Ali Marpet, a second-round draft pick in 2015. He played 76.24 percent of snaps and will receive an extra $230,062.01 to earn $863.678.