Andrew Bogut has two million more reasons to be thrilled at being named as one of the NBA's premier defensive players.
The Melbourne-born centre is set to pocket a $US2 million ($A2.54 million) bonus after he got the nod for the All-Defensive Second Team on Wednesday.
A bonus clause in his contract, requiring him to play in at least 65 regular-season games for Golden State as well as get selected for either All-Defensive team, is worth 15 per cent on top of his $US12.97 million ($A16.47 million) annual salary, according to The Bay Area News Group.
So often beset by injury in the past, Bogut managed 67 of 82 regular season games and will earn an extra $US1.94 million ($A2.46 million) for his stellar defensive campaign.
Bogut was joined by Warrior teammate Draymond Green, runner-up for the Defensive Player of the Year award, who was named to the All-Defensive First Team.
It's the first time both players were named in the teams.
The two big men were rewarded after leading the Warriors to not only the NBA's best regular season record, but also the best defensive record.
The Warriors defeated the Houston Rockets in game one of the Western Conference Finals in Oakland on Tuesday and are favourites to go on and win the NBA Championship.
Joining Green in the NBA defensive first team are Kawhi Leonard (San Antonio), Tony Allen (Memphis), DeAndre Jordan (Los Angeles Clippers) and Chris Paul (LA Clippers).
In the second team Bogut is joined by Anthony Davis (New Orleans), Jimmy Butler (Chicago), John Wall (Washington) and Tim Duncan (San Antonio).
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