Basketball Australia can confirm that Andrew Bogut will sign a five-year extension with the Milwaukee Bucks tomorrow worth $US60 million in guaranteed money, with bonuses that could take the deal to $US72.5 million.
The extension does not affect the $6.9 million that Bogut will be paid on the final year of his rookie contract in 2008/09, and it keeps the Australian Boomer with the Bucks through the 2013/14 season.
Bogut departed Melbourne this morning and will arrive in Milwaukee tomorrow, where he will sign the deal and hold a press conference before boarding a return flight home.
“I’m very happy to be remaining in Milwaukee for another six years,” Bogut said.
“The people, the city and the culture made it an easy decision for me to stay and I look forward to being a part of the process of working hard to become a consistently winning franchise once again.
“I’d like to thank [Bucks owner] Senator Kohl and General Manager John Hammond for believing in my abilities and having the confidence to offer me a great long term deal.
“Lastly, I’d like to thank my parents, family and friends as well as my trainer Sinisa Markovic.”
Bogut will arrive back in Melbourne on Monday afternoon and join the Australia Post Boomers at their final training camp on the Gold Coast on Tuesday, where a press conference for local media will be held.
Bogut’s Australian agent, Bruce Kaider, was thrilled with the news and the respect that a contract like this will generate among the NBA player group.
“A deal of this size is a clear indication of Andrew’s value to the Milwaukee franchise and it puts him among the elite players of the NBA in monetary terms,” Kaider said.
“He has had three great seasons with the Bucks and he deserved an extension of this magnitude.
“He is extremely happy with the agreement and relieved to be able to join the Boomers in the lead up to Beijing.”
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