The Milwaukee Bucks will be one of four teams to meet with DeAndre Jordan, a dominant defender and powerful dunker of a center from the Clippers, when free agency commences on July 1 according to the LA Times:
When free agency starts at 9:01 p.m. PDT Tuesday, Jordan will be home in Houston. The officials said four teams will visit Jordan at home — the Clippers, Lakers, Mavericks and Milwaukee Bucks.
Well, this is surprising. Jordan is sure to command a max deal from one of the teams listed above. The Bucks can offer him $80 million over four years, as can the Lakers and Mavericks. The Clippers can offer $108 million over five years.
The Bucks will have around $16 million available when free agency begins, according to this helpful cap sheet from BrewHoop. They could use that $16 million and then sign Khris Middleton to whatever sort of deal they work out with him, as teams are allowed to exceed the cap to re-sign their players. Unless the Bucks are able to unload four or five million in salaries early in free agency, Jordan seems out of the question.
Meeting with a player the caliber of Jordan represents progress for the Bucks and shows the influence Jason Kidd has around the league. A year ago, the Bucks were the worst team in the league. Milwaukee was nothing but a small, cold, unsuccessful market, and the idea of meeting with one of the summer’s premier free agents was absurd. But the Bucks made significant progress last year, both on and off the court.
As a developing team in a feeble conference where an impact player could have his impact amplified, suddenly Milwaukee may not look like such an outpost anymore.