The Seahawks' three-year deal with safety Bradley McDougald is worth $13.5 million with $6.5 million guaranteed, according to a source. It includes a $4 million signing bonus and a fully guaranteed $1.5 million base salary for 2018. Each year, McDougald can earn $500,000 in per-game active roster bonuses. He'll count $3.33 million against Seattle's 2018 cap. That should put the Seahawks in the neighborhood of $27 million in available 2018 cap space, not factoring in restricted-free-agent tenders.