A question about semaphore.release() in vllm_server_task()

[keontang]

olmocr/olmocr/pipeline.py

Line 669 in be1f845

if server_printed_ready_message and last_queue_req == 0 and time.time() - last_semaphore_release > 30 and semaphore.locked():

When starting up vllm_server, why we need 30s timeout? If 'server_printed_ready_message == True', can we free up semaphore directly?

[jakep-allenai]

Hmm, the semaphore starts with 1 so the first worker can always start right away. I think it's to allow the server to start up normally before you start freeing it again.