Question 1446

Thirty Open Windows

A job computes a 30 minute sliding window with a 1 minute slide over 90,000 active keys. Each key holds about 240 bytes of aggregate state per window.

Estimate the steady-state window state the job must keep.

A

About 1.30 GB, because a 30 minute window sliding by 1 minute keeps 60 windows open, one per minute of the window plus one per slide

B

About 648 MB: at any instant 30/1=3030 / 1 = 30 windows are open, so 90,000×30×240 bytes=648,000,00090{,}000 \times 30 \times 240\ \text{bytes} = 648{,}000{,}000

C

About 21.6 MB, because only the current window is being accumulated: 90,000×24090{,}000 \times 240

D

About 19.4 GB, because each of the 30 open windows retains the full 30 minutes of raw records: 90,000×30×30×24090{,}000 \times 30 \times 30 \times 240