Code School of Analytics
ExpertDates & Time50 points

SLA breaches by ticket priority

A managed-services company promises resolution within a number of hours per priority. The monthly review needs the median resolution time and the number of breaches. Data: support.csv, tickets opened between April and June 2025.

  • ticket_id: unique code
  • priority: P1 (most urgent) to P4
  • opened: timestamp YYYY-MM-DD HH:MM
  • closed: timestamp in the same format, empty while the ticket is still open

Task: line 1 holds four integer SLA limits in hours for P1, P2, P3, P4 (separated by spaces). Line 2 is an opening month YYYY-MM or ALL. Resolution time is closed - opened in hours (use datetime). A ticket is breached if its resolution time is strictly greater than the limit. For P1, P2, P3, P4 in this order print P1: closed=C open=O median=Mh breached=B where M is the median resolution time of closed tickets (statistics.median, 1 decimal) or n/a if none closed; or P1: no tickets when the priority has no tickets in the selection.

Files your program can read

support.csv221 lines
ticket_id,priority,opened,closed
S9001,P3,2025-06-03 12:39,2025-06-05 00:12
S9002,P1,2025-05-15 15:54,2025-05-15 22:41
S9003,P3,2025-05-18 15:27,2025-05-19 20:21
S9004,P2,2025-05-05 18:44,2025-05-06 04:05
S9005,P3,2025-06-24 16:04,2025-06-28 00:24
S9006,P3,2025-05-26 10:54,2025-05-29 17:02
S9007,P3,2025-06-26 08:57,2025-06-29 03:05
...

Open them by name, for example open("support.csv").

Sample test

Input

4 24 72 120
ALL

Expected output

P1: closed=33 open=2 median=3.5h breached=14
P2: closed=57 open=9 median=13.2h breached=9
P3: closed=79 open=5 median=49.0h breached=18
P4: closed=32 open=3 median=97.1h breached=10

Your answer is also checked against 5 hidden tests.

Need a hint?

Subtracting two datetime objects gives a timedelta; total_seconds() / 3600 turns it into hours.