A systemd host has no atd installed and no package installation is permitted. You must run /usr/local/bin/report.sh exactly once, fifteen minutes from now, without writing any unit files. Which command does it?

LPIC-1 Exam 102-500, objective 107. Administrative tasks hard

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The options

Correct systemd-run --on-active=15m /usr/local/bin/report.sh

Correct. --on-active creates a transient timer unit plus a transient service unit, with the timer set to elapse the given interval after it is started.

Not correct systemd-run --on-calendar=15m /usr/local/bin/report.sh

Wrong. --on-calendar expects a calendar event expression such as `*-*-* 02:00:00` or `Mon *-*-* 09:00`, not a relative delay, so this is rejected as an invalid specification.

Not correct systemctl start --timer=15m /usr/local/bin/report.sh

Wrong. systemctl start activates an existing unit by name; it has no --timer option and takes no executable paths.

Not correct echo /usr/local/bin/report.sh | at now + 15 minutes

Wrong under the stated constraints. This is the correct classic answer, but at is a client for the atd daemon, and the scenario says atd is not present, so the job would never be executed.

Why

systemd-run runs a command inside a transient unit created on the fly. With no timer option it starts immediately; with --on-active=, --on-boot=, --on-startup=, --on-unit-active= or --on-calendar= it instead creates a transient .timer that activates the transient service. Monotonic options take a time span such as 15m or 90s, while --on-calendar takes systemd calendar syntax, and confusing the two is the usual mistake.

Where this comes from

Cited
manual page systemd-run(1)

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