Oracle SQL remove Sunday and consider only 6 working days

I tried this SQL CASE to get number of days left for Over KPI(OKPI).

There are 2 main conditions which are confusing me,

1- Remove or don't consider Sunday as a working day.

2- Actual bkg_date is after 1 day. So for example if bkg_date = '08 Jul 2023' then bkg_date will be = '10 Jul 2023'

CASE
    WHEN SYSDATE - (x.bkg_date + 1) <= x.dlvry_kpi THEN
        CASE
            WHEN x.dlvry_kpi - (
                -- Full weeks from Monday of start week to Monday of current week
                (TRUNC(SYSDATE, 'IW') - TRUNC(x.bkg_date + 1, 'IW')) * 6 / 7
                -- Add extra days in the current week excluding Sunday
                + CASE WHEN TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW') + 1 <= 6 THEN TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW') + 1 ELSE 6 END
                -- Subtract days in the week before the start date excluding Sunday
                - CASE WHEN TRUNC(x.bkg_date + 1) - TRUNC(x.bkg_date + 1, 'IW') <= 6 THEN TRUNC(x.bkg_date + 1) - TRUNC(x.bkg_date + 1, 'IW') ELSE 6 END
            ) <= 0 THEN '00 days left'
            ELSE
                TO_CHAR(
                    x.dlvry_kpi - (
                        -- Full weeks from Monday of start week to Monday of current week
                        (TRUNC(SYSDATE, 'IW') - TRUNC(x.bkg_date + 1, 'IW')) * 6 / 7
                        -- Add extra days in the current week excluding Sunday
                        + CASE WHEN TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW') + 1 <= 6 THEN TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW') + 1 ELSE 6 END
                        -- Subtract days in the week before the start date excluding Sunday
                        - CASE WHEN TRUNC(x.bkg_date + 1) - TRUNC(x.bkg_date + 1, 'IW') <= 6 THEN TRUNC(x.bkg_date + 1) - TRUNC(x.bkg_date + 1, 'IW') ELSE 6 END
                    )
                ) || ' days left'
        END
    ELSE 'OKPI'
END AS AGING

So, this is what I've come up with and gives the result OKPI for following inputs;

x.bkg_date = '08 Jul 2023'

sysdate = '17 Jul 2023'

x.dlvry_kpi = 7

You can find the next business day (skipping Sundays) using:

bkg_date + CASE TRUNC(bkg_date) - TRUNC(bkg_date, 'IW') WHEN 5 THEN 2 ELSE 1 END

For the sample data:

CREATE TABLE table_name (bkg_date, dlvry_kpi) AS
  SELECT TRUNC(SYSDATE, 'IW') + 8 - LEVEL, 1 FROM DUAL CONNECT BY LEVEL <= 15;

Then:

SELECT bkg_date,
       next_bkg_date,
       dlvry_kpi,
       TRUNC(SYSDATE) AS today,
       (TRUNC(next_bkg_date, 'IW') - TRUNC(SYSDATE, 'IW')) * 6 / 7 AS full_weeks,
       LEAST(TRUNC(next_bkg_date) - TRUNC(next_bkg_date, 'IW'), 5) AS days_at_end,
       LEAST(TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW'), 5) AS day_at_start,
       -- Full weeks from Monday of current week to Monday of end week
       (TRUNC(next_bkg_date, 'IW') - TRUNC(SYSDATE, 'IW')) * 6 / 7
       -- Add extra days in the end week excluding Sunday
       + LEAST(TRUNC(next_bkg_date) - TRUNC(next_bkg_date, 'IW'), 5)
       -- Subtract days in the current week before the today excluding Sunday
       - LEAST(TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW'), 5) AS days
FROM   (
  SELECT bkg_date,
         bkg_date + CASE TRUNC(bkg_date) - TRUNC(bkg_date, 'IW') WHEN 5 THEN 2 ELSE 1 END as next_bkg_date,
         dlvry_kpi
  FROM   table_name
)

Outputs:

And to get the number of days left then:

SELECT bkg_date,
       dlvry_kpi,
       days,
       CASE
       WHEN days < -dlvry_kpi
       THEN 'OKPI'
       ELSE TO_CHAR(dlvry_kpi + days, 'FM90') || ' days left'
       END AS aging
FROM   (
  SELECT bkg_date,
         dlvry_kpi,
         -- Full weeks from Monday of current week to Monday of end week
         (TRUNC(next_bkg_date, 'IW') - TRUNC(SYSDATE, 'IW')) * 6 / 7
         -- Add extra days in the end week excluding Sunday
         + LEAST(TRUNC(next_bkg_date) - TRUNC(next_bkg_date, 'IW'), 5)
         -- Subtract days in the current week before the today excluding Sunday
         - LEAST(TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW'), 5) AS days
  FROM   (
    SELECT bkg_date,
           bkg_date + CASE TRUNC(bkg_date) - TRUNC(bkg_date, 'IW') WHEN 5 THEN 2 ELSE 1 END as next_bkg_date,
           dlvry_kpi
    FROM   table_name
  )
)

Which outputs:

fiddle