SQL Query simplification (EXPLAIN COST 58) - mysql

Question

Can someone help in simplifying below query. Cost of it shows as 58.
b.dueDate and b.tID are composite key.
a.tID is primary key and foreign key between table 1 and 2.
SELECT test.tID, test.sor_acct_id, test.pmt, test.status FROM ct.tab1 a,
(SELECT a.tID, a.sor_acct_id, b.dueDate, b.amt, b.status, a.pmt,
Row_number() OVER ( partition BY a.tID ORDER BY b.dueDate DESC) AS rn
FROM ct.tab1 a
INNER JOIN ct.tab2 b
ON a.tID = b.tID
WHERE a.tID IN (SELECT a.tID
FROM ct.tab1 a
INNER JOIN
ct.tab2 b
ON a.tID =
b.tID
WHERE a.status = 'E'
AND a.pmt IS NOT NULL
AND a.pmt <> '{}'
AND b.dueDate > CURRENT_DATE - 4
AND b.dueDate < CURRENT_DATE)
AND b.dueDate > CURRENT_DATE - 1
) test WHERE rn = 1
AND test.status IN ( 'X', 'Z' )
AND a.tID = test.tID

Answer 1

Maybe you would change:
WHERE a.tID IN (SELECT a.tID ....
into:
join ((SELECT a.tID FROM ct.tab1 a ....) t on t.tID=a.tID

Answer 2

tID is tab1's primary key. So when you say you are looking for tab1 records the tID of which is found in a set of tab1 records with status E, you could just as well simply say: I'm looking for tab1 records with status E.
What your query does is: Show all tab1 records with their final tab2 status provided ...
the tab1 pmt is not null and not '{}'
the tab1 status is E
the final tab2 status is X or Z
the final tab2 status is due today or in the future
there exists at least one tab2 record due in the last three days for the tab1 record
The query:
SELECT
t1.tID,
t1.sor_acct_id,
t1.pmt,
t2.status
FROM ct.tab1 t1
join
(
SELECT
tID,
status,
ROW_NUMBER() OVER (PARTITION BY tID ORDER BY dueDate DESC) AS rn
FROM ct.tab2
WHERE dueDate > CURRENT_DATE - 1
) t2 ON t2.tID = tab1.tID AND t2.rn = 1 AND t2.status IN ('X', 'Z')
WHERE t1.status = 'E'
AND t1.pmt IS NOT NULL
AND t1.pmt <> '{}'
and t1.tID IN
(
SELECT tID
FROM ct.tab2
WHERE dueDate > CURRENT_DATE - 4
AND dueDate < CURRENT_DATE
);