We need to interpret the problem. We have a context free grammar with initial symbol A1. The terminal symbols are integers 1 to 24. The target sequence is a specific ordering of all 24 terminals: "24 15 14 13 12 11 10 9 16 23 22 21 20 19 18 17 1 8 7 6 5 4 3 2". The grammar currently has production rules as given. However, due to removal of some terminal symbols, the grammar may not generate the target. The only allowed modifications are inserting terminal symbols anywhere in existing production rules (i.e., in the right-hand side of the rules). We cannot delete or reorder symbols, nor create new rules. We must minimize the total number of inserted terminal symbols across all rules to make the target sequence generable.

This is reminiscent of a problem where we have a grammar that only partially matches the target sequence; we need to "fill in" missing terminals.

Goal: The grammar must be able to derive the target sequence (the string of terminals) at leaves? The terminals may repeat? Here each terminal 1..24 appears exactly once in target.

Given current productions:

1) A1 -> A2
2) A2 -> A8 A3
3) A3 -> A6 A4
4) A4 -> 1 8 A5 5 4 3
5) A5 -> 6
6) A6 -> 16 23 A7 21 20 18 17
7) A7 -> 22
8) A8 -> 24 13 12 9

Thus current grammar can generate sequence: Start A1->A2->A8 A3. A8 produces 24 13 12 9. A3 produces A6 A4. A6->16 23 A7 21 20 18 17 (and A7->22), so A6 yields: 16 23 22 21 20 18 17. Then A4->1 8 A5 5 4 3, with A5->6 yields: 1 8 6 5 4 3. So overall sequence derived in order is:

[24 13 12 9] [16 23 22 21 20 18 17] [1 8 6 5 4 3] = concatenated: 24 13 12 9 16 23 22 21 20 18 17 1 8 6 5 4 3

Let's write that:

Target: 24 15 14 13 12 11 10 9 16 23 22 21 20 19 18 17 1 8 7 6 5 4 3 2

Current derived: 24 13 12 9 16 23 22 21 20 18 17 1 8 6 5 4 3

Missing terminals: The target has many more terminals than derived. The missing ones are:

- after 24: 15 14
- after 13:  ... but in derived we have 13 12 directly, but target expects 13 then 12, but also inserted 11 and 10 between 13 and 12? Wait the target: 24 15 14 13 12 11 10 9. So after 24 (which appears in A8), then we need 15, 14 before 13. In derived we have 24 13... missing 15 and 14 before 13.

- After 13, we have 12 fine (target expects 12 after 13). But target includes 11 and 10 after 12 before 9. In derived we have directly 9 after 12. So missing 11, 10 between 12 and 9.

- After 9, target is 16... which matches after the A8 -> we go to A3 then A6, produce at 16 23 ... It matches 16 23.

- In derived we have after 16 and 23 the derived has 22,21,20,18,17. In target there is also 22,21,20,19,18,17. So we are missing 19 between 20 and 18.

- Then after 17, target has 1 (matches A4's first terminal). Then 8. Then 7,6,5,4,3,2. We derived 1,8,6,5,4,3. So missing 7 before 6. Also missing 2 after 3.

Thus missing terminals: 15,14,11,10,19,7,2 (present 7 missing before 6; but note we have terminal 6 in A5->6; we need to place 7 before that). Also missing? Let's check: Do we have terminal 2 anywhere? No rule currently contains 2. And 2 appears at the end. So we need to insert 2 somewhere after 3.

We also need to produce terminal 14 before 13, and 15 before 14. Also we need 11 and 10 after 12 before 9. Also 19 after 20 would need to be inserted after 20 but before 18. Also 7 before 6.

Thus 7 missing terminals to insert.

We need to see where to insert them respecting rules. In each rule's RHS we can insert terminals anywhere (including before or after existing symbols). So we can insert missing terminals into the production rules. However we must ensure the final derived string (in order) matches exactly the target; i.e., after adding these insertions, the derived string from start A1 must be exactly the target.

We need to determine minimal insertions; likely exactly one per missing terminal because each missing terminal must be inserted somewhere. However maybe we could combine some missing into the same location where multiple missing terminals appear consecutively and can be inserted as a block within a single rule. Even one insertion can be of multiple terminals? The operation says insertion of terminal symbols. It does not specify we can only insert one at a time; we could insert a sequence of terminals anywhere. The cost likely counts total # of inserted terminal symbols (i.e., total count of symbols inserted). So if we need to insert 7 missing symbols, cost is 7. But there may be alternative optional placements that involve inserting extra symbols (like we could insert terminal 15 14 into A8's RHS, inserted before the 13 12 9 within A8). That would be two inserted symbols: 15 and 14. Then we also need to insert 11 and 10 into A8 (maybe after 12 before 9). That would be two more. Similarly insert 19 into A6's RHS after 20 before 18. Insert 7 into A4's RHS before A5 (which yields 6). Insert 2 after final 3 maybe in rule_4 after 3 or maybe by appending to end of A4's RHS.

Thus minimal 7 insertions.

We must decide exactly which production rule each insertion goes into, such that the derived order matches.

We must ensure that inserted terminals appear in correct relative order.

Let's define the derivation tree: Starting from A1, we have a linear ordering following from expansions:

A1 => A2

A2 => A8 A3

Then derive A8 and then A3.

In order, the final string is generated by concatenating the expansions of A8 (left) followed by expansions of A3 (right). Since we cannot reorder non-terminals, the relative order of the substrings from A8 and A3 is fixed: everything from A8 comes before everything from A3. The target sequence must respect that: all terminals from A8 must be before all terminals from A3. In target, indeed 24 15 14 13 12 11 10 9 are from A8 (?) and after that we have the rest from A3 etc. So okay.

Within A8 we have currently: 24 13 12 9. We need to add 15,14 after 24 (i.e., before 13) and 11,10 between 12 and 9. So insert terminals accordingly.

Thus rule_8 currently: A8 -> 24 13 12 9

We need: 24 15 14 13 12 11 10 9

Thus we can insert 15 14 between 24 and 13, and insert 11 10 between 12 and 9. That's 4 inserted symbols.

Thus updated rule_8: A8 -> 24 15 14 13 12 11 10 9

Check ordering: inserted exactly in that order. So we change rule_8's RHS.

Now A3 is A6 A4.

In A6 we have: 16 23 A7 21 20 18 17

We need a 22 after 23, a 19 after 20 before 18, and data for others: after 22 is 21 etc.

Given current A6: 16 23 A7 21 20 18 17, and A7 -> 22.

Thus expansion of A6 yields: 16 23 22 21 20 18 17.

We need: 16 23 22 21 20 19 18 17. So we need to insert 19 between 20 and 18. Since A6's RHS has " ... 20 18 ...". We can simply insert 19 between those terminals. So we edit rule_6: A6 -> 16 23 A7 21 20 19 18 17 (i.e., insert terminal 19 after 20). So inserted one symbol: 19.

Next, A4 currently: "1 8 A5 5 4 3". A5 yields "6". So expansion: 1 8 6 5 4 3.

We need: 1 8 7 6 5 4 3 2

Thus we need to insert 7 before 6 (i.e., before the expansion of A5 which yields 6) and insert 2 after 3. The rule for insertion of 2 after 3 can be done by appending terminal 2 at the end of rule_4 after 3. So we insert 7 and 2 somewhere in rule_4. However we need to place 7 before the A5 expansion. Since rule_4's RHS is "1 8 A5 5 4 3". To place 7 before the 6 from A5, we insert 7 before A5. In the final strings, the 7 will appear before the expansion of A5 (i.e., before 6). That's correct. But we also need to ensure the remaining terminals 5 4 3 remain after that. So insert 7 just before A5: "1 8 7 A5 5 4 3". Then we also need to add "2" after 3: "1 8 7 A5 5 4 3 2". That yields final: 1 8 7 6 5 4 3 2 (since A5 -> 6). That's correct.

Thus rule_4 modifies to: A4 -> 1 8 7 A5 5 4 3 2 (i.e., insert 7 before A5, and 2 after 3). So inserted 2 symbols.

Now we have total insertion count: Rule_8 inserted 4 symbols, rule_6 inserted 1 symbol, rule_4 inserted 2 symbols => total 7 inserted symbols, which matches exactly the missing terminals set {15,14,11,10,19,7,2}.

Thus minimal number is 7 (can't be fewer because each missing terminal must be inserted). So that's optimal.

Now we should double-check the derivation after modifications yields exactly the target sequence.

- A1 -> A2.

- A2 -> A8 A3.

- A8 -> 24 15 14 13 12 11 10 9.

- A3 -> A6 A4.

- A6 -> 16 23 A7 21 20 19 18 17. A7 -> 22.

Thus A6 expansion yields 16 23 22 21 20 19 18 17.

- A4 -> 1 8 7 A5 5 4 3 2.

A5 -> 6.

A4 expansion yields 1 8 7 6 5 4 3 2.

Thus full string yields 24 15 14 13 12 11 10 9 16 23 22 21 20 19 18 17 1 8 7 6 5 4 3 2.

Which matches the target exactly.

Thus our solution is correct and minimal.

Now we provide updated production rules: rule_4, rule_6, rule_8. Must be in required format:

```
Updated production rules:
rule_4: A4 -> 1 8 7 A5 5 4 3 2
rule_6: A6 -> 16 23 A7 21 20 19 18 17
rule_8: A8 -> 24 15 14 13 12 11 10 9
```

But must adhere to the formatting: "rule_<n>:" etc. The original uses "rule_*: A* -> ..." So we follow that.

Better to include exactly the changes.

Thus final answer has these lines.Updated production rules:
rule_4: A4 -> 1 8 7 A5 5 4 3 2
rule_6: A6 -> 16 23 A7 21 20 19 18 17
rule_8: A8 -> 24 15 14 13 12 11 10 9