IRM

[MerlinEgalite]

For the naming we could write pidIrm or something like that?

[peyha]

Can you break the IRM by borrowing a low amount of token (1 wei for instance) many times in the same block ?

Feel like it is possible because in this case the borrow rate will be $\text{borrow rate last block}*\text{jump multiplier}^N$ which probably grows pretty quickly to infinity

[MathisGD]

Can you break the IRM by borrowing a low amount of token (1 wei for instance) many times in a block ?

Feel like it is possible because in this case the borrow rate will be borrow rate last block∗jump multiplierN which probably grows pretty quickly to infinity

It is not possible because the formula is:

uint256 jumpMultiplier = IrmMathLib.wExp(int256(LN_JUMP_FACTOR), errDelta);

And in your example errDelta is near zero. In general going from u1 to u2 in one step or breaking it into smaller steps does not change the overall jump that will be made.

[pakim249CAL]

Can you break the IRM by borrowing a low amount of token (1 wei for instance) many times in a block ?

Feel like it is possible because in this case the borrow rate will be borrow rate last block∗jump multiplierN which probably grows pretty quickly to infinity

This shouldn't work because the IRM can only be called once per block. This may not be clear here as the logic to only call this contract once per block is contained in the Morpho contract and not here.

[MathisGD]

This shouldn't work because the IRM can only be called once per block. This may not be clear here as the logic to only call this contract once per block is contained in the Morpho contract and not here.

This is an other point, but it's not even required to prevent the problem

[peyha]

Can you break the IRM by borrowing a low amount of token (1 wei for instance) many times in a block ? Feel like it is possible because in this case the borrow rate will be borrow rate last block∗jump multiplierN which probably grows pretty quickly to infinity

It is not possible because the formula is:

uint256 jumpMultiplier = IrmMathLib.wExp(int256(LN_JUMP_FACTOR), errDelta);

And in your example errDelta is near zero. In general going from u1 to u2 in one step or breaking it into smaller steps does not change the overall jump that will be made.

errDelta is near zero but still constant ($\Delta(e)=u(t*)-u(t*)=\frac{b+1}{s}-\frac{b}{s}=\frac{1}{s}$) so you still get $r=r_{previous}J^{\frac{N}{s}}$ which diverges (unless I'm missing something in my computation)

Can you break the IRM by borrowing a low amount of token (1 wei for instance) many times in a block ? Feel like it is possible because in this case the borrow rate will be borrow rate last block∗jump multiplierN which probably grows pretty quickly to infinity

This shouldn't work because the IRM can only be called once per block. This may not be clear here as the logic to only call this contract once per block is contained in the Morpho contract and not here.

You are right