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What is redemption

Any btUSD holder could always redeem X btUSD against the system for $X worth of collateral. However, there would be a dynamic fee applied to redemption, corresponding to the redemption size related to the total supply of btUSD: the more btUSD redeemed, the more fee charged.

For example, if the current redemption fee is 1% and the collateral price is $50000, then you redeem 1000 btUSD, you would get 0.0198 collateral (0.02 minus a redemption fee of 0.0002).

All active Troves in the system will take a "socialized" share when redemption happens. This means each Trove would expect a decrease in BOTH collateral and debt but with a higher(healthier) ICR after redemption.

This mechanism of redemption would bring less gas cost and encourage trove usage for better capital efficiency.

Note if the system TCR is lower than MCR, then redemption is not allowed until TCR return to a higher level.

Why redemption is needed?

Redemption enforce a low pricing bound for btUSD, i.e., when btUSD is traded below $1 peg, redemption incentivizes arbitrageur to help fixing the peg.

For example, if btUSD is traded at $0.98 in the market, then arbitrageur could buy 100 btUSD from market using $98 then redeem it to receive $99 worth of collateral (considering 1% redemption fee), thus resulting a $1 net profit for arbitrageur. Since less btUSD exist in the market after redemption, the price peg is expected to restore efficiently.

As a Trove owner, do I expect a loss if redemption happens?

If redemption happens, troves do NOT suffer any loss in face value since the redemption is executed at exactly the reported collateral price. However, a trove will lose part of its original collateral exposure.

As a Trove owner, what will happen if my trove's debt is reduced by redemption to the level below minimum requirement?

When a trove's debt is below minimum requirement (200 btUSD), anyone could close the trove by repaying any remaining debt and take a "scavenge" reward from the trove's collateral worth up to 10% of the minimum debt, i.e., 20 btUSD.

Math for redemption

Following are some math proof provided for interested users. Feel free to skip this section.

M0: Redemption fee calculation and decaying

The redemption fee is given by the formula \((baseRate + 0.5\%) * CollateralWorthOfRedeemed\)

The baseRate will increase by each redemption and decay to 0 over time at a half-life rate of 12 hours. Specifically, upon each redemption:

  • baseRate is decayed based on how many minutes are passed since the last fee event (either borrowing or redemption)
  • baseRate is incremented by an amount proportional to the fraction of the total btUSD supply that was redeemed
  • \(baseRateNew = baseRateOld + \frac{btUSDRedeemed}{2 * btUSDTotalSupply}\)

Where baseRateOld is the value just prior to this redemption, and baseRateNew is the new value (and gets applied to this redemption).

M1: ICR will increase after redemption

Suppose a trove (not liquidatable) with original collateral \(C\) and debt \(D\) when the collateral price is \(p\), now a redemption happens and bring socialized reduction to trove's collateral and debt: \(\Delta{c}\) and \(\Delta{d}\) respectively. Then we could compare the trove's ICR before and after redemption:

  • Before redemption, \(ICR_{before} = \frac{C * p}{D} \gt 110\%\)
  • After redemption, \(ICR_{after} = \frac{(C - \Delta{c}) * p}{D - \Delta{d}} = \frac{(C - \Delta{c}) * p}{D - \Delta{c} * p}\)
  • The trove's ICR would increase: \(ICR_{after} - ICR_{before} = \frac{(C - \Delta{c}) * p}{D - \Delta{c} * p} - \frac{C * p}{D} = \frac{\Delta{c}*p*(C*p - D)}{(D - \Delta{c} * p)*D} \gt 0\)

M2: ICR relation between existing Troves will remain the same

Suppose two existing troves (not liquidatable) with same original collateral \(C\) but different debts \(D_{1} < D_{2}\) when the collateral price is \(p\), now a redemption happens and bring socialized reduction to both two trove's collaterals and debts: \(\Delta{c}\) and \(\Delta{d}\) respectively. Then we could compare these two trove's ICRs after redemption:

  • Before redemption, \(ICR_{1Before} = \frac{C * p}{D_{1}} \gt ICR_{2Before} = \frac{C * p}{D_{2}} \gt 110\%\)
  • After redemption, \(ICR_{1After} = \frac{(C - \Delta{c}) * p}{D_{1} - \Delta{d}}\)
  • After redemption, \(ICR_{2After} = \frac{(C - \Delta{c}) * p}{D_{2} - \Delta{d}}\)
  • The two trove's ICRs ratio would be: \(\frac{ICR_{1After}}{ICR_{2After}} = \frac{D_{2} - \Delta{d}}{D_{1} - \Delta{d}} \gt \frac{D_{2}}{D_{1}}\)

M3: ICR relation between new and old Troves will remain the same

Suppose an existing trove A (not liquidatable) with original collateral \(C\), debt \(D\) and stake \(S_{A} = C * \frac{S_{snapshot}}{C_{snapshot}}\) when the collateral price is \(p\), now a redemption of worth-collateral size \(c_{R1}\) happens and bring socialized reduction to the trove's collateral and debt: \(\Delta{c_{A1}}\) and \(\Delta{d_{A1}}\) respectively.

  • After this redemption, total collateral snapshot of the system decrease: \(C_{snapshot} = C_{snapshot} - c_{R1}\)
  • After this redemption, trove A's ICR: \(ICR_{A} = \frac{(C - \Delta{c_{A1}})* p}{D - \Delta{d_{A1}}}\)

Now another trove B is created with same collateral \(C\) and debt \(D\) but its stake is calcualted as \(S_{B} = C * \frac{S_{snapshot}}{C_{snapshot} - c_{R1}} \gt S_{A}\), meaning it will get more share reduction for future redemption than trove A. At this moment, trove B's \(ICR_{B} = \frac{C*p}{D}\) is less than \(ICR_{A}\)

Some time passes. Then another redemption of worth-collateral size \(c_{R2}\) happens and brings socialized reduction to trove A's collateral and debt: \(\Delta{c_{A2}}\) and \(\Delta{d_{A2}}\) respectively. Note \(c_{R2} \lt (C_{snapshot} - c_{R1})\) since you can't redeem all collateral in the system.

Considering how redemption share is calculated with respect to trove' stake, we could get trove B's reduction share for this redemption: \(\Delta{c_{A2}} * \frac{S_{B}}{S_{A}}\) and \(\Delta{d_{A2}} * \frac{S_{B}}{S_{A}}\) respectively. So now we could compare both troves' ICR:

  • trove A's \(ICR_{A2} = \frac{(C - \Delta{c_{A1}} - \Delta{c_{A2}})* p}{D - \Delta{d_{A1}} - \Delta{d_{A2}}} = \frac{(C - \Delta{c_{A2}})*p - \Delta_{c_{A1}}*p}{D - \Delta_{c_{A2}}*p- \Delta_{c_{A1}}*p}\)
  • trove B's \(ICR_{B2} = \frac{(C - \Delta{c_{A2}} * \frac{S_{B}}{S_{A}}) * p}{D - \Delta{d_{A2}} * \frac{S_{B}}{S_{A}}}\), we could note \(\frac{S_{B}}{S_{A}}\) as \(K \gt 1\)
  • With a bit rearrange, trove B's ICR could be written as \(ICR_{B2} = \frac{(C - \Delta{c_{A2}})*p - (K - 1) * \Delta_{c_{A2}}*p}{D - \Delta_{c_{A2}}*p - (K - 1) * \Delta_{c_{A2}}*p}\)
  • By observing the formula of trove A's and trove B's ICR, the problem now comes down to a question: which is larger? \(\Delta_{c_{A1}}\) or \((K - 1) * \Delta_{c_{A2}}\)
  • \(\frac{(K - 1)*\Delta_{c_{A2}}}{\Delta_{c_{A1}}} = (\frac{C_{snapshot}}{C_{snapshot} - c_{R1}}-\frac{C_{snapshot - c_{R1}}}{C_{snapshot - c_{R1}}})*\frac{\Delta{c_{A2}}}{\Delta_{c_{A1}}}=\frac{c_{R1}}{C_{snapshot} - c_{R1}}*\frac{\Delta{c_{A2}}}{\Delta_{c_{A1}}} \le \frac{c_{R1}}{C_{snapshot} - c_{R1}}*\frac{c_{R2}}{c_{R1}} \lt 1\)
  • Applying similar proof in above M1 section, we could know that \(ICR_{A2} > ICR_{B2}\) still holds