This research note presents explicit formulas for debt and repayment sizing. It provides a direct method to determine the target Debt Service Coverage Ratio (DSCR) and the corresponding sculpted repayment schedule based on a specific debt size. Conversely, it also offers a formula to calculate the maximum debt size for a given target DSCR. The availability of these explicit formulas streamlines the construction of a robust and light-weight project financial model, eliminating the need for macros, circular references, or goal seek functions.
Keywords: Project Finance, Debt Sizing, Sculpted Repayment, Target DSCR
Notation:
Denote by
- C: Cash Flows Available for Debt Service (CFADS)
- O: Outstanding Bond (Debt)
- g: Guarantee fee rate (if quarterly payment, it's a quarterly rate)
- E: Other bond issuance cost
- D: Debt Service
- P: Principal repayment
- N: Number of (re)payments
- r: Bond interest rate at each payment date (if quarterly payment, it's quarterly rate)
- x: Inverse of target DSCR
- n: Moratorium period of non-repayment
Assumption:
- (1) Other bond issuance cost Et is assumed independent of debt outstanding amount Dt. In practice, the bond issuance cost may depend on the bond outstanding amount; however, we assume the variation due to any change in outstanding amount is negligible.
- (2) Guarantee fee gt=gˉ is constant.
Formulas:
Given a debt amount of O0, the target DSCR is given by:
DSRC=O0∏i=n+1NRi∗+∑i=n+1NEi∗∏j=i+1NRj∗∑i=n+1NCi∗∏j=i+1NRj∗,
where Rt∗:=1−gˉ1+rt;Et∗:=1−gˉEt;Ct∗:=1−gˉCt.
Given a target DSCR, the maximum debt amount can be determined as follow:
O0=∏i=1NRi∗x∑i=1NCi∗∏j=i+1NRj∗−∑i=1NEi∗∏j=i+1NRj∗
where Rt∗:=1−gˉ1+rt;Et∗:=1−gˉEt;Ct∗:=1−gˉCt.
Proof:
The debt service at payment date t is given by
Dt=Ot−1rt+Otgt+Et+Pt
Also, we have
Ot=Ot−1−Pt
Noting that Dt=xCt, we add (1) and (2) to get
xCt=Ot−1(rt+1)−Ot(1−gˉ)+Et
Re-arranging, we obtain
Ot=Ot−11−gˉ1+rt+1−gˉEt−1−gˉCtx
Therefore,
Ot=Ot−1Rt∗+Et∗−Ct∗x.
For t=1,
O1=O0R1∗+E1∗−C1∗x.
Substituting it into O2, we get
O2=O0R1∗R2∗+(E1∗R2∗+E2∗)−(C1∗R2∗+C2∗)x.
Subsituting it into O3, we get
O3=O0R1∗R2∗R3∗+(E1∗R2∗R3∗+E2∗R3∗+E3∗)−(C1∗R2∗R3∗+C2∗R3∗+C3∗)x.
By continuing to substitute progressively, we obtain
ON=O0i=1∏NRi∗+i=1∑NEi∗j=i+1∏NRj∗−xi=1∑NCi∗j=i+1∏NRj∗
Knowing that ON=0, we can derive x as below:
x=∑i=1NCi∗∏j=i+1NRj∗O0∏i=1NRi∗+∑i=1NEi∗∏j=i+1NRj∗.
In case of a moratorium period of n periods, On=On−1=...=O0, and therefore,
x=∑i=n+1NCi∗∏j=i+1NRj∗O0∏i=n+1NRi∗+∑i=n+1NEi∗∏j=i+1NRj∗.
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