Cost-effectiveness of restricted donations

By Vasco Grilo @ 2024-02-03T09:36 (+10)

TLDR. You can use this Sheet to calculate the cost-effectiveness of restricted donations.

Even if a donor restricts a donation to some target projects, part or all of the donation may be allocated to others due to funging. Unrestricted funds initially allocated to the target projects may be directed to others in order to minimise or offset the change in the relative allocation of funds across projects. Here is a simple model:

Before receiving the restricted donation, the organisation has total funds , of which $f_{0} C_{0}$ are allocated to the target projects, and $(1 - f_{0}) C_{0}$ to others, where $f_{0}$ represents the funds initially allocated to the target projects as a fraction of the total^[1].
After receiving the restricted donation, the organisation will have total funds $C_{0} + D$ , of which $f_{0} C_{0} + D - F$ will be allocated to the target projects, and $(1 - f_{0}) C_{0} + F$ to others, where $D$ is the donation size, and $F$ is the amount of funging.

The funging $F_{o f f s e t}$ to offset the effect of the restricted donation on the relative allocation of funds across projects is such that the funds allocated to the target projects as a fraction of the total are maintained, i.e. $\frac{f_{0} C_{0} + D - F_{o f f s e t}}{C_{0} + D} = f_{0} \Leftrightarrow F_{o f f s e t} = (1 - f_{0}) D$ . In reality, the organisation may not want to be maximally adversarial, aiming for a target funging which is only $k$ times as large, $F_{t a r g e t} = k F_{o f f s e t}$ , where the adversarial factor $k$ can range from 0 to 1. If ${f_{u}}_{0}$ is the fraction of the funds initially allocated to the target projects which are unrestricted, the amount of funging is also limited to $F_{m a x} = {f_{u}}_{0} f_{0} C_{0}$ . So the funging will be $F = min (F_{m a x}, F_{t a r g e t})$ .

Denoting the cost-effectiveness of the increase in funds allocated to the target and other projects by $E_{t a r g e t}$ and $E_{o t h e r}$ ^[2], the respective additional benefits will be $B_{t a r g e t} = (D - F) E_{A}$ and $B_{o t h e r} = F E_{B}$ . Consequently, the donation will have cost-effectiveness $E = \frac{B_{t a r g e t} + B_{o t h e r}}{D} = (1 - \frac{F}{D}) E_{t a r g e t} + \frac{F}{D} E_{o t h e r}$ , where the funging as a fraction of the donation size is $\frac{F}{D} = min ({f_{u}}_{0} f_{0} \frac{C_{0}}{D}, k (1 - f_{0}))$ .

Some sanity checks:

If there are no unrestricted funds initially allocated to the target projects, ${f_{u}}_{0} = 0$ , therefore $F = F_{m a x} = 0$ and $E = E_{t a r g e t}$ . In other words, there is no funging, and the cost-effectiveness of the donation equals the cost-effectiveness of the increase in funds allocated to A, which makes sense.
If the unrestricted funds initially allocated to the target projects are larger than the donation size, and the organisation is maximally adversarial, ${f_{u}}_{0} f_{0} C_{0} > D$ and $k = 1$ , therefore $F = F_{t a r g e t} = F_{o f f s e t}$ and $E = f_{0} E_{t a r g e t} + (1 - f_{0}) E_{o t h e r}$ . In other words, the funging offsets the effect of the restricted donation, the relative allocation of funds across projects is maintained, and the cost-effectiveness of the donation equals the cost-effectiveness of an unrestricted donation, which checks out.
If the donation size tends to infinity, $D \to \infty$ , therefore $\frac{F}{D} \to 0$ and $E \to E_{t a r g e t}$ . In other words, funging will tend to be negligible relative to the donation size, and the cost-effectiveness tends to that of the increase in funds allocated to the target projects, as expected.

You can make a copy of this Sheet to use the model.

^{^}
The funds allocated to the target and other projects add up to the total funds, so they should include overhead, i.e. funds not strictly allocated to any project.
^{^}
These are equal to the marginal cost-effectiveness of the target and other projects for a marginal increase of the funds allocated to them.