Capital Allocation Across Development Programs

From Program Value to a Budgeted Portfolio

A development portfolio is not simply a ranked list of programs. A program with high expected value may require so much capital that it prevents funding several other attractive programs. Capital allocation therefore asks a joint question: which feasible combination of programs produces the greatest expected incremental value within the available budget?

The analysis below treats each program as an indivisible choice at a common decision date. It uses expected value as the objective, as in the earlier material on expected value of continuing a program and risk-adjusted NPV with technical success. The focus here is how to allocate a fixed budget across multiple candidate programs rather than whether one program is worthwhile in isolation.

Definition: For program $i$, let $x_i$ equal 1 if the program is funded and 0 otherwise; let $C_i$ be the capital required at the decision date; let $p_i$ be its probability of technical success; and let $R_i^S$ and $R_i^F$ be the values, at that date, conditional on success and failure, respectively, excluding $C_i$. The program’s incremental expected value is $$ e_i = p_iR_i^S + (1-p_i)R_i^F - C_i. $$ If the total available budget is $B$, the basic capital-allocation problem is to maximize $\sum_i x_i e_i$, subject to $\sum_i x_i C_i \leq B$ and $x_i\in\{0,1\}$.

All values must be measured in the same currency and at the same decision date. The conditional values $R_i^S$ and $R_i^F$ should represent the value of each outcome after any relevant future costs have been accounted for. The cost $C_i$ is the capital committed by the current allocation decision. This separation makes it possible to distinguish the outcome value from the funding required now.

Why the Objective Is Additive

Suppose the total portfolio value is the sum of the values of its funded programs, with no additional portfolio-level payoff or interaction. Let $Y_i$ denote the uncertain outcome value for program $i$, excluding its allocated cost. The portfolio’s incremental value is then $\sum_i x_i(Y_i-C_i)$. By linearity of expectation:

E\!\left[\sum_i x_i(Y_i-C_i)\right] = \sum_i x_i\bigl(E[Y_i]-C_i\bigr) = \sum_i x_i e_i.

This equality does not require program outcomes to be independent; it follows from additivity of the portfolio value and linearity of expectation. However, the basic objective does not describe the distribution of total portfolio outcomes. If a decision-maker wants to penalize portfolio downside, impose a probability-of-success target, or account for interactions among programs, maximizing expected value alone is not enough. Those are different decision objectives.

The formulation also assumes that each program is either funded or not funded, the budget is a hard constraint, and the cost and outcome values are appropriate for the same decision. In particular, the expected-value objective reflects a decision-maker who is willing to compare uncertain alternatives by their expected monetary values. An organization with a binding risk limit, strategic priorities, or nonfinancial objectives may need additional constraints or a different objective.

Worked Example: Selecting Programs Under a Fixed Budget

A hypothetical sponsor has a budget of $20 million and is choosing among four indivisible programs. Each success value is net of future costs but excludes the current funding cost. Failure value is zero. All figures below are hypothetical and in millions of dollars.

Program	Cost $C_i$	Success probability $p_i$	Success value $R_i^S$
A	12	0.35	50
B	8	0.55	26
C	6	0.25	40
D	5	0.70	14

For each program, use $e_i=p_iR_i^S+(1-p_i)R_i^F-C_i$. Since failure value is zero:

Program A: $e_A=(0.35)(50)+(0.65)(0)-12=17.5-12=\$5.5$ million.
Program B: $e_B=(0.55)(26)+(0.45)(0)-8=14.3-8=\$6.3$ million.
Program C: $e_C=(0.25)(40)+(0.75)(0)-6=10-6=\$4.0$ million.
Program D: $e_D=(0.70)(14)+(0.30)(0)-5=9.8-5=\$4.8$ million.

All four programs have positive incremental expected value, but their combined cost is $12+8+6+5=\$31$ million, which exceeds the budget. Compare feasible combinations. If A is selected, the remaining budget is $8 million: the best addition is B, producing a total expected value of $5.5+6.3=\$11.8$ million. If A is not selected, B, C, and D together cost $8+6+5=\$19$ million and have expected value $6.3+4.0+4.8=\$15.1$ million. That value exceeds the best feasible combination without A that omits at least one of B, C, or D: the pair B and D is worth $11.1 million, B and C $10.3 million, and C and D $8.8 million. Thus funding B, C, and D is optimal under the stated objective and budget, leaving $1 million unspent.

Why a Value-per-Dollar Ranking Can Fail

A tempting shortcut is to divide each program’s incremental expected value by its cost and fund programs in descending order of that ratio. This can be a useful screening heuristic, but it does not generally solve the problem when programs are indivisible. A high ratio may lead to unused budget that could have funded a more valuable combination of lower-ratio programs.

Worked Example: A Ratio Ranking Misses the Best Combination

Consider three hypothetical programs and a budget of $10 million. Their costs and incremental expected values are: A, cost $6 million and value $12 million; B, cost $5 million and value $9 million; and C, cost $5 million and value $9 million. Their value-per-dollar ratios are $12/6=2.00$ for A and $9/5=1.80$ for both B and C.

The ratio ranking selects A first. The remaining budget is $10-6=\$4$ million, so neither B nor C can be funded. The resulting expected value is $12 million. But selecting B and C costs $5+5=\$10$ million and gives total expected value $9+9=\$18$ million. The ratio-based choice is therefore not optimal. The correct method compares feasible combinations, or solves the corresponding binary optimization problem, rather than treating a ratio ranking as a proof of optimality.

The ratio can be exact in special cases, such as a divisible-resource allocation under suitable assumptions. That is not the situation in these examples: a program cannot be partly funded to obtain a proportional share of its expected value. For discrete program selection, explicit combination comparison or an integer optimization method is the appropriate general approach.

Probability Assumptions and Decision Boundaries

Expected value depends on the probability and outcome values used. The calculation is only as relevant as those inputs are to the actual decision: probabilities should describe technical success under the specified development plan, and payoffs should reflect the consequences of the corresponding outcomes. A probability estimate from a different population, endpoint, or development strategy may not be a suitable input without adjustment.

Worked Example: Finding the Probability at Which the Allocation Changes

Return to the $20 million budget and the hypothetical programs in the first example. Hold all inputs fixed except program C’s success probability $p_C$. Its expected value becomes $e_C=40p_C-6$, in millions of dollars. One leading alternative is B and D, which costs $13 million and has expected value $6.3+4.8=\$11.1$ million. Funding B, C, and D costs $19 million and has expected value $6.3+e_C+4.8=11.1+e_C$ million.

The B-C-D portfolio is preferable to B-D when $11.1+e_C>11.1$, or $e_C>0$. But the most valuable competing portfolio in the original example is A-B, worth $11.8 million. The B-C-D portfolio exceeds that alternative when

11.1+e_C > 11.8 \quad\Longleftrightarrow\quad 40p_C-6 > 0.7 \quad\Longleftrightarrow\quad p_C > \frac{6.7}{40}=0.1675.

Thus, with the other inputs unchanged, B-C-D exceeds A-B when $p_C$ is above 0.1675. At the original value $p_C=0.25$, $e_C=40(0.25)-6=\$4.0$ million, so B-C-D is worth $11.1+4.0=\$15.1$ million. This boundary is a comparison of the two stated alternatives; it does not establish that an estimated probability is known without uncertainty.

Checking the Allocation and Its Limits

Before treating an optimized allocation as a decision recommendation, check the inputs and the model structure. In particular, confirm that costs and values use a common currency and valuation date; verify which costs are included in each conditional payoff; and ensure that the probability refers to the same program definition used to estimate its outcome values. A mismatch between the funded scope and the probability or payoff assumptions can make a precise calculation irrelevant.

Also distinguish money already spent from money that the current decision can still allocate. A past, unrecoverable expenditure does not normally belong in the comparison of future choices: include the cost that changes depending on whether the program is funded now. Conversely, omitting a prospective cost that is necessary to achieve the modeled outcome will overstate incremental value. These checks help prevent double-counting costs or treating sunk costs as if they could be recovered through a new allocation.

Set a common basis.
Express costs and outcome values at the same decision date and in consistent monetary units.

Calculate each program’s incremental expected value.
Apply the conditional outcome values and success probability, then subtract the capital cost relevant to the funding decision.

Enforce the constraint.
Compare only combinations whose total funding cost does not exceed the fixed budget.

Test the result.
Check competing combinations and examine whether plausible changes in important inputs alter the preferred allocation.

There are several reasons the basic answer may not be the organization’s final choice. Expected monetary value does not express the chance or severity of a low-value outcome. A hard budget may be an approximation if capital can be raised or carried forward. Costs or payoffs may interact across programs, and strategic or scientific considerations may not be captured in monetary values. In those situations, the analyst should state which assumptions fail and what additional objective or constraint is needed, rather than presenting the expected-value optimum as universally best.

Other approaches answer different questions. Ranking by standalone expected value ignores whether the resulting set fits the budget. Ranking by value per dollar is a heuristic for discrete programs, not a general optimum. Equal allocation ignores differences in cost and expected value. Maximizing the number of funded programs is appropriate only if program count is itself the objective. A constrained expected-value comparison is the appropriate starting point when the decision objective is additive expected value and the budget is fixed.

Key takeaway: Calculate each program’s incremental expected value, then choose the feasible combination with the greatest total expected value. A program’s standalone attractiveness—or its expected value per dollar—does not by itself determine the best portfolio.

Check Your Understanding

Answer each question using the stated objective and assumptions. Solutions follow the questions.

A program costs $4 million, has success probability 0.30, success value $20 million, and failure value zero. What is its incremental expected value?
Two indivisible programs each cost $6 million and each has incremental expected value $8 million. With a $10 million budget, is it feasible to fund both? What is the best expected value if there are no other programs?
Why does maximizing the sum of expected program values not require independence of outcomes when total portfolio value is additive?
In the first worked example, what is the expected value of A and B together, and why is that combination not the optimal allocation under a $20 million budget?

Solutions. 1. The incremental expected value is $(0.30)(20)+(0.70)(0)-4=6-4=\$2$ million. 2. Both programs would cost $6+6=\$12$ million, which exceeds the $10 million budget. Funding either one gives the best feasible expected value, $8 million. 3. Linearity of expectation gives $E[Y_1+Y_2]=E[Y_1]+E[Y_2]$ whether or not the outcomes are independent, provided the total value is their sum. 4. A and B cost $12+8=\$20$ million and have expected value $5.5+6.3=\$11.8$ million. B, C, and D cost $19 million and have expected value $6.3+4.0+4.8=\$15.1$ million, so A and B are feasible but not optimal.