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Optimal investment in HIV prevention programs: more is not always better

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Abstract

This paper develops a mathematical/economic framework to address the following question: Given a particular population, a specific HIV prevention program, and a fixed amount of funds that could be invested in the program, how much money should be invested? We consider the impact of investment in a prevention program on the HIV sufficient contact rate (defined via production functions that describe the change in the sufficient contact rate as a function of expenditure on a prevention program), and the impact of changes in the sufficient contact rate on the spread of HIV (via an epidemic model). In general, the cost per HIV infection averted is not constant as the level of investment changes, so the fact that some investment in a program is cost effective does not mean that more investment in the program is cost effective. Our framework provides a formal means for determining how the cost per infection averted changes with the level of expenditure. We can use this information as follows: When the program has decreasing marginal cost per infection averted (which occurs, for example, with a growing epidemic and a prevention program with increasing returns to scale), it is optimal either to spend nothing on the program or to spend the entire budget. When the program has increasing marginal cost per infection averted (which occurs, for example, with a shrinking epidemic and a prevention program with decreasing returns to scale), it may be optimal to spend some but not all of the budget. The amount that should be spent depends on both the rate of disease spread and the production function for the prevention program. We illustrate our ideas with two examples: that of a needle exchange program, and that of a methadone maintenance program.

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Acknowledgment

This work was supported by a grant from the National Institute on Drug Abuse (NIDA), National Institutes of Health (NIH; grant DA-R01-15612).

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Authors and Affiliations

  1. Department of Management Science and Engineering, Stanford University, Stanford, CA, 94305, USA

    Margaret L. Brandeau

  2. Ivey School of Business, University of Western Ontario, London, ON, N6A 3K7, Canada

    Gregory S. Zaric

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Correspondence to Gregory S. Zaric.

Appendix: Derivation of optimal expenditure results in Table 1

Appendix: Derivation of optimal expenditure results in Table 1

Case 1: IA(x) has increasing returns to scale for 0 ≤ x ≤ B.

If IA″(x) > 0, then IA(x) has increasing returns to scale. The optimal solution is specified as follows:

If W × IA(B) − B ≥ 0, then x* = B; otherwise, x* = 0.

Since IA(x) is increasing and has increasing returns to scale, the largest possible value of W × IA(x) − x will occur at the largest possible value of x, x = B. If this value is negative, then it must also be negative for all smaller values of x, x < B, and it is optimal to invest nothing. Note that if W × IA(B) − B = 0, then a decision maker would be indifferent between investment of B versus none since the return on investment is exactly equal to the decision maker’s willingness to pay.

Case 2: IA(x) has decreasing returns to scale for 0 ≤ x ≤ B.

If IA″(x) < 0, then IA(x) has decreasing returns to scale. We define the point x D (if it exists) as the point that sets IA′(x D) = 1/W. The optimal solution is specified as follows:

  1. 1.

    If IA′(0) < 1/W, then x* = 0;

  2. 2.

    If x D ≥ B, then x* = B;

  3. 3.

    If x D < B, then x* = x D.

In case (1), the benefits of investment are not justified by the costs even for the smallest possible investments. Since IA(x) has decreasing returns to scale, additional investments will yield proportionally fewer benefits, so it is optimal to invest nothing. In case (2), the point at which marginal cost equals marginal benefit involves expenditure greater than available funds, so it is optimal to invest the entire budget. In case (3), the point at which marginal cost equals marginal benefit involves an expenditure that is within the available budget, so it is optimal to invest up to this point. Since IA(x) has decreasing returns to scale, funds invested beyond x D will generate less benefit than 1/W per dollar invested, and thus would represent wasted opportunities elsewhere.

Case 3: IA(x) has constant returns to scale for 0 ≤ x ≤ B.

If IA″(x) = 0, then IA(x) has constant returns to scale. The optimal solution is specified as follows:

If IA′(0) ≥ 1/W then x* = B; otherwise x* = 0.

In this case, if the marginal benefit of investment is greater than the marginal cost, it is worth investing the entire budget; otherwise, no investment should be made. If IA′(x) = 1/W, a decision maker would be indifferent between investing anything in the intervention and not investing since the return on investment is exactly equal to the decision maker’s willingness to pay.

Case 4: IA(x) is “s-shaped” in the interval 0 ≤ x ≤ B.

Let x F be a point of inflection (0 < x F < B) where IA(x) switches from having increasing to decreasing returns to scale. IA(x) is “s-shaped” if IA″(x) > 0 for x < x F and IA″(x) < 0 for x ≥ x F. The optimal solution is determined by solving the two sub-problems corresponding to 0 ≤ x ≤ x F [the region over which IA(x) has increasing returns to scale] and x F < x ≤ B [the region over which IA(x) has decreasing returns to scale]. The first problem is easily solved by comparing W × IA(x F) − x F and W × IA(0) − 0 = 0. The second problem is formed by defining λ 0 = λ(x F) and reducing the budget to B − x F. The solution to the second problem will indicate whether any additional investment beyond x F is required. The solution to this second problem is found using the methods outlined above for the case of decreasing returns to scale.

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Brandeau, M.L., Zaric, G.S. Optimal investment in HIV prevention programs: more is not always better. Health Care Manag Sci 12, 27–37 (2009). https://doi.org/10.1007/s10729-008-9074-7

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