Global Surveillance and the Value of Information: The Case of the Global Polio Laboratory Network
by Esther de Gourville, Radboud J. Duintjer Tebbens, Nalinee Sangrujee, Mark A. Pallansch, and Kimberly M. Thompson, Risk Analysis 2006;26(6):1557-1569. PDF

Abstract

Effective control and eradication of diseases requires reliable information from surveillance activities, including laboratories, which typically incur real financial costs. This article presents data from a survey we conducted to estimate the costs of the Global Polio Laboratory Network (GPLN), which currently supports aggressive global surveillance for acute flaccid paralysis (AFP) to detect circulating polioviruses. The Global Polio Eradication Initiative (GPEI) of the World Health Organization (WHO) provides resources for some of the laboratory network costs, but the total cost of the network remains relatively poorly characterized given the limited documentation of national contributions. We surveyed network laboratories to quantify AFP surveillance support costs and provide data for cost estimates of potential posteradication surveillance policies related to the laboratories. We estimate that the GPLN currently requires millions (US dollars 2002) in total support annually, and that half of the support for national and regional reference laboratories comes from external donors through the WHO or bilateral agreements and half from within nations that host those laboratories. The article also presents the framework for considering the value of information from this global surveillance network and suggests that the expected value of surveillance information from the GPLN currently exceeds its costs. We also provided important insights about how the value of information may change after successful eradication of wild polioviruses.

Corrections to printed article: Page 1564, right column, lines 14-15 should say: "Thus, E(N)=pC4+(1-p)C1 and E(Y)=pC2+(1-p)C3,..." and page 1565, first equations on the page should be: "E(N)=pC4+(1-p)C1 (choose N if p<p*) E(Y)=pC2+(1-p)C3 (choose Y if p>p*)."

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