Friday, November 23, 2007

Exploring Power Distribution in a Global Assembly

[NOTE: Though I still consider the underlying arguments below sound, I have determined that the actual procedure described below for assigning seats is less than optimal. For the new method that produces an assembly of an exact size (800 individuals), please go to]

Dear citizens of the world,

In my previous post, I introduced the following equation for deciding how many seats a constituency should get in an assembly:

s = CEIL[ SQRT(160000 p n / SUM(p[1],...p[n]) n^2 )
+ SQRT(160000 p n / SUM(p[1],...p[n]) n^2 )]

The equation is here used to calculate seats. It could, however, be used to simply weigh votes of the current General Assembly of the UN. I don't think a General Assembly with weighted votes constitutes sufficient reforms for the UN. The fact that the representatives are not elected directly by their constituencies breaks the rule that the evaluation of the appropriateness of an agent's action should be at a minimal distance from the constituency. And a nation is hardly a monolithic constituency, especially when we talk about nations like India, the U.S, Brazil or Nigeria. Having one single agent to represent them is simply insufficient.

Nonetheless, we can use the principles and result of the equation in both cases. It expresses the power a nation's voting block would carry, regardless of whether the block consists of 1 or several agents. Instead of seats, we will call it just "weight", or w for short. The equation remains the same but we eliminate the ceiling part since we are dealing with just abstract weights (and not individuals that can't be quartered):

w = SQRT(160000 p n / SUM(p[1],...p[n] n^2 ))
+ SQRT(160000 p n / SUM(p[1],...p[n] n^2 ))

We will use GDP in millions of US dollars for the e factor instead of contributions since it represents what a nation could potentially give if there were a healthier system for supplying the commons with resources. Our sample will include an array of populous and poor to populous and rich, and small and poor to small and rich:

  • U.S.

  • Germany

  • Russia

  • Brazil

  • India

  • China

  • Nigeria

  • Egypt

  • Iran

  • Sweden

  • Guinea

  • Luxembourg

  • Suriname

  • Tuvalu

Let's plug in the numbers for the U.S. (where, according to Wikipedia, the population is 303,158,700 and the GDP, in millions of U.S. dollars, is 13,194,700). The world population, according to the UN is 6,671,226,000. And the global GDP according to the IMF, 6,671,226,000. For the number of countries, we will use 221:

w =

SQRT( (160000 * 303158700 * 221) / (6671226000 * 221^2) )
SQRT( (160000 * 13194700 * 221) / (48245198 * 221^2) ) =

SQRT( 10719691632000000 / 325829349066000) +
SQRT( 466564592000000 / 2356343715518 ) =

SQRT( 32.8997 ) + SQRT ( 198.0036) =

5.7358 + 14.0714 = 19.8072

The U.S. voting block would carry a weight of 19.8072 (or 20 if assigning seats with 1 vote each to actual individuals).

Plugging in the numbers for the previously listed countries we get:

U.S. - 19.8072
China - 18.2816
India - 14.6998
Germany - 9.6025
Brazil - 8.4873
Russia - 7.7682
Nigeria - 5.3311
Iran - 4.6067
Egypt - 4.1318
Sweden - 3.3982
Guinea - 1.2255
Luxembourg - 1.0143
Suriname - 0.4010
Tuvalu - 0.0481

Based on this sample, it would seem that the algorithm I propose using for distributing power in a global assembly strikes a balance between the status quo and emerging changes. China and the U.S. receive nearly equal votes, meeting current expectations within the international community. It is also important to note that the power distribution would be automatically redistributed as new powers like India fulfill their potential.

At the same time, since a nation's power grows at exponentially smaller degrees as it grows larger, smaller nations retain appropriate leverage in the global assembly.

It does however become appearant that the European Union will wield great power in the assembly since each member country is assigned its own power. Counting only Germany and Sweden (2 of the EU's 27 members), the EU would already have a weight in the assembly of over 13. Whether this is appropriate or not depends on how cohesive a political unit the EU is. If we consider the EU to be tightly integrated in its international politics, the problem can be easily resolved by assigning weight not to the EU's individual members but the EU as a whole. We would treat it as a true federation with a common international agenda. In this case, the EU would get a weight of 22.1370, putting it on par with the U.S. and China.

The issue raises a question: how do we decide what constitutes one single constituency at the various levels of governance? The problem is not easily resolved. The old and common politcal problem of gerrymandering begins to rear its uggly face.

1 comment:

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