Wednesday, 19 September 2012

Numbers in Coalition Governments: How Math works?

‘Mamata withdraws support to UPA government’ this is the front page column in today’s news paper ‘The Hindu’. The part of the news attracted me more is a table named ‘How the arithmetic works’ right at the centre of the article.

UPA – Trinamool
UPA – Trinamool + SP + BSP + RJD
UPA – Trinamool - SP + BSP + RJD
UPA – Trinamool - SP - BSP + RJD
UPA – Trinamool + SP + BSP + RJD +JD(S)
UPA – Trinamool - SP - BSP + RJD +JD(S)
 *original value given in the article is 250. It is changed for the ease of calculations

Here I do not want to contemplate on the fate of UPA in case Mamata Deedi withdraws support or Lalluji & Malluji extend it or what so ever. What I see is this: six relationships are given between member-strengths of six parties. Can’t we find the member-strengths of each party? Let us give a try..
First let us write down these in the form of equations:
U-T = 246                                --(1)
U-T+S+B+R = 293                    --(2)
U-T-S+B+R = 271                     --(3)
U-T-S-B+R = 251                      --(4)
U-T+S+B+R+J = 296                 --(5)
U-T-S-B+R+J = 253                  --(6)
Now let us solve:
(2)-(1) gives => S+B+R = 47    --(7)
(3)-(1) gives => -S+B+R = 25   --(8)
(4)-(1) gives => -S-B+R = 5      --(9)
(7)-(8) gives => 2S = 22 => S = 11      
(8)-(9) gives => 2B = 20 => B = 10     
(7)+(9) gives => 2R = 52 => R = 26    
(5) - (2) gives => J = 3

By now, we have got strengths of four parties SP, BSP, RJD and JD(S). But can we do it for UPA and Trinamool?
We can’t. With the given set of data it is not possible. Even though six equations in six unknowns are given, we cannot solve for all the six variables. Why so?
Here U and T are represented together as (U-T) in all the equations. It is represented as a single entity and the value of that entity (U-T) is only known. Even though we are provided with n number of such equations we can’t find the individual value of either U or T. And one more thing, only one of the equations 5 or 6 is enough to find the value of J and the other equation is useless (‘redundant’ in mathematical sense). So effectively there are less than six equations.

(To prove this mathematically, math-savvies can try out one of the standard methods: Cramer’s or Matrix or Rank method. Here Rank method is suitable as in the other methods it involves dealing with determinant of 6X6 matrix, which is difficult)
Click on the compressed sheet down here for the mathematical proof by Rank method:

No comments:

Post a Comment