 I am welcome to our session. Let us discuss the following question. The question says, to Goddams, A. N. V., have a grain storage capacity of 100 quintals and 50 quintals, respectively. This applies to three Russian shops, D. E. N. F., whose requirements are 60, 50, and 40 quintals, respectively. The cost of transportation for a quintal from Goddams to the shops are given in the point table. This is the table given to us. How much the supply is being transported in order that the transportation cost is made up? Let us now begin with the solution. Let quantity of grain supplied from Goddams A to Russian shop B X quintals quantity Russian shop E by quintals. The storage capacity of Goddams A 100 quintals quantity which can be supplied from Goddams A minus X minus Y quintals. Now as the supply cannot be negative, this is greater than equal to 0 and Y is also greater than equal to 0 and 100 minus X minus Y is also greater than equal to 0. This implies 100 is greater than equal to X plus Y and this implies X plus Y is less than equal to 100. So, this is our first in equation. Now requirement shop B, 60 quintals being transported from Goddams A 60 minus X quintals will be supplied on Goddams B which implies B minus X is greater than equal to 0 or we can say that X is less than equal to 60. So, this is our second in equation. Similarly, requirement shop E quintals have already been supplied from Goddams A. Therefore, the remaining 50 minus Y quintals will be transported from Goddams B. Therefore, 50 minus is greater than equal to 0 or we can say that Y is less than equal to 50. So, this is our third in equation. Requirement of shop F is 40 quintals and 100 minus X minus Y quintals have already been transported from Goddams. The remaining 40 minus 100 minus X minus Y 40 is the requirement of shop F and 100 minus X minus Y quintals have already been transported from Goddams A. Therefore, remaining 40 minus 100 minus X minus Y which is equal to minus 60 quintals will be transported from Goddams B. Therefore, X plus Y minus 60 is greater than equal to 0. This implies X plus Y is greater than equal to 60. So, this is our fourth in equation. We will find objective function Z. Look at the table. Transportation cost per quintal from Goddams A to shop B is 6 rupees. Therefore, transportation cost of X quintal will cost rupees 6X from Goddams A to shop B. Z is equal to 6X plus we know that requirement of shop B is 60 quintals. Now, X quintals have already been transported from Goddams A. Therefore, remaining 60 minus X quintals will be supplied from Goddams B and transportation cost per quintal from Goddams B to shop D is 4 rupees. Therefore, transportation cost of X minus X quintals will be 4 into 60 minus X. Assume that quantity supplied from Goddams A to ration shop E as Y quintals. Now, transportation cost per quintal from Goddams A to shop E is rupees 3. Therefore, transportation cost of Y quintals will be 3X. So, we have plus 3X. We know that requirement of shop E is 50 quintals and Y quintals have already been supplied from Goddams A. Therefore, the remaining 50 minus Y quintals will be transported from Goddams B. Now, transportation cost per quintal from Goddams B to shop E is 2 rupees. Therefore, transportation cost of 15 minus Y quintals will be 2 into 15 minus Y. So, now we have 2 into 15 minus Y. Similarly, we can find transportation cost of grains from Goddams A to shop F. Transportation cost of grains per quintal from Goddams A to shop F is 2.5 into 100 minus X minus Y and transportation cost from Goddams B to shop F is 3 into X plus Y minus 60. On simplifying list, we get 2.5X plus 1.5Y plus 410. Now, this is our objective function which is to be minimized. So, mathematical formulation of LPP is minimize Z equals to 2.5X plus 1.5Y plus 410 subject to the constraints X plus Y less than equal to 100, X plus Y greater than equal to 60, X is less than equal to 60, Y is less than equal to 50, X is greater than equal to 0 and Y is also greater than equal to 0. Equation corresponding to this in equation is X plus Y equals to 100, to this in equation is X plus Y equals to 60, to this equation is X equals to 60 and to this in equation is Y equals to 50. The points lying on this line are 0, 100, 100, 0, points lying on this line are 60, 0, 0, 60, points on this line are 60, 0, 60, 50, points lying on this line are 0, 50, 60, 50. Now, we will draw the lines on the graph. First draw the line X plus Y equals to 100, points are 0, 100, 100, 0. This is the line X plus Y equals to 100. Second line is X plus Y equals to 60, points are 0, 60 and 60, 0. Third line is X equals to 60, points are 60, 0. This is the point and then is 60, 50. This is the point 60, 50. Now, this line is parallel to Y axis. Fourth line is Y equals to 50, points are 0, 50. This is the point 0, 50, 60, 50. This is the point 60, 50. Let us now join these points. So, this is the line Y equals to 50. Now, we will find the feasible region. This is the common region determined by all the constraints. So, let us now shape this region. On the rest of this polygon R, this point we can obtain by solving these two equations simultaneously. On solving them, we get this point as 10, 50. This point is 60, 0. This point we can obtain by solving equation X plus Y equals to 100 and X equals to 60. On solving these two equations, we get this point as 60, 40. This point is 50, 50. We have obtained this point by solving equation X plus Y equals to 100 and Y equals to 50. According to fundamental extreme point theorem, the minimum value of Z will occur at one of these points. So, now we will calculate value of Z at each of these points. Z is equal to 2.5X plus 1.55 plus 410. Now, the first point is 10, 50. Value of Z at this point is 510. Second point is 60, 0. Value of Z at this point is 560. Third point is 50, 50. Value of Z at this point is 610. Fourth point is 60, 40. Value of Z at this point is 620. So, minimum value of Z is 510 at the point 10, 50. So, this means from go down A, quantity of grains applied to shops D, E and F should be 10 quintals, 50 quintals and 40 quintals respectively and quantity of grains applied to shops D, E and F from go down B should be 50 quintals, 0 quintals, 0 quintals. Hence, our required answer is minimum cost is rupees 510, quantity of grain transported from go down A from D, E and F, 10 quintals, 10 quintals, 30 quintals, quantity of grain transported from go down B, 10 quintals, 0 quintals, 0 quintals respectively. This is our required answer. So, this completes the session. I in take care.