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hello and welcome operations research again transportation problem and now we are going to discuss about the diene and be generous in transportation problem degeneracy faced while solving the transportation problem as we know that when we get initial solution of a transportation problem by any of the methods out of northwest corner method least cost method or Vogel's approximation method one if the number of occupied cells are less than M plus n minus 1 that is number of columns plus number of rows minus 1 then the situation is called degeneracy and when there is the situation of degeneracy it is not possible us to write all the UI and VJ values then what - yes we are going to get the answer of this question what to do in case of the situation of degeneracy this is a transportation problem and one of the clerks in the transportation company has suggested a transportation plan as follows some experienced person is dead and he or she has suggested a plan that 30 units to be transported from blonde one to market a 10 units from plant one to market D the 20 units from plant to to market D 20 units from plan to to market he and so on in this way he or she has suggested total 7 different routes to transport 220 units now as we can see that there are for markets and for plants but the total demand for four markets is 200 only and the total supply available from the four plants is 220 so we have to create a dummy column of fifth market II where we should transfer or and get 20 units and in this way the transportation problem becomes a bearish transportation problem from the unbalance vil now we need to check whether this say suggested solution is optimal or not for that we have to follow the UV method or mo di method we have to calculate D IJ values etc but to calculate the area values first we have we should have you inv devil use to have UI and VJ values first we should check the I am condition whether the number of a locations are [Applause] equivalent to or more than M plus n minus 1 or not here M plus n equals to 5 plus 4 minus 1 equals to 8 on the other hand number of occupied cells equals to 7 1 2 3 4 5 6 7 that means it is less than the I am condition the meaning of this is if we go for checking the optimality of distribution LEDs we cannot have all the UI and VJ values some values of UI and or VJ may be see we cannot be able to write all the values because of this condition this is called degeneracy this situation is called the generous e in solving the transportation problem now if we cannot write all the UI and VJ values we cannot calculate all the DHA values then we cannot check out commodity and ultimately we cannot arrive at the conclusion whether this or any other initial solution is optimal or not and if it is not optimal we cannot reach the optimal solution by the path of closed loop so it is at more at most necessary to have the number of occupied cells at least equivalent to this number and as we can see in this case they are not so now the question is what to do under the situation of degeneracy the only way is to have this much number of occupied cells but here the reality is something different then how can we have this number of occupied cells when the actual number of occupied cells is less than this the simple meaning is we should have the required number of occupied cells to be able to these number of occupied cells this sound or rather this statement sound somewhat unusual we can say that we should have one or more that means in required number the dim the occupied cells now the thing is how to select this kind of cells first of all out of all the empty cells we have to see at one or more required number of cells to be treated as deemed occupied cells now let's discuss the technique of that thing the technique is to select one empty cell which is independent and then treat is so retained their cell loq myself let's do everything on board and on paper first of all which are the empty cells 1 2 3 4 5 6 7 8 9 10 11 12 and 13 we must total there are 5 into 4 20 cells out of which 7 are occupied so remaining 13 are empty cells now let's identify these 30 cells 1 be the cost is 4 1 see the cost is 9 1e the post is 0 2 a the cost is 22 be the cost is 6 to see the cost is 11 3 a the cost is 7 3 me the cost is 1 3 D the cost is 14 3 E the cost is 0 for C the cost is 12 4 D the cost is 6 and for e the force is 0 1 2 3 4 5 6 7 8 9 10 11 12 13 13 empty cells and 7 occupied cells the total is 20 and yes we have 20 cells fireworms into four rows now out of these empty cells we have to select in this case one cell because we already 7oq five cells and the minimum number of occupations we need is 8 so we just require to select only one empty cell to be treated as deemed occupied cell which one the say rule is to select independent empty cell now what do you mean by independent empty set if we refer to reference book if we go through the lectures and seminars of some seniors they always speak this statement select independent empty cell and put the sign of epsilon into it to treat it as and occupy itself but what do we mean by this word independent the meaning of the word independent is the cell we select should be such that from where we can not trace a close to look yes the empty cell we select should be such that from where we cannot start the closed loop yes okay now let's check all this 13 if we want to start a closed loop from 1d is it possible if we start the closed loop from one be a very easy route is available to have a closed loop to draw a closed loop that means one B is not independent because from that cell we can easy a closed loop so it does not qualify one see if we start closed loop from here either we have to go here or we have to go there if we start in clockwise first time second time third turn that's all first turn second turn that's all first and second and third and that's all that means from this cell we cannot press a closed loop so this cell qualifies 1c qualifies yes this is independent when we solve this kind of problem we have to check all the empty cells for this purpose it is not as easy as to write a sentence that select the independent set as a student if I think as a student it is a time consuming job we how do consume required time for this every time we face a degenerate solution okay one e can we draw up your loop yes easy so this doesn't qualify to a can we draw a closed loop 1 2 3 4 yes so do a doesn't qualify to be can we draw a closed loop from to be 1 2 3 no 1 2 3 4 5 no we cannot reach here I mean which one is that to be again let's check again is it possible to come again here yes it is possible 1 2 3 4 5 & 6 yes closed loop is possible from to be so it doesn't v 1c can we from one see yes one two three four five no one see can we know from one seat we can't so it is implemented three a guess three a can we draw a closed loop from three a is it possible one two that's all one two three four no we cannot draw a closed loop from three a so it is also independent 3 B is it possible one one two three four five but we can't reach here because we cannot draw a closed loop air with diagonal line no it can be let's try a lane one two three four five six no we cannot reach again here so 3 B is also independent whatever 3 C 3 C's occupied whatever 3d let's check 3 D is it possible to draw one to know one two three four no we cannot reach again here so it is also independent whatever 3e can we draw 1 2 3 4 5 6 but still if this stereo view by then only we can complete the closed loop so 3e is also independent 3e what about 4c can we draw a loop from four see one two three four okay oh this is the critical cell from there it is not possible to take down on this side and hence the closed loop can not be drawn from 3c so 3c is also independent what are the three four see what about 4d can we draw yes easily one two three and four we can draw a closed loop from 40 so that cell doesn't qualify and now for ii-if we start drawing closed loop from here can be one two three four five and six yes we can go so for e also cannot qualify so out of 30 we have one two three four five six and seven independent cells from where we cannot create a closed loop we can select any one out of this seven now the question is which which cells should we select to treat it as an occupied cell or which cells should be selected as deemed occupied cell the error is from our seniors is to select the independent empty cell with the least cost yes in this case cost is nine eleven seven one 14-0 and well obviously the least is zero so we are going to select three e 3e s if it is occupied and we substitute or rather we are going to follow tradition of writing Epsilon the side epsilon into that epsilon is the lowest possible quantity of a location it is actually said being that this cell is occupied we were according to our aim of minimizing the cost if we go for a location out of these cells the first condition is it must be an independent set so we selected seven independent sets now if we have to allocate some quantity to any of these seven cells obviously we want to select the cell with the least burst that's why we have to select the independent cell with zero cost yes even if it falls in the dummy column or row it equally qualifies to be selected yes now we have one two three four five six seven eight occupied cells and yes we are very occupied cells so now the condition of M plus n minus one occupied number of cells is satisfied that means now we can have all the UI and VJ values now before going forward I want to give you an assignment slide this case separately in your rough notebook or in your rough page and don't go for this process and just try to write UI and VJ will use having only seven occupied cells and see what happens to you this is your assignment now let's move forward and let's first of all determine all the UI and I'm very sorry in a hurry I started to rub it but first of all now we are going to substitute rather determine all UI and VJ values and as we know that any one of you our V J values we can take a zero we have all rows with two where locations and we have no other row or column B 3 and occasions so let's select u1 and take its value as 0 and we know that C IJ equals to UI plus VJ therefore UI equal to CIJ minus VJ + VJ equal to CIJ minus UI so according to this formula we can have these two values V 1 can be 6 minus 0 6 and V 4 can be 1 minus 0 1 v 4 3 now on the basis of these two and relevant occupied cells we can help to more UI value 6 and this occupied cell for a C minus V 7 minus 6 that will be u 1 this is u 4 and similarly on the basis of this - we can have u 2 C 3 minus V 1 so u comes to 2 that is YouTube on the basis of this value of U and this occupied cell we can help VJ value in this cell that is v 5c minus u 2 0 minus 2 minus 2 will be the value of V 5 now on the basis of this V value and this occupies any other occupied cell can we go for another thing is it yes zero minus minus 2 that will be positive 2 value of U 3 value of U 3 positive 2 is it yes then how can we arrive at this two way use yes on the basis of these two values we can have these two values 0 minus 2 so this value of V will be minus 2 it is V 3 and similarly 1 minus 1 C minus u 1 minus 1 this value of V is 0 that is V 2 in this way we have all UI and VJ values now only we can calculate D as a values for all empty cells except this