1 00:00:16,390 --> 00:00:25,430 in the last class we saw the graphical method to solve a linear programming problem we considered 2 00:00:25,430 --> 00:00:32,110 the example that is shown which is a maximization problem that maximizes ten x one plus nine 3 00:00:32,110 --> 00:00:38,650 x two subject to three x one plus three x two less than or equal to twenty one four 4 00:00:38,650 --> 00:00:44,370 x one plus three x two less than or equal to twenty four x one x two greater than or 5 00:00:44,370 --> 00:00:52,550 equal to zero we do the graph corresponding to the constraints we identified the feasible 6 00:00:52,550 --> 00:01:00,830 region with is shown in a different color and we also computed the corner points and 7 00:01:00,830 --> 00:01:12,729 then we observed that it is enough to evaluate up the corner point and we also found out 8 00:01:12,729 --> 00:01:20,979 that this corner point the corner point three comma four the optimal solution lies in this 9 00:01:20,979 --> 00:01:28,189 when we evaluated the objective function value at the four corner points zero comma zero 10 00:01:28,189 --> 00:01:38,000 gave us z equal to zero six comma zero gave us z equal to sixty zero comma seven gave 11 00:01:38,000 --> 00:01:48,410 us z equal to sixty three and three comma gave us z equal to sixty six since we are 12 00:01:48,410 --> 00:02:01,369 maximizing this function we said this is the corner point so this is the corner point three 13 00:02:01,369 --> 00:02:09,600 comma four that has the best value of the objective function and hence is the optimal 14 00:02:09,600 --> 00:02:18,120 solution now let us look at the same thing from a slightly different way 15 00:02:18,120 --> 00:02:27,480 now what we do now is we draw objective function lines i have shown a few objective function 16 00:02:27,480 --> 00:02:33,610 lines so what we have essentially done is the objective function is ten x one plus nine 17 00:02:33,610 --> 00:02:43,050 x two and we keep ten x one plus nine x two equal to a constant for example we could keep 18 00:02:43,050 --> 00:02:52,920 ten x one plus nine x two equal to fifty so this line could represent this line could 19 00:02:52,920 --> 00:03:04,980 represent 20 00:03:04,980 --> 00:03:12,959 this line could represent ten x one plus nine x two equal to fifty and this line is drawn 21 00:03:12,959 --> 00:03:20,510 this straight line is drawn and then we objective function lines which are parallel to this 22 00:03:20,510 --> 00:03:27,690 for example this could represent ten x one plus nine x two equal to sixty and so on so 23 00:03:27,690 --> 00:03:35,500 as we increase ten x one plus nine x two to a constant and keep increasing this constant 24 00:03:35,500 --> 00:03:44,360 the objective function lines become parallel to each other and move in this direction you 25 00:03:44,360 --> 00:03:50,370 can see that the objective function is moving in this direction and as it moves the objective 26 00:03:50,370 --> 00:03:57,790 function line finally it touches the feasible region at the point three comma four the corner 27 00:03:57,790 --> 00:04:04,739 point where the objective function lines last touch before leaving the feasible region is 28 00:04:04,739 --> 00:04:16,090 this point 29 00:04:16,090 --> 00:04:20,799 is this point three comma four which is the optimum solution 30 00:04:20,799 --> 00:04:28,030 so there are two ways of of getting to the optimum solution one of which is to evaluate 31 00:04:28,030 --> 00:04:33,390 the four corner points in this example there are four corner points in other examples there 32 00:04:33,390 --> 00:04:38,469 could be a different number of corner points evaluate the objective function at all the 33 00:04:38,469 --> 00:04:45,010 corner points and find out which corner point has the best value the other is to draw a 34 00:04:45,010 --> 00:04:50,590 series of objective function lines where the objective function is a constant and if it 35 00:04:50,590 --> 00:04:56,610 is maximization increase the constant the lines move in the direction as shown the lines 36 00:04:56,610 --> 00:05:19,770 move in this direction as shown 37 00:05:19,770 --> 00:05:26,010 the lines move in this direction and because we are maximizing the lines finally leave 38 00:05:26,010 --> 00:05:33,940 the feasible region at this corner point which happens to be the optimum solution now we 39 00:05:33,940 --> 00:05:41,899 will understand how we solve a minimization problem using the graphical method the method 40 00:05:41,899 --> 00:05:47,630 is similar however we would explain it through an example 41 00:05:47,630 --> 00:05:54,280 now we look at this minimization problem which minimizes seven x one plus five x two subject 42 00:05:54,280 --> 00:06:00,419 to two constraints x one plus x two greater than or equal to four and five x one plus 43 00:06:00,419 --> 00:06:08,410 two x two greater than or equal to ten x one x two greater than or equal to zero we draw 44 00:06:08,410 --> 00:06:15,710 the graph as we did for the maximization problem and then as since x one and x two are greater 45 00:06:15,710 --> 00:06:21,099 than or equal to zero we will look at only the first quadrant we will not look at the 46 00:06:21,099 --> 00:06:26,380 other quadrants because in these quadrants which are here in these quadrants which are 47 00:06:26,380 --> 00:06:34,180 here or here or in this place you'll get x one or x two or both will become negative 48 00:06:34,180 --> 00:06:44,200 and therefore we will be looking at only this quadrant which is the first quadrant and we 49 00:06:44,200 --> 00:06:54,500 also have x one here as the x axis and we have x two here as the y axis we now draw 50 00:06:54,500 --> 00:07:01,810 x one plus x two greater than or equal to four to do that we first draw x one plus x 51 00:07:01,810 --> 00:07:09,560 two equal to four so x one plus x two equal to four is this line which passes through 52 00:07:09,560 --> 00:07:16,560 four comma zero and zero comma four which are shown here so this is the first line is 53 00:07:16,560 --> 00:07:23,599 x one plus x two equal to four which passes through four comma zero and zero comma four 54 00:07:23,599 --> 00:07:29,469 which are here and this line is shown here x one plus x two equal to four 55 00:07:29,469 --> 00:07:36,190 now this line divides the graph into two regions one of which is less than four and the other 56 00:07:36,190 --> 00:07:44,440 is greater than four so to find out which one it is we look at a convenient point which 57 00:07:44,440 --> 00:07:56,830 is the origin you look at the origins which is here and then check whether zero comma 58 00:07:56,830 --> 00:08:04,860 zero satisfies x one plus x two greater than or equal to four or not now when we substitute 59 00:08:04,860 --> 00:08:12,159 zero comma zero into the left hand side of this inequality to the left hand side of this 60 00:08:12,159 --> 00:08:24,779 inequality left hand side of this inequality we realize 61 00:08:24,779 --> 00:08:31,860 the left hand side value is zero and therefore it does not satisfy the inequality therefore 62 00:08:31,860 --> 00:08:53,700 the region that is x one plus x two greater than four is the region on this side 63 00:08:53,700 --> 00:09:01,150 so all this region is the region x one plus x two greater than or equal to four 64 00:09:01,150 --> 00:09:07,530 we now draw the next constraint which is five x one plus two x two greater than or equal 65 00:09:07,530 --> 00:09:14,120 to ten so we first draw the line five x one plus two x two equal to ten and that line 66 00:09:14,120 --> 00:09:32,920 is shown here so that line is shown here so this is the line five x one plus two two 67 00:09:32,920 --> 00:09:40,150 x two equal to ten and to draw this line we have two points which is two comma zero and 68 00:09:40,150 --> 00:09:48,650 zero comma five now once again this line divides a graph into two regions one of which is less 69 00:09:48,650 --> 00:09:55,710 than ten and the other is greater than ten so once again we look at the origin which 70 00:09:55,710 --> 00:10:02,750 is here and check whether the origin satisfies the condition or violates the condition 71 00:10:02,750 --> 00:10:12,120 so when we substitute five x one plus two x two to the left hand side of of this inequality 72 00:10:12,120 --> 00:10:17,870 we get zero on the left hand side which is less than ten therefore the other one is the 73 00:10:17,870 --> 00:10:26,240 one that satisfies greater than or equal to ten so now looking at this you will realize 74 00:10:26,240 --> 00:10:35,930 that the the region goes like this so all of these is the region that we are talking 75 00:10:35,930 --> 00:10:42,870 about so the entire thing is the region that we are talking about and that is shown the 76 00:10:42,870 --> 00:10:47,800 line is five x one plus two x two equal to ten and this is the region which is now shown 77 00:10:47,800 --> 00:11:09,820 in a different color which 78 00:11:09,820 --> 00:11:14,480 now this is the region which is shown in a in a light green color which is the feasible 79 00:11:14,480 --> 00:11:26,180 region for this particular example we also observed that points in this space 80 00:11:26,180 --> 00:11:31,370 satisfy one of the constraints they satisfy the first constraint but they violate the 81 00:11:31,370 --> 00:11:42,100 second constraint and therefore they do not lie in the feasible region 82 00:11:42,100 --> 00:11:53,610 now our feasible region has three corner points the first corner point is given here which 83 00:11:53,610 --> 00:12:03,290 is four comma zero the other corner point is here and the third corner point is zero 84 00:12:03,290 --> 00:12:10,470 comma five so at four comma zero we evaluate the objective function and the objective function 85 00:12:10,470 --> 00:12:21,230 value is given as z is equal to twenty eight the next corner point is zero comma five which 86 00:12:21,230 --> 00:12:27,430 is here and when we evaluate the objective function seven x one plus five x two we get 87 00:12:27,430 --> 00:12:38,860 z is equal to twenty five seven into zero plus five into five gives us twenty five 88 00:12:38,860 --> 00:12:46,570 we now look at the third corner point which is two by three comma ten by three now this 89 00:12:46,570 --> 00:12:52,950 two by three comma ten by three can be found in the graph if we draw this graph to scale 90 00:12:52,950 --> 00:12:57,940 and we draw the graph correctly as we have attempted to do now you realize that this 91 00:12:57,940 --> 00:13:03,870 point has an x coordinate of two by three and has a y coordinate of ten by three so 92 00:13:03,870 --> 00:13:18,540 this point is two by three comma ten by three so this is two by three comma ten by three 93 00:13:18,540 --> 00:13:25,150 that i have shown here and the objective function value when we substitute we get seven into 94 00:13:25,150 --> 00:13:31,930 two by three which is fourteen by three plus ten into five by three which is fifty by three 95 00:13:31,930 --> 00:13:39,360 so fourteen by three plus fifty by three is sixty four by three which is what we have 96 00:13:39,360 --> 00:13:49,980 here 97 00:13:49,980 --> 00:13:56,240 now we have twenty eight for this corner point we have twenty five for this corner point 98 00:13:56,240 --> 00:14:06,230 and we have sixty four by three which is twenty one point three three for the other corner 99 00:14:06,230 --> 00:14:14,990 point which is two by three comma ten by three therefore the optimum solution to this minimization 100 00:14:14,990 --> 00:14:22,060 problem is that corner point which has the smallest value of objective function and out 101 00:14:22,060 --> 00:14:29,080 of these three we realize that the point two by three comma ten by three with sixty four 102 00:14:29,080 --> 00:14:37,200 by three has the smallest value and therefore is the optimum solution so this point is the 103 00:14:37,200 --> 00:14:42,910 optimum solution with value sixty four by three 104 00:14:42,910 --> 00:14:51,130 like we did for the maximization problem we can also draw what are called iso objective 105 00:14:51,130 --> 00:15:04,910 function lines where we fix seven x one plus five x two to to a constant so we fix seven 106 00:15:04,910 --> 00:15:17,820 x one so these are 107 00:15:17,820 --> 00:15:35,360 seven x one seven x one plus five x two equal to a constant 108 00:15:35,360 --> 00:15:49,940 for example this would mean seven x one plus five x two would be equal to about thirty 109 00:15:49,940 --> 00:15:55,160 and since it's a minimization problem we draw the objective function lines in the other 110 00:15:55,160 --> 00:16:01,730 direction where seven x one plus five x two gets progressively reduced 111 00:16:01,730 --> 00:16:08,560 so the direction of movement of the objective function is this way for a minimization problem 112 00:16:08,560 --> 00:16:15,180 unlike in a maximization problem so it moves in this direction and as it moves downwards 113 00:16:15,180 --> 00:16:22,250 or in the direction in this direction it finally touches this corner point finally touches 114 00:16:22,250 --> 00:16:36,370 this corner point 115 00:16:36,370 --> 00:16:42,510 it finally touches this corner point before it leaves the feasible region so the last 116 00:16:42,510 --> 00:16:52,130 feasible point is given by two by three comma ten by three is given by two by three comma 117 00:16:52,130 --> 00:17:00,460 ten by three which is the optimum solution to the minimization problem so this way we 118 00:17:00,460 --> 00:17:11,659 solve minimization problems we solve minimization problems using the graphical method 119 00:17:11,659 --> 00:17:17,560 now the only difference between the maximization and the minimization is many times it will 120 00:17:17,560 --> 00:17:23,970 happen that in a minimization problem the feasible region will not be bounded as in 121 00:17:23,970 --> 00:17:29,890 this case even though i have shown a certain portion of the feasible region one would realize 122 00:17:29,890 --> 00:17:46,880 that a point here a point here is also feasible and a feasible region is not bounded or it's 123 00:17:46,880 --> 00:17:53,309 not a closed region whereas in a maximization problem many times we would get a feasible 124 00:17:53,309 --> 00:17:59,740 region that is a closed region so that is one of the things that we need to understand 125 00:17:59,740 --> 00:18:08,130 the other most important thing where the reason for which we evaluate only the corner points 126 00:18:08,130 --> 00:18:14,330 is that we have shown that every point inside the region there will be a corner point which 127 00:18:14,330 --> 00:18:20,470 will have a better value of the objective function now that happened due to what is 128 00:18:20,470 --> 00:18:28,980 called the convexity property of the feasible region if we take two points in the feasible 129 00:18:28,980 --> 00:18:36,030 region and join them by a line we will realize that all the points in the line will lie in 130 00:18:36,030 --> 00:18:43,030 the feasible region and because this feasible region is convex we we can easily show that 131 00:18:43,030 --> 00:18:49,110 for every point inside the feasible region there is always a point on the boundary there 132 00:18:49,110 --> 00:18:54,340 is always a point on the boundary and then there is always a corner point which has a 133 00:18:54,340 --> 00:19:00,660 better value of the objective function so these are the basic ideas and principles that 134 00:19:00,660 --> 00:19:09,850 we use to solve the linear programming problem using the graphical method while the graphical 135 00:19:09,850 --> 00:19:18,150 method is simple and easily understandable we are limited by the fact that we can only 136 00:19:18,150 --> 00:19:26,700 represent two variables in the graphical method the x one and x two we have already seen formulations 137 00:19:26,700 --> 00:19:32,890 that involve more than two variables and several linear programming problems have a large number 138 00:19:32,890 --> 00:19:39,510 of variables so we need other methods other than the graphical to solve linear programming 139 00:19:39,510 --> 00:19:46,760 problems when we have more than two variables we will see the algebraic method in the next 140 00:19:46,760 --> 00:19:47,030 class