1
00:00:17,990 --> 00:00:27,220
in this class we summarize the graphical and
algebraic methods and we relate the two methods
2
00:00:27,220 --> 00:00:43,830
which are popular to solve linear programming
problems we started with a maximization problem
3
00:00:43,830 --> 00:00:51,019
and we solved it using the graphical method
and then in another class we to a minimization
4
00:00:51,019 --> 00:00:57,530
problem and solved it using the graphical
method in the last two classes we have solved
5
00:00:57,530 --> 00:01:04,980
the same two examples one for maximization
and one for minimization using the algebraic
6
00:01:04,980 --> 00:01:12,020
method now let us go back to the graphical
method and look at what we did and then we
7
00:01:12,020 --> 00:01:17,670
go back to the algebraic method and try to
see whether there is a relationship between
8
00:01:17,670 --> 00:01:23,520
what we did in the graphical method and what
we did in the algebraic method
9
00:01:23,520 --> 00:01:35,490
now in the graphical method we found out the
feasible region and then we explained that
10
00:01:35,490 --> 00:01:43,570
for every point inside the feasible region
there is always a boundary point or a corner
11
00:01:43,570 --> 00:01:51,299
point which will give a better solution a
better value of the objective function and
12
00:01:51,299 --> 00:01:58,560
then we restricted ourselves to evaluating
only the corner point solutions and we found
13
00:01:58,560 --> 00:02:11,630
the optimum solution from the best corner
point solution in this example we had these
14
00:02:11,630 --> 00:02:23,300
four corner points zero comma zero we had
these four corners points zero comma zero
15
00:02:23,300 --> 00:02:33,110
six comma zero zero comma seven and three
comma four which are our four corner points
16
00:02:33,110 --> 00:02:40,470
and we said that the optimum solution is this
corner point three comma for which gave us
17
00:02:40,470 --> 00:02:53,390
a value of sixty six gave us a value of sixty
six now let us go back to the corresponding
18
00:02:53,390 --> 00:03:08,620
algebraic method and let us see what we did
19
00:03:08,620 --> 00:03:16,080
now this is the table corresponding to the
six solutions for the same linear programming
20
00:03:16,080 --> 00:03:24,000
problem which maximize ten x one plus nine
x two plus zero x three plus zero x four subject
21
00:03:24,000 --> 00:03:32,180
to these constraints now as i mention with
the inclusion of this slack variable there
22
00:03:32,180 --> 00:03:39,810
are four variables out of which we can only
solve for two at a time and therefore we had
23
00:03:39,810 --> 00:03:50,240
four c two or six ways of solving these solutions
now these six solutions are given here for
24
00:03:50,240 --> 00:04:00,010
the six problem instances and let us look
at these solutions only from the point of
25
00:04:00,010 --> 00:04:07,010
view of x one and x two and relate it to x
one and x two because the graphical method
26
00:04:07,010 --> 00:04:13,260
we used only x one and x two
so the first solution that we have here is
27
00:04:13,260 --> 00:04:20,459
x one equal to zero and x two equal to zero
so the first solution that we have here is
28
00:04:20,459 --> 00:04:36,639
x one equal to zero
and x two equal to zero because x three and
29
00:04:36,639 --> 00:04:44,620
x four have some values the second solution
that we have here is x one equal to six x
30
00:04:44,620 --> 00:04:52,879
two equal zero which is six comma zero the
third solution that we have here is x one
31
00:04:52,879 --> 00:05:00,569
equal to seven and x two equal to zero which
is a seven comma zero the fourth solution
32
00:05:00,569 --> 00:05:07,150
that we have is x one equal to zero x two
equal to eight the fifth solution that we
33
00:05:07,150 --> 00:05:13,610
have is x one equal to zero x two equal to
seven and the sixth solution that we have
34
00:05:13,610 --> 00:05:20,250
is x one equal to three and x two equal to
four so these are the six solutions that we
35
00:05:20,250 --> 00:05:25,180
have
now out of these six solutions we also said
36
00:05:25,180 --> 00:05:33,360
that situations three and four are infeasible
now we also have to remember that all the
37
00:05:33,360 --> 00:05:41,029
six solutions are basic solutions which means
that the non basic variables which are here
38
00:05:41,029 --> 00:05:49,060
which are x one x two x two x four extra are
fixed to zero even though they could have
39
00:05:49,060 --> 00:05:56,810
been fixed to any other value these six solutions
have been obtained by fixing the non basic
40
00:05:56,810 --> 00:06:02,900
variables to zero so now let us go back to
the graphical method and try to see where
41
00:06:02,900 --> 00:06:10,949
these six solutions lie in the graph that
we actually have so let us do that
42
00:06:10,949 --> 00:06:18,729
the first solution is x one equal to zero
and x two equal to zero so we denote this
43
00:06:18,729 --> 00:06:25,840
first solution which is x one equal to zero
and x two equal to zero as the first solution
44
00:06:25,840 --> 00:06:36,169
that we have the second solution out of the
six is x one equal to six and x two equal
45
00:06:36,169 --> 00:06:50,430
to zero so x one equal to six x two equal
to zero is our second solution the third solution
46
00:06:50,430 --> 00:06:56,520
is x one equal to seven and x two equal to
zero which happened to be infeasible in the
47
00:06:56,520 --> 00:07:11,559
algebraic method so x one equal to seven x
two equal to zero is the third solution
48
00:07:11,559 --> 00:07:18,199
which is here and you can observe from the
graph that this solution seven comma zero
49
00:07:18,199 --> 00:07:28,889
is not within the feasible region and therefore
it is infeasible the fourth solution is x
50
00:07:28,889 --> 00:07:36,000
one equal to zero and x two equal to eight
which was also infeasible now we go back and
51
00:07:36,000 --> 00:07:43,499
mark it in the graph and we realize that this
point this is the fourth solution so this
52
00:07:43,499 --> 00:07:55,159
is marked as four the fifth solution was x
one equal to zero and x two equal to seven
53
00:07:55,159 --> 00:08:02,449
and we mark this as the solution which is
here this is the solution x one equal to zero
54
00:08:02,449 --> 00:08:12,479
and x two equal to seven is marked as the
fifth solution and the sixth solution is x
55
00:08:12,479 --> 00:08:20,219
one equal to three and x two equal to four
which is marked as the sixth solution marked
56
00:08:20,219 --> 00:08:28,360
as the sixth solution
now we observe that the six basic solutions
57
00:08:28,360 --> 00:08:38,190
that we obtained in the algebraic method are
actually six corner points in the graph and
58
00:08:38,190 --> 00:08:43,669
since two of these corner points the points
where these lines intersect two of them are
59
00:08:43,669 --> 00:08:51,250
infeasible these two are now shown as infeasible
solutions now when we look at the algebraic
60
00:08:51,250 --> 00:08:58,460
method and when we look at these six solutions
we have to leave out the infeasible ones in
61
00:08:58,460 --> 00:09:05,310
this example there are two infeasible solutions
we leave out the infeasible solutions we consider
62
00:09:05,310 --> 00:09:12,329
only the basic feasible solutions and we considered
only the basic feasible solutions and found
63
00:09:12,329 --> 00:09:20,310
out the best basic feasible solution out of
the four now this is exactly the same as we
64
00:09:20,310 --> 00:09:28,320
now realize that the four feasible or four
basic feasible solutions are actually the
65
00:09:28,320 --> 00:09:37,560
four corner points which are points one two
five and six therefore what we did in the
66
00:09:37,560 --> 00:09:44,089
algebraic method is the same as what we did
in the graphical method
67
00:09:44,089 --> 00:09:51,320
now in the graphical method we were restricted
to two variables and we were able to clearly
68
00:09:51,320 --> 00:09:58,620
look at the graph and say that these two solutions
are infeasible by looking at the graph we
69
00:09:58,620 --> 00:10:05,660
can do that and therefore we did not consider
them for evaluating the objective function
70
00:10:05,660 --> 00:10:11,260
we looked only at the corner points and we
defined those corner points which are basic
71
00:10:11,260 --> 00:10:18,839
feasible solutions so the first relationship
is that the algebraic method evaluates the
72
00:10:18,839 --> 00:10:26,009
intersection points that the graphical method
has now out of these six intersection points
73
00:10:26,009 --> 00:10:32,900
that we have two of them are infeasible and
four are basic feasible the algebraic method
74
00:10:32,900 --> 00:10:39,790
also evaluates the same six points and declares
two of them that are infeasible and the rest
75
00:10:39,790 --> 00:10:45,160
of them are basic feasible
the graphical method looks at the four corner
76
00:10:45,160 --> 00:10:52,910
points which happened to be the four basic
feasible solutions so every corner point in
77
00:10:52,910 --> 00:11:01,110
the graphical solution corresponds to a basic
feasible solution in the algebraic method
78
00:11:01,110 --> 00:11:08,010
so when we say that the graphical method has
four corner points and the algebraic method
79
00:11:08,010 --> 00:11:15,380
has four basic feasible solutions it is enough
they are one and the same let us now go back
80
00:11:15,380 --> 00:11:20,680
to another aspect in the algebraic method
now let us take this particular solution where
81
00:11:20,680 --> 00:11:27,959
we are solving for x one and x two now we
are fixing x three and x four to zero and
82
00:11:27,959 --> 00:11:35,089
we are solving now when we are fixing x three
and x four to zero then the equations become
83
00:11:35,089 --> 00:11:41,009
three x one plus three x two equal to twenty
one and four x one plus three x two equal
84
00:11:41,009 --> 00:11:47,899
to twenty four so x one and x two is the solution
of three x one plus three x two equal to twenty
85
00:11:47,899 --> 00:11:55,620
one four x one plus three x two equal to twenty
four now that is exactly the solution here
86
00:11:55,620 --> 00:12:00,620
which is between three x one plus three x
two equal to twenty one that gives us the
87
00:12:00,620 --> 00:12:06,529
three comma four that is the solution between
three x one plus three x two equal to twenty
88
00:12:06,529 --> 00:12:11,170
one and four x one plus three x two equal
to twenty four
89
00:12:11,170 --> 00:12:19,259
so fixing the non basic variable to zero gives
us the point of intersection of the lines
90
00:12:19,259 --> 00:12:27,149
which is the corner point if we had fixed
x three and x four to one say just for the
91
00:12:27,149 --> 00:12:34,029
sake of illustration and not fixed it to zero
then we would have solved for three x one
92
00:12:34,029 --> 00:12:40,500
plus three x two equal to twenty and four
x one plus three x two equal to twenty three
93
00:12:40,500 --> 00:12:45,420
and we would have got a different solution
and we would have got a point inside the feasible
94
00:12:45,420 --> 00:12:56,110
region so fixing the non basic variables to
any positive value is equivalent of evaluating
95
00:12:56,110 --> 00:13:05,230
a point inside the feasible region while fixing
the non basic variables to zero is equivalent
96
00:13:05,230 --> 00:13:11,839
of evaluating the corner point and from the
graphical method we have already learnt that
97
00:13:11,839 --> 00:13:19,170
points inside the feasible region need not
be evaluated now this implies that when we
98
00:13:19,170 --> 00:13:26,129
do the algebraic method it is enough to fix
the non basic variables only to zero and not
99
00:13:26,129 --> 00:13:33,470
fix them to any other value fixing them to
any other value would end up evaluating points
100
00:13:33,470 --> 00:13:41,339
which are inside the feasible region or fixing
them to negative values would mean evaluating
101
00:13:41,339 --> 00:13:45,870
infeasible points
therefore the understanding and the equivalent
102
00:13:45,870 --> 00:13:52,480
between the graphical and the algebraic method
tells us that since corner point solutions
103
00:13:52,480 --> 00:14:00,199
are optimal it is enough to fix the non basic
variables to zero in the algebraic method
104
00:14:00,199 --> 00:14:07,600
and not to fix them at any other positive
value so now we understand by considering
105
00:14:07,600 --> 00:14:14,379
the same example we understand that the graphical
method and the algebraic method essentially
106
00:14:14,379 --> 00:14:20,279
try to do the same thing but the graphical
method is simpler because we are used to drawing
107
00:14:20,279 --> 00:14:27,389
graphs and noti[cing]-noticing the corner
points but the graphical method has a restriction
108
00:14:27,389 --> 00:14:34,129
that it can be used only upto two variables
now when we have a large number of variables
109
00:14:34,129 --> 00:14:42,019
we can use the algebraic method to solve we
have illustrated the algebraic method by considering
110
00:14:42,019 --> 00:14:48,360
a two variable problem but we can also illustrate
the algebraic method by considering more variables
111
00:14:48,360 --> 00:14:55,540
if the original problem itself had three variables
and two constraints then when we add the slack
112
00:14:55,540 --> 00:15:01,220
variables we would have got five variables
and then two constraints so we have to solve
113
00:15:01,220 --> 00:15:06,850
for two at a time so we will have five c three
possible solutions we'll have more number
114
00:15:06,850 --> 00:15:14,189
of solutions as the number of variables and
constraints increase now out of these solutions
115
00:15:14,189 --> 00:15:20,279
we have to evaluate all of them and we have
to leave out the infeasible ones and look
116
00:15:20,279 --> 00:15:25,589
only at the basic feasible ones
to begin with we have to understand that we
117
00:15:25,589 --> 00:15:33,129
do not have a way to identify that something
is going to result in an infeasible solution
118
00:15:33,129 --> 00:15:43,860
so we have to evaluate all the solutions and
if we have n variables and m equations we
119
00:15:43,860 --> 00:15:58,279
end up creating n c m solutions in this example
n is equal to four there are four variables
120
00:15:58,279 --> 00:16:05,769
m is equal to two there are two constraints
four c two is equal to six now if we go back
121
00:16:05,769 --> 00:16:17,699
to
122
00:16:17,699 --> 00:16:24,749
the minimization problem now when we solved
the minimization problem we also had six solutions
123
00:16:24,749 --> 00:16:31,060
in the algebraic method and three of these
solutions were infeasible while three of these
124
00:16:31,060 --> 00:16:36,600
solutions were basic feasible
if we go back and look at the graph corresponding
125
00:16:36,600 --> 00:16:58,949
to the minimization problem
126
00:16:58,949 --> 00:17:05,750
now we realize that these are the three solutions
that are feasible four comma zero zero comma
127
00:17:05,750 --> 00:17:12,400
five and two by three comma ten by three these
were the solutions that are feasible the other
128
00:17:12,400 --> 00:17:19,400
three solutions that the algebraic method
evaluated were here zero comma zero two comma
129
00:17:19,400 --> 00:17:27,350
zero and zero comma four it evaluated and
these three became infeasible so out of these
130
00:17:27,350 --> 00:17:35,190
six solutions we looked at only three solutions
in this and those three solutions are the
131
00:17:35,190 --> 00:17:49,350
three corner points in the feasible region
now in summary we observe that the graphical
132
00:17:49,350 --> 00:17:56,289
method can be used only for a two variable
problem the algebraic method can evaluate
133
00:17:56,289 --> 00:18:03,730
a large number of solutions and sometimes
evaluates infeasible solutions we observed
134
00:18:03,730 --> 00:18:09,690
that for the maximization problem two out
of the six were infeasible for the minimization
135
00:18:09,690 --> 00:18:15,180
problem three out of the six were infeasible
so that leads to the next question do we have
136
00:18:15,180 --> 00:18:22,370
a method that evaluates only feasible solutions
does not evaluate infeasible solutions do
137
00:18:22,370 --> 00:18:30,120
we have a method that progressively evaluates
better solutions and do we have a method that
138
00:18:30,120 --> 00:18:37,390
can stop and tell us that the best solution
has been obtained even before evaluating all
139
00:18:37,390 --> 00:18:42,900
possible basic solutions
so we end this discussion on graphical and
140
00:18:42,900 --> 00:18:48,630
algebraic with this question that do we have
a method that can do these three things the
141
00:18:48,630 --> 00:18:52,249
answer to this question we will see in the
next class