1. SOLVING LINEAR PROGRAMMING PROBLEMS: The Simplex Method

2. Simplex Method Used for solving LP problems will be presented
Put into the form of a table, and then a number of mathematical steps are performed on the table
Moves from one extreme point on the solution boundary to another until the best one is found, and then it stops
A lengthy and tedious process but computer software programs are now used easily instead
Programs do not provide an in-depth understanding of how those solutions are derived
Can greatly enhance one's understanding of LP

3. First Step First step is to convert model into standard form
s1 and s2, represent amount of unused labor and wood
No chairs and tables are produced, s1=40 and s2=120
Unused resources contribute nothing to profit, Z=0
Decision variables as well as profit at origin are:

4. Assigning (n-m) Variables Equal to Zero
Determine values of variables at every possible solution point
Have two equations and four unknowns, which makes direct simultaneous solution impossible
Assigns n-m variables=0
n=number of variables
m=number of constraints
Have n = 4 variables and m = 2 constraints
x1 = 0 and s1 = 0 and substituting them results in x2 = 20 and s2=60

5. Feasible and Basic Feasible Solution
Solution corresponds to A
Referred to as a basic feasible solution
Feasible solution is any solution that satisfies the constraints
Basic feasible solution not only satisfies the constraints but also contains as many variables with nonnegative values as there are model constraints
m variables with nonnegative values and n-m values equal to zero
X1=0 chairs
S1=0
X2=20 tables
S2=60 A B C

6. Feasible and Basic Feasible Solution
x2=0 and s2=0, results in x1=30 and s1=10
Corresponds to point C
s1=0 and s2=0, results in two equations with two unknown variables
Get a solution which corresponds to point B, where x1=24, x2=8, s1=0, and s2= 0
Previously identified as optimal solution point A B
X1=30 chairs
S1=10
X2=0 tables
S2=0 C

7. Two Questions
Two questions can be raised by the identification of solutions at points O, A, B, and C
How was it known which variables to set equal to zero?
How is the optimal solution identified?
Answers these questions by performing a set of mathematical steps
Determines at each step which variables should equal zero and when an optimal solution has been reached
X1=0 chairs
S1=0
X2=20 tables
S2=60 A
X1=24 chairs
S1=0
X2=8 tables
S2=0 B
X1=30 chairs
S1=10
X2=0 tables
S2=0 C

8. Initial Tableau Model is put into the form of a table, or tableau
Our maximization model
Z = 40x1 + 50x2 + 0s1 + 0s2
Subject to
x1 + 2x2+ s = 40
4x1 + 3x s2 =120
x1, x2 , s1, s2 ≥ 0
Initial simplex tableau with various column and row headings
Record the model decision variables, followed by the slack variables
Bottom rows represent constraint equations whose right-hand sides are given in the “solution” column
z-row is obtained from
z -40x1-50x2=0
Basic Z x1 x2 s1 s2
Solution 1
-40
-50 2 40 4 3
120

9. Determining Basic Feasible Solution
Determine a basic feasible solution
Know which two variables will form the basic feasible solution and which will be assigned a value of zero?
Selects the origin as the initial basic feasible solution
x1= 0 and x2= 0; variables in basic feasible solution are s1 and s2 which are listed under column "Basic" and their values, 40 and 120, are listed under column “solution“
Z=0 under column “solution”
Basic Z x1 x2 s1 s2
Solution 1
-40
-50 2 40 4 3
120

10. Entering Variable
Interested in giving us either some chairs or some tables
Want nonbasic variables will enter solution and become basic
z-row values represent increase per unit of entering a nonbasic variable into the basic solution
Make as much money as possible
Select variable x2 as the entering basic variable
has the greatest increase in profit per unit, $50—the most negative value in the z-row
x2 column is highlighted and is referred to as the pivot column
Referred to in mathematical terminology as pivot operations
Basic Z x1 x2 s1 s2
Solution 1
-40
-50 2 40 4 3
120

11. Entering Variable by Graphical Solution
Demonstrated by the graph
At the origin nothing is produced
Moves from one solution point to an adjacent point
Move along either the x1 axis or the x2 axis
Chooses x2 as the entering variable
X1=0 chairs
X2=20 tables A
X1=24 chairs
X2=8 tables B
X1=30 chairs
X2=0 tables O C

12. Leaving Basic Variable
One of s1 or s2 has to leave and become zero
Produce tables as many as possible
Check availability of our resources
Using labor constraint, maximum we can produce table is 20
Enough labor is available to produce 20 tables
Using wood, maximum we can produce is 40
Limited to 20 tables
Point A is feasible, whereas point R is infeasible
Wood constraint A
Labor constraint B C

13. Leaving Variable
Determined by dividing RHS values by pivot column values
Leaving basic variable is variable with minimum nonnegative ratio
s1 is leaving variable
s1 row is referred to as pivot row
Value of 2 is called pivot number or pivot element
Entering variable, x2, in the new solution also equals 20
Z also increases from 0 to 1000
Basic Z x1 x2 s1 s2
Solution 1
-40
-50 2 40 4 3
120
Basic x2
Solution
Ratio S1 2 40
40/2=20 s2 3
120
120/3=40

14. Second Tableau Nonbasic and basic variables at solution point
Nonbasic: x1 and s1
Basic: x2 and s2
Various values are computed using pivot row and all other rows
First, x2 row, is computed by using every value in pivot row of first (old) tableau and divide it by pivot number, or
(0/2 1/2 2/2 1/2 0/2 40/2)
New row values are shown in Table
Basic Z x1 x2 s1 s2
Solution 1
-40
-50
1/2 20 4 3
120

15. Second Tableau-Cont. For the remaining row values, use
0-(3)*0=0
4-(3)*1/2=-5/2
3-(3)*1=0
0-(3)*1/2=--3/2
1-(3)*0=1
120-(3)*20=60
Second Tableau-Cont.
1-(-50)*0=1
-40-(-50)*1/2=--15
-50-(-50)*1=0
0-(-50)*1/2=25
0-(-50)*0=0
0-(-50)*20=1000
For the remaining row values, use
New row=(current row)-(its pivot column coefficient) x (New pivot row)
Requires use of both old tableau and new one
Various values for new z-row and s2-row:
Z-row= current z-row – (-50)x new pivot row
S2-row=current s2-row – (3)x new pivot row
Values are inserted in Table
Basic Z x1 x2 s1 s2
Solution 1
-15 25
1000
1/2 20
5/2
-3/2 60
Basic Z x1 x2 s1 s2
Solution 1
-40
-50 2 40 4 3
120

16. Checking Optimal Solution
Set nonbasic variables x1 and s1 to zero, solution column provides new basic solution x2=20, s2=60 and z=1000
Is not optimal
x1 has a negative coefficient
Select variable x1 as entering basic variable
Ratio computations, show that s2 is the leaving variable
Steps are repeated to develop the third tableau
Pivot row, pivot column, and pivot number are indicated in Table
Basic Z x1 x2 s1 s2
Solution 1
-15 25
1000
1/2 20
5/2
-3/2 60
Basic x1
Solution
Ratio x2
1/2 20
20/0.5=40 s2
5/2 60
60/2.5=24

17. Third Tableau New pivot row, x1, is computed using the same formula
All old pivot row values are divided by 5/2
Values for z-row and s2-row are computed here and shown in Table
Basic Z x1 x2 s1 s2
Solution 1 16 6
1360
4/5
-1/5 8
-3/5
2/5 24
Basic Z x1 x2 s1 s2
Solution 1
-15 25
1000
1/2 20
5/2
-3/2 60

18. Checking Optimal Solution
Coefficient of nonbasic variables are positive
Optimal solution has been reached
x1 = 24 chairs, x2= 8 tables and z= $1,360 profit
Corresponds to point B shown previously
Basic Z x1 x2 s1 s2
Solution 1 16 6
1360
4/5
-1/5 8
-3/5
2/5 24

19. Summary of the Simplex
Transform constraint inequalities into equations
Set up the initial tableau
Determine the pivot column (entering nonbasic solution variable)
Determine the pivot row (leaving basic solution variable)
Compute the new pivot row values
Compute all other row values
Determine whether or not the new solution is optimal
If these coefficients are zero or positive, the solution is optimal
If a negative value exists, return to step 3 and repeat the simplex steps

20. Simplex Tableaus by TORA
Used manually to solve relatively small problems
Can be so time-consuming and subject to error that manual computation
Computer solution has become so important
Many computer programs use the simplex method
Computer output includes option to display simplex tableaus
Simplex tableaus provided by TORA software

21. Minimization Problem
Demonstrated simplex method for a maximization problem
A minimization problem requires a few changes
Recall the minimization model
minimize Z = 6 x1 + 3 x2
subject to
2 x1 + 4 x2 ≥ 16
4 x1 + 3 x2 ≥ 24
x1 and x2 ≥ 0
Transformed this model into standard form by subtracting surplus variables
2 x1 + 4 x2 – s1 = 16
4 x1 + 3 x2 – s2 = 24

22. Introducing Artificial Variable
Simplex method requires initial basic solution at the origin
Test this solution at origin
Violates the non-negativity restriction
Reason is shown in Figure
Solution at origin is outside feasible solution space
Add an artificial variable (R1) to the constraint equation
Create an artificial positive solution at the origin
Artificial solution helps get the simplex process started
Not want it to end up in the optimal solution
Our phosphate constraint becomes
2 x1 + 4 x2 – s1 + R1= 16 A
Feasible area B C

23. Effect of Surplus and Artificial Variables on Objective Function
Surplus variable has no effect on objective function
0 is assigned to each surplus variable
Must also ensure that an artificial variable is not in the final solution
Achieved by assigning a very large cost
Assign a value of M, which represents a large positive cost
Produces objective function:
Minimization model can now be summarized as
minimize Z = 6 x1 + 3 x2 + MR1 + MR2
subject to 2 x1 + 4 x2 – s1 + R1 = 16
4 x1 + 3 x2 – s2 + R2 = 24
x1, x2, s1, s2 , R1, R2 ≥ 0.

24. Initial Tableau Developed in the same way
Use R1 and R2 as a starting basic solution
x1=x2=s1=s2=0 and R1=16 and R2=24
Notice that the origin is not in the feasible solution area
Artificially created solution
Simplex process moves toward feasibility in subsequent tableaus
Decision variables are listed first, then surplus variables, and finally artificial variables
z=24xM+16xM=40M instead of 0
Explained in the subsequent slides
Basic Z x1 x2 s1 s2 R1 R2
Solution 1 -6 -3 -M 2 4 -1 16 3 24

25. Modification in Initial Tableau
Inconsistency exists because R1 and R2 coefficient have nonzero coefficient
Eliminate this inconsistency by substituting out R1 and R2 in the z-row
We can write R1 =16-2 x1 -4 x2 +s1
R2 = 24-4 x1 -3 x2 + s2
Substituting R1 and R2 in objective function
Z = 6 x1 + 3 x2 + M (16-2 x1 -4 x2 +s1 )+ M(24-4 x1 -3 x2 + s2 )
Z = 40M+x1 ( 6-6M)+x2 (3-7M) +Ms1+ Ms2
Transforming objective function into standard form
Z +x1 ( -6+6M)+x2 (-3+7M)-Ms1- Ms2= 40M
Must be used as the z-row

26. Entering and Leaving Variables
Modified tableau thus becomes
z=40M, which is consistent now with values of the starting basic feasible solution R1=24 and R2=16
x2 column is selected as the entering variable because 7M - 3 is the largest value in z-row
R1 is selected as leaving basic variable because the ratio of 4
Second tableau is developed using simplex formulas
Basic Z x1 x2 s1 s2 R1 R2
Solution 1
6M-6
7M-3 -M
40M 2 4 -1 16 3 24

27. Second Tableau Second tableau is shown in Table
R1 row has been eliminated
Once an artificial variable leaves basic feasible
Will never return
Basic Z x1 x2 s1 s2 R1 R2
Solution 1
2.5M-4.5
0.75M-0.75 -M
-1.75M+0.75
12M+12
0.5
-0.25
0.25 4
2.5
0.75 -1
-0.75 12

28. Third Tableau Third tableau starts replacing R2 with x1
R1 and R2 rows have been now eliminated
x1 row was selected as the pivot row
-4 value for the x2 row was not considered

29. x1 x2 s1 s2 R1 R2 2.5 x1 x2 s1 s2 R1 R2 Basic Z Solution 1 2.5M-4.5
0.75M-0.75 -M
-1.75M+0.75
12M+12
0.5
-0.25
0.25 4
2.5
0.75 -1
-0.75 12
Basic Z x1 x2 s1 s2 R1 R2
Solution 1
0.6
-1.8
-M-0.6
M-1.8
33.6
-0.4
0.2
0.4
-0.2
1.6
0.3
-0.3
4.80

30. Forth Tableau (Optimal Solution)
Starts replacing s1 with x1
Turns out into final tableau which gives optimal solution
z-row contains no positive values
Optimal solution:
x1=0, s1=16, x2=8, s2= 0, and Z = $24

31. x1 x2 s1 s2 R1 R2 0.3 x1 x2 s1 s2 R1 R2 Basic Z Solution 1 0.6 -1.8
0.6
-1.8
-M-0.6
-M+1.8
33.6
-0.4
0.2
0.4
-0.2
1.6
0.3
-0.3
4.80
Basic Z x1 x2 s1 s2 R1 R2
Solution 1 -2 -1 -M
-M+1 24
1.33
-0.33
0.33 8
3.33
-1.33 16

32. Mixed Constraints LP Problems
Discussed maximization problems with all “≥” constraints and minimization problems with all “≤” constraints
Solve a problem with a mixture of “≤”, “≥”, and “=“ constraints
A maximization problem with “≤”, “≥”, and “=“ constraints
Max z=$400x1+200x2
Subject to
x1+x2=30
2x1+8x2≥80
x1≤20
x1, x2≥0

33. First Step Transform the inequalities into equations
First constraint is an equation
Not necessary to add a slack variable
Test it at the origin
Constraint is not feasible in this form
“≥” constraint did not work at origin either
Same thing can be done here
At the origin, x1=0 and x2=0, R1= 30; R1=30
Artificial variable cannot be assigned a positive value of M in objective function
A positive M value would definitely end up in final solution
Must give artificial variable a large negative value, or —M
Other constraints are transformed into equations
Completely transformed LP problem
Max z=$400x1+200x2-MR1-MR2
s.t. x1+x2+R1=30
2x1+8x2-s1+R2=80
x1+s2=20

34. Initial Tableau Initial tableau is shown in Table
Basic solution variables are a mix of artificial and slack variables
Second, third, and optimal tableaus are shown in following slides

35. Initial tableau Basic Z x1 x2 s1 R1 R2 s2 solution -1 -3M-400 -9M-200
-110M 1 30 2 8 80 20

36. Second and Third Tableau
Basic Z x1 x2 s1 R1 R2 s2
solution -1
-0.75M-350
-1.125M-25
-20M+2000
0.75
0.125 1
-0.125 20
0.25 10
Basic Z x1 x2 s1 R1 R2 s2
solution -1
-0.125M-25
1.125M-25
0.75M+350
-5M+9000
0.125 1
-0.125
-0.75 5
-0.25 20

37. Optimal Tableau Solution is: x1=20, x2=10, s1=40 Z=10,000 Basic Z x1R1 R2 s2
solution -1
M+200 M
200
10000 1 8 -6 40 10 20

38. Transforming Rules
Rules for transforming all three types of model constraints
Are as follows
Objective Function Coefficient
Constraint
Adjustment
Max
Min ≤
Add a slack variable =
Add an artificial variable -M M ≥
Subtract a surplus variable
add an artificial variable

39. Simplex Method of the 2nd Example
Steps:
Starts at origin, point A
Moves to an adjacent corner point, either B or F
Because of higher coefficient of x1, it moves toward corner point B
At B, the process is repeated for higher z value
Finds point C (the final one)
Three iterations E D C F A B
Max z=3x1+2x2
s.t.
x1 +2x2 6
2x1+x2 8
-x1+x2 1
x2 2
x1, x20 A B C

40. Basic and Nonbasic Variables
Basic and nonbasic variables associated with two points A and B
At point B, s2 enters as a nonbasic variable and x1 leaves as a basic variable
Corner point
Nonbasic variables
Basic variables A B
x1,x2
s2,x2
s1,s2,s3,s4
s1, x1,s3,s4 A B

41. The Standard Form and the General Concept
m equations (m=4)
n unknowns (n=6)
Sets (n-m) variables to zero and then solves for the remaining m unknowns
The variables set equal to zero are called nonbasic variables. The remaining ones are called basic variables
At point A, x1 =x2=0 which yields s1 =6, s2 =8, s3=1, and s4=2
At point B, s2 =x2=0, which yields x1 =4, s1 =2, s3=5, and s4=2
Max
z=3x1 +2x2+0s1 +0s2 +0s3 +0s4
s.t.
x1 +2x2+s1 =6
2x1 +x2 +s2 =8
-x1 +x s3 =1
x s4=2
x1,x2,s1,s2,s3,s40 A B

42. The Simplex Method in Tabular Form
z-3x1 -2x2+0s1 +0s2 +0s3 +0 s4 =0 Obj fcn
x1 +2x2+s1 =6
2x1 +x s2 =8
-x1 +x2 +s3 =1
x2 +s4 =2
Constraint
Equations

43. Initial Tableau s2 s3 s4 z -3 -2 1 2 -1 6 8
Basic x1 x2 s1 s2 s3 s4
Solution z -3 -2 1 2 -1 6 8
The basic column identifies the current basic variables s1, s2, s3, and s4
The nonbasic variables x1 and x2 are not present in the basic column and are equal to zero
The value of Obj Fcn is z=3*0+2*0+6*0+8*0+1*0+2*0=0

44. Example: Basic x1 x2 s1 s2 s3 s4 Solution z -3 -2 0 Ratio 1 2 -1
Ratio 1 2 -1
/1=6
/2=4 2
The column associated with the entering variable is called entering column
The row associated with leaving variable is called pivot equation
The element at the intersection of the entering column and pivot row is called pivot element

45. Example: Type 1 Computation
Basic x1 x2 s1 s2 s3 s4
Solution z 1
1/2
8/2=4

46. Complete New Tableau x1 =4, x2=0, and z=12 (point B) Basic x1 x2 s1 s2
Solution z
-1/2
3/2
Ratio 1 -0
1/2 /3 /3
3/2
x1 =4, x2=0, and z=12 (point B)
Nonbasic variables: x2 and s2 (zero variables)
Basic variables: s1, x1, s3, and s4
(nonzero variables)
x2 enters the solution
s1 leaves the solution
The obj fcn can be improved by 4/3*1/2=2/3

47. Iteration 3-Optimal s4 x1 =4/3, x2=10/3, and z=38/3 (point C) Basic x1
Solution z
1/3
4/3
12 2/3 1
2/3
-1/3 -1
-2/3
10/3 3
New pivot s1-Equation=old s1-Equation /1.5
New z-Equation=old z-Equation – (-1/2) *new pivot equation
New x1 -Equation=old x1 -Equation – (1/2) * new pivot equation
New s3 -Equation=old s3 -Equation – (3/2) * new pivot equation
New s4
-Equation=old s4-Equation – (1) * new pivot equation
x1 =4/3, x2=10/3, and z=38/3 (point C)

48. Interpreting the Simplex Tableau
The optimum solution
The status of the resources
The unit worth of the resources
The sensitivity of the optimum solution to changes in the availability of resources, coefficients of the obj fcn, and usage of resources by activities
The first three are in optimal solution of the simplex tableau
Forth one needs additional computations

49. Summary
Simplex method was introduced as algebraic procedure for solving LP problems
Described how initial tableau of a linear program is a necessary step in the simplex solution procedure, including (≤), (≥), and (=) constraints into tableau form
Discussed how the special cases of infeasibility, unboundedness, multiple optimal solution, and degeneracy can occur with the simplex method

50. Special Cases in the Simplex Method
Special types of LP problems need special attentions
Special types:
More than one optimal solution
Infeasible problems
Unbounded solutions
Ties for pivot column and/or ties for the pivot row
Constraints with negative quantity values

51. Multiple Optimal Solutions
Recall maximization problem
Alter objective function z=40x1+50x2 To z=40x1+30x2
Makes it parallel to the constraint line
4x1+ 3x2 = 120
Endpoints B and C, are referred to as alternate optimal solutions
Objective function A B C

52. Multiple Solutions from the Simplex Tableau
Not in solution
But has a 0 coef.
The optimal simplex tableau is:
How do we know that multiple optimal solution exist?
Determined from the row Z
Negative row Z values represent profit
Values of zero indicate no increase or loss
Coefficients of x1and s1 have zero values
Coefficient of x2 has a value of zero and it is not part of basic solution
Basic Z x1 x2 s1 s2
Solution 1 10
1200
1.25
-0.25
0.75
0.25 30

53. Multiple Solutions from the Simplex Tableau (cont.)
If x2 enters solution, we will have a new solution without changing z
Multiple optimal solution is indicated by a value of zero in row Z for a nonbasic variable
To determine alternate solution, select x2 as the entering variable and then determine pivot row as usual
Basic Z x1 x2 s1 s2
Solution 1 10
1200
1.25
-0.25
0.75
0.25 30
Basic Z x1 x2 s1 s2
Solution 1 10
1200
0.8
-0.2 8
-0.6
0.4 24

54. The Infeasible Solution
Special case where a LP has not feasible solution area
Max Z=5x1+3x2
s.t. 4x1+2x2<=8
x1>=4
X>=6
Because no point satisfies all three constraints simultaneously, there is no solution.

55. The Infeasible Solution (cont.)
All values are zero or positive indicating that it is optimal
Solution: x2=4, R1=4, and R2=2
Existence of artificial variables in final tableau makes the solution meaningless
Occur as a result of making errors in defining problem correctly or in formulating the LP problem
Basic Z x1 x2 s1 s2 s3 R1 R2
Solution 1
M+1 M
0.5M+1.5
-0.6M+12 2
0.5 4 -1 -2
-0.5

56. Unbounded Problems Occur where feasible solution area is not closed.
Objective function increases indefinitely without ever reaching a maximum value
Never reaches boundary of the feasible solution area
Z=4x1+2x2
Subject to
x1≥4
x2≤2
x1, x2≥0

57. Unbounded Problems (cont.)
First tableau
Basic Z x1 x2 s1 s2 R1
Solution 1
M-4 -2 M
-4M -1 4 2
In second tableau, s1 is chosen as the entering variable because it has most negative coefficient in row Z
There is no leaving variable because one row value is -4 and the other is undefined
Indicates that solution is unbounded
A solution is unbounded if row value ratios are all negative or undefined
Occurs as a result of making errors in defining problem or in formulating LP problem
Second tableau
Basic Z x1 x2 s1 s2 R1
Solution 1 -2 -4
M+4 16 X1 -1 4 2

58. Degeneracy Possible to have a tie for pivot row
A tie should be broken arbitrarily
Other choice for leaving variable will posses a zero value in next tableau
Solution is said to be degenerate
Z value never improves
An example of a degeneracy condition
Maximize Z = 4x1+ 6x2
subject to
6x1+ 4x2 ≤ 24
x2 ≤ 3
5x1+10x2 ≤ 40

59. Tie for the Pivot Row
In second tableau, we will see a tie for pivot row between s1, and s3 rows
S3 row is selected arbitrarily, the resulting third tableau will be shown
Basic Z x1 x2 s1 s2 s3
Solution 1 -4 6 18 12 3 5
-10 10
S1 ratio: 12/6 =2
S3 ratio: 10/5 =2

60. Loop in Solutions A value of zero
for s1 indicates a degeneracy condition
s2 is entering variable and s1 as pivot row
Basic Z x1 x2 s1 s2 s3
Solution 1 -2
4/5 26 8
-6/5 3
1/5 2

61. Loops in Solutions (Cont.)
Optimal solution did not change
Graphical analysis of this problem is shown in the next slide
Basic Z x1 x2 s1 s2 s3
Solution 1
1/4
1/2 26
1/8
-3/20
-1/8
3/20 3
-1/10 2

62. Graphical Analysis
The intersection causes the tie for pivot row and the degeneracy
Simplex process stays at point B
Degeneracy occurs when a problem continually loops back to same solution
Two tableaus represent two different basic feasible solutions with two different model constraint equations B

63. HW Assignments 1. Problem set 3.2A, problem 2, part a, b, and c.
2. Given the following linear programming model:
Min Z=4x1+x2
s.t.
3x1+6x2>=15
8x1+2x2>=12
x1, x2>=0
Solve graphically and using the simplex method.
What type of special case is this problem? Explain?
3. Transform the following linear programming model into proper form and setup the initial tableau. Do not solve it.
Min Z=40x1+55x2+30x3
x1+2x2+3x3 <=60
2x1+x2+x3 = 40
x1+3x2+x3>=50
5x2+x3>=100
x1, x2, x3>=0
4. Given the following linear programming model:
Max Z=x1+2x2-x3
4x2+x3<=40
x1-x2<=20
2x1+4x2+3x3<=60
Solve this problem using the simplex method.