MTH 242/HANDOUT 2/CSMJACOB " 16
MTH 242 – OPERATIONS RESEARCH CSMJACOB
HANDOUT #2 – Linear Programming: The Simplex Method
Simplex algorithm – a general procedure for solving linear programming (LP) problems developed by George Dantzig in
1947
– requires the LP model to be in standard form
Advantages and Characteristics of the simplex method
i. More realistic approach as it is not limited to problems with two decision variables
ii. Systematically examines basic feasible solutions for an optimal solution.
iii. Based on the solutions of linear equations (equalities) using slack variables to achieve equality.
Properties of the LP model in standard form:
i. All the constraints are equations with non-negative right-hand-side (RHS) values.
ii. All the variables are non-negative.
iii. The objective of the problem is to either maximize or minimize the objective function value.
Constraints
1. A constraint of the type can be converted into an equation by adding a slack variable to the left side of the constraint.
Ex.4x+5y 8 is converted into 4x+5y+s1=8 in the standard form, with s1 0.
If the constraint represents the limit on the availability of a resource, s1 will represent the slack or unused amount of that resource.
2. A constraint of the type can be converted into an equation by adding a surplus variable to the left side of the constraint.
Ex. 2x+4y-6z 9is converted into 2x+4y-6z-s2=9 in the standard form, with s2 0.
If the constraint represents the minimum production requirement, s2 will represent the items produced beyond the minimum required.
3. The RHS can be made non-negative by multiplying both sides by -1, and the direction of an inequality is reversed when both sides are multiplied by a negative number.
Ex.-3x+7y -15 is equivalent to 3x-7y 15 and is then converted into 3x-7y-s3=15 in the standard form with s3 0.
Objective Function
The maximization of a function is equivalent to the minimization of the negative of the same function. Equivalence means that for the same set of constraints, the optimum values of the variables are the same in both cases, although the values of the objective functions will appear with opposite signs but the same magnitude.
The Simplex Method
The simplex algorithm considers the corner point feasible (CPF) solutions as the candidates for the optimal mix. It starts at the origin, referred to as the starting solution, then moves to an adjacent corner point, the choice dependent on which point will improve the value of the objective function more. The iterative process is repeated by moving to another adjacent corner point, until the corner point where the value of the objective function is optimized, is finally reached.
The simplex algorithm is governed by two general rules:
1. The next corner point must be adjacent to the current one. Thus, moving from one corner point to the next involves moving along the edges of the solution area, and never across it.
2. The solution can never go back to a previously considered extreme point.
I. Maximization
Recall the example of the furniture manufacturer that produces wooden tables and chairs, formulated below:
Maximize: Z=6X+8Y
Subject to: 30X+20Y 300
5X+10Y 110
X, Y 0
The standard form of the LP model is given by:
Maximize: Z=6X+8Y+0S1+0S2
Subject to: 30X+20Y+S1=300
5X+10Y+S2=110
X, Y,S1,S2 0
The solution area is shown below, with the corner points denoted by A, B, C, and D, and every point in the solution space represented in terms of the variables X, Y, S1, and S2 of the standard form.
Y
30X + 20Y + S1 = 300
C
5X + 10Y +S2 = 110
D
X
B
A
Identifying the corner points algebraically:
Corner Point
Zero Variables
Nonzero Variables
A
B
C
D
X, Y
X, S2
S1, S2
Y, S1
S1, S2
Y, S1
X, Y
X, S2
From the table above, it can be observed that:
i. Since the standard form has two equations and four unknowns, each corner point must have two (4-2=2) variables at zero level. In general, given m equations and n variables, all feasible corner points are determined by considering all the unique non-negative solutions of the m equations in which n-m are set equal to zero. Such unique solutions are called basic solutions. If a basic solution satisfies the non-negativity constraints, it is called a basic feasible solution. The variables set equal to zero are called nonbasic variables, the remaining ones are called basic variables.
ii. Adjacent corner points differ only in one variable. Thus, movement from a current corner point to an adjacent one would involve interchanging a current nonbasic (zero) variable into a basic one. The basic-to-nonbasic interchange results in the following terms: the entering variable is a current nonbasic variable that will "enter" the set of basic variables at the next (adjacent corner point); the leaving variable is a current basic variable that will "leave" the basic solution in the next iteration.
Exercise:
Complete the constraints inequalities based on the shaded solution space below.
Maximize: Z = 3X + 2Y
Subject to: X + 2Y 6 1
2X + Y 8 2
-X + Y 1 3
Y 2 4
X, Y, 0
2
1
4
3
C
D
E
F
X
Y
B
A
2. Give the standard form of the LP model above.
3. Complete the table below.
Corner Point
Nonbasic Variables
Basic Variables
A
B
C
D
E
F
X, Y
5. Based on the graph of the solution space, determine the entering and leaving variables when the solution moves between the pairs of adjacent corner points shown below.
a. B C ___________ enters and ____________ leaves
b. C D ___________ enters and ____________ leaves
c. E F ___________ enters and ____________ leaves
Solving an LP Problem by the Simplex Algorithm
Step 0. Using the standard form, determine a starting basic feasible solution by setting n – m appropriate variables at
zero level.
Step 1. Select an entering variable from among the current nonbasic variables which, when increased above zero, can
improve the value of the objective function. If none exists, stop; the current basic solution is optimal. Otherwise,
go to step 2.
Step 2. Select a leaving variable from among the current basic variables that must be set to zero when the entering
variable becomes basic.
Step 3. Determine the new basic solution by making the entering variable basic and the leaving variable nonbasic. Go to
step 1.
Illustration:
A manufacturing firm produces two products, A and B. Each of these products must be processed through two different machines. One machine has 12 hours and the second machine has 8 hours of available capacity. Each unit of product A requires two hours of time on both machines. Each unit of product B requires three hours of time on the first machine and one hour on the second machine. The incremental profit is P300 per unit of product A and P350 per unit of product B, and the firm can sell as many units of each product as it can manufacture.
The objective of the firm is to maximize profits. The problem is to determine how many units of product A and product B should be produced within the limits of available machine capacities.
Formulation
Let X1 – number of units of product A to be produced
X2 – number of units of product B to be produced
Maximize: Z = 300X1 + 350X2 (objective function/profit function –computes the total profit with any given value of the basic variables X1 and X2)
Subject to: 2X1 + 3X2 12 (constraint on the limited availability of Machine I)
2X1 + X2 8 (constraint on the limited availability of Machine II)
X1 , X2 0 (non-negativity constraints)
The standard form of the model is:
Maximize: Z = 300X1 + 350X2 + 0S1 + 0S2 (Since any unused machine time does not contribute to profit, the coefficients of the slack variables in the objective function are zeros.)
Subject to: 2X1 + 3X2 + S1 = 12
2X1 + X2 + S2 = 8
X1 , X2, S1, S2 0
Illustration of the effect of the different production levels on the slack variable:
Quantity of Items Produced
Number of hours of Machine I used
Excess Available Time of Machine I
Product A
Product B
Product A
Product B
0
0
0
0
12 – 0 = 12 (S1 = 12)
1
1
2
3
12 – 5 = 7 (S1 = 7)
2
2
4
6
12 – 10 = 2 (S1 = 2)
Note: Only the functional constraints need to be transformed into equations. The non-negativity constraints are left as inequalities, since they are not affected by the simplex solution method.
Setting up the initial tableau
In the simplex method, an initial solution needs to be established. The simplex starting solution for the firm is to manufacture none of either product, that is, X1 = 0 and X2 = 0. This solution is technically feasible but not financially attractive. It is represented in the graphical solution by the origin with coordinates (0,0). Thus,
X1 = 0
X2 = 0
S1 = 12 – 2(0) – 3(0) = 12 hours unused for Machine I
S2 = 8 – 2(0) – 0 = 8 hours unused for Machine II
This solution contains only the slack variables S1 and S2. Substituting the values of the variables (X1 and X2) and the slack variables in the objective function gives the following profit:
Profit (Z) = 300X1 + 350X2 + 0S1 + 0S2
= 300(0) + 350(0) + 0(12) + 0(8)
= 0
This first feasible solution is shown in the initial simplex tableau as
Basic Variables
Quantity
X1
X2
S1
S2
S1
12
2
3
1
0
S2
8
2
1
0
1
The entries in the coefficient rows
2X1 + 3X2 + S1 = 12 come directly from the constraint
2X1 + X2 + S2 = 8 equations
Note that the first column contains the variables in the solution. The variables in the first solution are S1 and S2, the slack variables. In the quantity column are the quantities of the variables that are in the solution:
S1 = 12 hours of availability of Machine I
S2 = 8 hours of availability of Machine II
Since the variables X1 and X2 do not appear in the mix, they are equal to zero, and thus are nonbasic in the initial solution.
The partial initial simplex tableau is shown below:
1 2 3 4 5 6 7
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
0
S1
12
2
3
1
0
0
S2
8
2
1
0
1
- The Cj column contains the profit per unit for the variables in the solution. The initial solution includes only the slack variables S1 and S2, and since these contribute zero to profit, the entries in the Cj column of the initial tableau are zeros.
- Columns 6 and 7 consist of the coefficients of the slack variables that are added to the constraint inequalities to make them equations.
- Columns 4 and 5 consist of the coefficients of the nonbasic variables X1 and X2, and their elements represent rates of substitution. For example, the element 2 in the X1 column means that if one item of A is to be manufactured (to bring 1 item of A in the solution), then 2 hours of Machine 1 will have to be used (decreasing the value of S1 by 2). Similarly, the element 3 in the X2 column means that if one item of B is to be made (to bring 1 item of B in the solution), then 3 hours of Machine I will have to be used (decreasing the value of S1 by 3).
- The element 1 in the S1 column means that if 1 hour of Machine 1 is to be introduced into the solution, then 1 of the 12 hours of S1 in the solution will have to be given up.
- The element 0 in the S2 column means that introducing 1 hour of Machine II in the solution has no effect on the available hours of Machine I.
To find the profit for each solution and to determine whether the solution can still be improved, two more rows need to be added to the simplex tableau: a Zj row and a Cj – Zj row, as shown below.
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
0
S1
12
2
3
1
0
0
S2
8
2
1
0
1
Zj
0
0
0
0
0
Cj - Zj
300
350
0
0
The value in the Zj row under the quantity column represents the total profit from this particular solution - 0, in this case. In this solution, there are 12 hours of unused Machine I time (S1 = 12) and 8 hours of unused Machine II time (S2 = 8). The total profit from this solution is computed by multiplying the profit per unit of S1 (0) by the quantity of S1 in the solution (12 hours) plus the profit per unit of S2 (0) times the quantity of S2 in the solution (8 hours).
Total profit for the initial solution: (0)(12) + (0)(8) = 0
The four values for Zj under the variable columns (all 0) are amounts by which profit would be reduced if 1 unit of any of the variables (X1 , X2 , S1 , S2) were added to the mix. For example, if 1 unit of X1 is to be manufactured, the elements 22 under X1 mean that 2 hours of S1 and 2 hours of S2 must be used, decreasing the unused time. But unused time is worth 0 profit per hour; hence, there is no (0) reduction in profit. The computation for the amount of profit lost by introducing 1 unit of X1 into the product mix is shown below.
Number of hours of S1 given up = 2
times profit per unit of S1 × 0 = 0
Number of hours of S2 given up = 2
times profit per unit of S12 × 0 = 0
Total profit given up 0
The computation of Zj values are summarized below:
Zj (total profit)
=
0(12)
+
0(8)
=
0
Zj for column X1
=
0(2)
+
0(2)
=
0
Zj for column X2
=
0(3)
+
0(1)
=
0
Zj for column S1
=
0(1)
+
0(0)
=
0
Zj for column S2
=
0(0)
+
0(1)
=
0
Cj has been defined as profit per unit; for product A (X1), Cj is P300. Hence Cj – Zj is the net profit which will result from introducing (or adding) 1 unit of a variable to the production schedule (or the solution). For example, if 1 unit of product X1 adds P300 of profit to the solution and if its introduction causes 0 loss, then Cj – Zj = 300 – 0 = 300. Calculations of net profit per unit of each variable follow:
Variable
Profit per unit -
Profit lost per unit
= Net profit per unit
Cj -
Zj
= Cj - Zj
X1
300 -
0
= 300
X2
350 -
0
= 350
S1
0 -
0
= 0
S2
0 -
0
= 0
The values in the Cj – Zj indicate the amount by which profit would improve with the introduction of 1 units of a variable into the solution. A negative value gives the amount by which profits would decrease with the entry of 1 unit of a variable in the solution. It can be seen that the variable X2 will improve the profit the most.
Setting up the second tableau
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
0
S1
12
2
3
1
0
0
S2
8
2
1
0
1
Zj
0
0
0
0
0
Cj - Zj
300
350
0
0
Determine the entering variable from the Cj-Zj equation by choosing the variable with the most positive coefficient because this has the potential of improving the value of the objective function the most. The optimality condition of the simplex method states that, in the case of maximization, if all the nonbasic variables have negative coefficients in the Cj-Zj row of the current tableau, the current solution is optimal. Otherwise, the nonbasic variable with the largest coefficient is selected as the entering variable.
From the initial tableau above, it can be seen that 1 unit of product B (X2 = 1) contributes P350 to the profit, while 1 unit of product A (X1 = 1) contributes only P300 to profit. Thus, X2 should be introduced into the solution, making the X2 column the entering column. By definition, the optimal/entering column is the column which has the largest positive value in the Cj – Zj row, or the column whose product will contribute the most profit per unit.
To determine the leaving variable, the feasibility condition is used, which selects the leaving variable as the current basic variable that will be the first to reach zero level when the entering variable reaches its maximum value at the adjacent corner point. From the tableau, the leaving variable is the current basic variable associated with the smallest non-negative ratio of the RHS values to the corresponding values in the entering column. The row associated with the leaving variable is called the pivot equation and the element at the intersection of the entering column and the pivot equation is called the pivot element. Determining the smallest non-negative ratio involves dividing each variable's quantity by the corresponding coefficient in the entering column, and choosing the smallest quotient. For the tableau above:
S1 row: 12 hours of Machine I3 units of product B = 4 units of product B to finish up all of S1
S2 row: 8 hours of Machine II1 unit of product B = 8 units of product B to finish up all of S2
Since the S1 row has the smaller positive quotient, it is called the replaced row and it will be replaced by 4 units of X2 in the next solution. The elements of the optimal column are called the intersectional elements.
The second tableau, row by row (with solution) is shown below.
Cj
Basic Variables
Quantity
300
350
0
0
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
X1
X2
S1
S2
0
S1
12
2
3
1
0
350
X2
4
2/3
1
1/3
0
1 2
Figure 1 above shows the replaced row as it appears in the initial solution. Figure 2 shows the replacing row, as it appears in the second and improved solution, with the introduction of 4 units of product B (X2 = 4) in the solution. The values of the replacing row are computed by dividing each row element by the pivot, or the number common to both the entering column and the replaced row, in this case, 3. The Cj column now shows a value of 350, since that is the contribution to profit of 1 unit of X2.
Element in replaced row÷pivot=(Element in replacing row)
12÷3=4 2 ÷3=23 3÷3=1 1÷3=13 0÷3=0
Cj
Basic Variables
Quantity
300
350
0
0
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
X1
X2
S1
S2
350
X2
4
2/3
1
1/3
0
350
X2
4
2/3
1
1/3
0
0
S2
8
2
1
0
1
0
S2
a
b
C
d
e
Fig. 3 Fig. 4
Figure 3 above shows the old S2 row as it has not been developed yet for the second solution. The values in the new S2 row (a, b, c, d, e) in figure 4 will be computed using the formula:
Elements in new row= Elements in old row- Intersectional element ofold rowCorresponding element inreplacing row
a = 8 – (1)(4) = 8 – 4 = 4
b = 2 – (1)(2/3) = 2 – 2/3 = 4/3
c = 1 – (1)(1) = 1 – 1 = 0
d = 0 – (1)(1/3) = 0 – 1/3 = -1/3
e = 1 – (1)(0) = 1 – 0 = 1
The partial second tableau below shows the replaced row with the values computed above.
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
350
X2
4
2/3
1
1/3
0
0
S2
4
4/3
0
-1/3
1
As in the initial tableau, the computations of Zj values are summarized below:
Zj (total profit)
=
350(4)
+
0(4)
=
1400
= total profit of second solution
Zj for column X1
=
350(2/3)
+
0(4/3)
=
700/3
Profits given up by introducing
1 unit of each of these
Variables
Zj for column X2
=
350(1)
+
0(0)
=
350
Zj for column S1
=
350(1/3)
+
0(-1/3)
=
350/3
Zj for column S2
=
0(0)
+
0(1)
=
0
The computations above indicate that introducing a unit of product B would lose P350 for the firm. This is explained in the following statements:
1. The current solution shows a production level of 4 units of product B.
2. Since a unit of product B uses 3 hours of Machine I, 4 units of product B uses up all the available
hours of Machine I.
3. To introduce another unit of product B (that needs 3 hours of Machine I) means giving up 1 of the 4
units of product B currently in the solution.
4. Giving up a unit of product B means losing P350 in profit.
Based on the above statements, can you explain why introducing 1 unit of X1 (product A) would lose P700/3 in profit for the firm?
Calculations of net profit per unit of each variable follow:
Variable
Profit per unit -
Profit lost per unit
= Net profit per unit
Cj -
Zj
= Cj - Zj
X1
300 -
700/3
= 200/3
X2
350 -
350
= 0
S1
0 -
350/3
= -350/3
S2
0 -
0
= 0
The complete second tableau with the improved solution is shown below.
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
350
X2
4
2/3
1
1/3
0
0
S2
4
4/3
0
-1/3
1
Zj
1400
700/3
350
350/3
0
Cj - Zj
200/3
0
-350/3
0
Setting up the third tableau
A look at the Cj – Zj row of the second tableau shows that X1 (product A) contributes a net profit of P200/3 per unit. Thus, the entering column is the X1 column. Product A will now be added to the solution, replacing one of the variables X2 or S2.
The replaced row is again determined by dividing 4 and 4 in the quantity column with their corresponding elements in the entering column, and selecting the row with the smaller non-negative ratio as the replaced row. Since 4 ÷ 2/3 = 6 (in the X2 row) and 4 ÷ 4/3 = 3 (in the S2 row), the S2 row is designated as the replaced row.
The figure below shows the entering column, replaced row, and intersectional elements of the second tableau.
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
350
X2
2/3
300
S2
4
4/3
0
-1/3
1
Zj
Cj - Zj
Replacing the S2 row with X1 gives the replacing row below.
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
350
X2
a
b
c
d
e
300
X1
3
1
0
-1/4
3/4
Zj
Cj - Zj
The new values in the X2 row are
Elements in new row= Elements in old row- Intersectional element ofold rowCorresponding element inreplacing row
a = 4 – (2/3)(3) = 4 – 2 = 2
b = 2/3 – (2/3)(1) = 2/3 – 2/3 = 0
c = 1 – (2/3)(0) = 1 – 0 = 1
d = 1/3 – (2/3)(-1/4) = 1/3 + 1/6 = 1/2
e = 0 – (2/3)(3/4) = 0 – 1/2 = -1/2
The computations of Zj values for the third tableau are summarized below:
Zj (total profit)
=
350(2)
+
300(3)
=
1600
= total profit of third solution
Zj for column X1
=
350()
+
300(1)
=
300
Profits given up by introducing
1 unit of each of these
variables
Zj for column X2
=
350(1)
+
0(0)
=
350
Zj for column S1
=
350(1/2)
+
300(-1/4)
=
100
Zj for column S2
=
350(-1/2)
+
300(3/4)
=
50
Calculations of net profit per unit of each variable follow:
Variable
Profit per unit -
Profit lost per unit
= Net profit per unit
Cj -
Zj
= Cj - Zj
X1
300 -
300
= 0
X2
350 -
350
= 0
S1
0 -
100
= -100
S2
0 -
50
= -50
The third tableau below shows all the values computed above.
Cj
Basic Variables
Quantity
300
350
0
0
X1
X2
S1
S2
350
X2
2
0
1
1/2
-1/2
300
X1
3
1
0
-1/4
3/4
Zj
1600
300
350
100
50
Cj - Zj
-300
-350
-100
-50
Since the net profit row (Cj – Zj) no longer shows any positive value, indicating that the solution can no longer be improved upon, the third tableau shows the final optimum solution.
The final optimal solution is to produce X1 = 3 units of product A and X2 = 2 units of product B for a maximum profit of P1,600.
Illustration2:
Maximize: Z=6X1+8X2
Subject to: 30X1+20X2+S1=300
5X1+10X2+S2=110
X1, X2 0
Below is the initial tableau for the simplex algorithm.
Cj
Basic Variables
Quantity
X1
X2
S1
S2
Zj
Cj - Zj
The second iteration will result in the tableau below:
Cj
Basic Variables
Quantity
X1
X2
S1
S2
Zj
Cj - Zj
Complete tableau after iteration 3:
Cj
Basic Variables
Quantity
X1
X2
S1
S2
Zj
Cj - Zj
The tableau above is optimal because none of the nonbasic variables has a positive coefficient in the Cj – Zj row. This completes the simplex method computations after 3 iterations. Thus, the optimal solution is for Margan Furniture to produce X1 = 4 tables and X2= 9 chairs, for a maximum profit = $96.
For either maximization or minimization problems, the following conditions apply:
Optimality condition
The entering variable in maximization (minimization) is the nonbasic variable with the most positive (negative) coefficient in the Cj –Zj row. A tie is broken arbitrarily.
Feasibility condition
For both the maximization and minimization problems, the leaving variable is the basic variable having the smallest positive ratio. A tie is broken arbitrarily.
Exercise: Solve the following problem using the simplex method:
Maximize Z = 2X1– 4X2+ 5X3– 6X4
subject to
X1 + 4X2 – 2X3 + 8X4 2
-X1 + 2X2 + 3X3 + 4X4 1
X1 ,X2 , X3 , X4 0
I. Minimization
To illustrate the minimization procedure, the example below will be used:
Margan Furniture makes two products – tables and chairs, which must be processed through assembly and finishing departments. The assembly department is available for 60 hours in every production period, while the finishing department is available for 48 hours of work. Manufacturing one table requires 4 hours in assembly and 2 hours in finishing. Each chair requires 2 hours in assembly and 4 hours in finishing. The marketing director has promised customers that the company will make at least 2 tables and at least 4 chairs per production period. It has been determined that it costs P800 per unit to manufacture a table, and P320 per unit to manufacture a chair. It is desired to minimize the total manufacturing cost of tables and chairs.
Let X1 – the number of tables to be manufactured
X2 – the number of chairs to be manufactured
The objective is to minimize cost: Z = 800X1 + 320X2
Subject to: 4X1 + 2X2 60 (Maximum availability of the assembly department)
2X1 + 4X2 48 (Maximum availability of the finishing department)
X1 2 (Minimum supply of tables)
X2 4 (Minimum supply of chairs)
X1, X2 0
The standard forms of the constraints are:
4X1 + 2X2 + S1 = 60 (where S1 – excess hours in the assembly department) 2X1 + 4X2 + S2 = 48 (where S2 – excess hours in the finishing department)
X1 - S3 = 2 (where S3 - surplus supply of tables)
X2 - S4 = 4 (where S4 – surplus supply of chairs)
X1, X2, S1, S2, S3, S4 0
The initial solution where X1 = 0 and X2 = 0 produces the following equations:
S1 = 60
S2 = 48
- S3= 2 These violate the non-negativity constraints, as these are equivalent to
- S4 = 4 S3 = -2 and S4 = -4
To be able to provide an initial solution that will satisfy all constraints, artificial variables will be introduced. Since artificial variables have no feasible physical meaning from the standpoint of the original problem, adding them will be valid only if they are forced to be zero when the optimum is reached. Thus, they are used only to start the solution and must subsequently be forced to assume a zero value at the optimal solution. Consider the constraint:
X1 2.
Adding an artificial variable A3 will transform the constraint to:
X1 + A3 2
Transforming the constraint into an equation:
X1 + A3 – S3 =2
A3 serves as an artificial variable for the supply of tables. Hence, it can be thought of as a very expensive substitute table that the company can buy to satisfy the supply requirement, but is too expensive to include in the optimal solution. Thus, when X1 = 0, A3 – S3 = 2, and A3 = 2. (Why is S3 = 0? Interpret the value of A3 = 2.)
To ensure that the artificial variables do not appear in the optimal solution, they are given very large coefficients in the objective function for a minimization problem, and very small coefficients for a maximization problem.
The standard form of the LP model is shown below:
Minimize: Z = 800X1 + 320X2 + 0S1 + 0S2 +0S3+ 1000A3 + 0S4 + 1000A4
Subject to: 4X1 + 2X2 + S1 = 60 constraint 1
2X1 + 4X2 + S2 = 48 constraint 2
X1 + A3 - S3 = 2 constraint 3
X2 + A4 - S4 = 4 constraint 4
X1, X2, S1, S2, A3, A4, S3, S4 0
Since the solution always starts at the origin, where all decision variables are equal to zero, only the slack and artificial variables will be basic in the first iteration.
Note: For a constraint, add an artificial variable and subtract the surplus variable before changing it into an equation.
Iteration 1
Cj
Basic
Quantity
800
320
0
0
1000
0
1000
0
X1
X2
S1
S2
A3
S3
A4
S4
Ratios
0
S1
60
4
2
1
0
0
0
0
0
30
0
S2
48
2
4
0
1
0
0
0
0
12
1000
A3
2
1
0
0
0
1
-1
0
0
-
1000
A4
4
0
1
0
0
0
0
1
-1
4
Zj
6000
1000
1000
0
0
1000
-1000
1000
-1000
Cj - Zj
-200
-680
0
0
0
1000
0
1000
The optimal/entering column is the one that contains the most negative value, that is, the variable whose introduction to the solution will decrease costs most rapidly.
The optimal column in the initial tableau above is the X2 column (Cj - Zj = -680), and the replaced row will be theA2 row (ratio of Qty column and intersectional element = 4/1 = 4).
Iteration 2
Cj
Basic
Quantity
800
320
0
0
1000
0
1000
0
X1
X2
S1
S2
A3
S3
A4
S4
Ratios
0
S1
52
4
0
1
0
0
0
-2
2
13
0
S2
40
2
0
0
1
0
0
-4
4
20
1000
A3
2
1
0
0
0
1
-1
0
0
2
320
X2
4
0
1
0
0
0
0
1
-1
-
Zj
3280
1000
320
0
0
1000
-1000
320
-320
Cj – Zj
-200
0
0
0
0
1000
680
320
Iteration 3
Cj
Basic
Quantity
800
320
0
0
1000
0
1000
0
X1
X2
S1
S2
A3
S3
A4
S4
Ratios
0
S1
44
0
0
1
0
-4
4
-2
2
13
0
S2
36
0
0
0
1
-2
2
-4
4
20
800
X1
2
1
0
0
0
1
-1
0
0
2
320
X2
4
0
1
0
0
0
0
1
-1
-
Zj
2880
800
320
0
0
800
-800
320
-320
Cj – Zj
0
0
0
0
200
800
680
320
The above solution is optimal since the net profit row contains no more negative coefficients. Thus, the optimal decision is to manufacture 2 tables and 4 chairs for a minimum cost of P2,880. (Interpret the values of S1 and S2) .
PROBLEMS:
1. A steel producer makes two types of steel: regular and special steel. A ton of regular steel requires 2 hours in the open-hearth furnace and 3 hours in the soaking pit. A ton of special steel requires 2 hours in the open-hearth furnace and 5 hours in the soaking pit. The open-hearth furnace is available 8 hours per day and the soaking pit is available 15 hours per day. The profit on a ton of regular steel is P40,000 and it is P60,000 on a ton of special steel. Determine how many tons of each type of steel should be made daily to maximize the profit, considering that demand on regular steel is at least 1 ton per day.
2. A television producer designs a program that will include a comedy portion and advertisement. The advertiser insists on at least 15 minutes of advertising time. The producer insists on no more than 25 minutes of advertisement. The comedian insists on at least 70 minutes of the comedy portion. The total time allotted for the comedy and advertisement cannot exceed 2 hours. If it has been determined that each minute of advertising attracts one million viewers and each minute of the comedy program attracts two million viewers, how many minutes should be given to each of the comedy and advertisement in order to maximize the number of viewers?
3. The Latex Manufacturing Company produces aluminum frying pans and casserole dishes. Each frying pan and each casserole dish requires 40 ounces of aluminum. The company's daily supply of aluminum is limited to 560 ounces. Each frying pan requires 20 minutes on the casting machine and each casserole dish requires 40 minutes on the casting machine. The casting machine is available for 400 minutes daily. Each frying pan requires an insulated handle and only 12 of these are available each day. Each casserole dish requires two special pick-up handles and only 16 of these are available daily. Each frying pan contributes P250 to profit and each casserole dish contributes P200. The objective is to find the number of casserole dishes and frying pans to manufacture daily in order to maximize profit.
4. A poultry raiser wants to mix two types of grains: A and B. Each unit of grain A costs P400 and contains 29 grams of fat, 10 grams of protein, and 800 calories. Each unit of grain B costs P480 and contains 3o grams of fat, 30 grams of protein and 600 calories. Suppose that the poultry raiser wants each unit of the final product to yield at least 180 grams of fat, at least 120 grams of protein, and at least 4800 calories. How many units of each type of grain should the poultry raiser use to minimize his cost?
5. A trust fund is planning to invest up to P600,000 in two types of bonds: A and B. The preferred stock carries a dividend of 25%, and B 20%. Suppose the fund rules state that no more than P400,000 may be invested in bond B, while at least P150,000 must be invested in bond A, how much should be invested in each type of bond to maximize the fund's return?
6. Angeline's Snack Center makes ice milk and ice cream daily. Each quart of ice milk requires 0.2 quart of milk and 0.1 quart of cream, while each quart of ice cream requires 0.1 quart of milk and 0.2 quart of cream. The store has 5 quarts of milk and 7 quarts of cream each day. Assuming that it can sell all ice cream and ice milk on stock daily, how many quarts of ice cream and ice milk are to be made if a quart of ice milk gives 160 profit and a quart of ice cream gives 200, to realize a maximum profit?
7. Two machines A and B produce items at the rate of 500 per hour and 400 per hour, respectively. The production plan indicates that the total items to be produced by the two machines must number at least 10,000. The total number of hours that the two machines can be run is at most 24. Running machine A costs P300 per hour, while running machine B costs P450 per hour. How many hours should each machine be used to satisfy the production requirements, and at the same time, minimize the cost of operation?
8. In order to ensure optimal health (and thus accurate test results), a lab technician needs to feed the rabbits a daily diet containing a minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4 g of protein. But the rabbits should be fed no more than five ounces of food a day. Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y, and blend them for an optimal mix. Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of protein per ounce, and costs P8.00 per ounce. Food Y contains 12 g of fat, 12 g of carbohydrates, and 1 g of protein per ounce, at a cost of P12.00 per ounce. What is the optimal blend?
9. You have P480,000 to invest, and three different funds from which to choose. The municipal bond fund has a 7% return, the local bank's CDs have an 8% return, and the high-risk account has an expected (hoped-for) 12% return. To minimize risk, you decide not to invest any more than P80,000 in the high-risk account. For tax reasons, you need to invest at least three times as much in the municipal bonds as in the bank CDs. Assuming the year-end yields are as expected, what are the optimal investment amounts?
10. A tennis-playing golfer has $15 to spend on golf balls (x) costing $1 each and tennis balls (y) costing 60 c each. He must buy at least 16 altogether and he must buy more golf balls than tennis balls. (a) What is the greatest number of balls he can buy? (b) After using them, he can sell golf balls for 10c each and tennis balls for 20 c each. What is his maximum possible income from sales?
11. A man has a spare time job spraying cars and vans. Vans take 2 hours each and cars take 1 hour each. He has 14 hours available per week. He has an agreement with one firm to do 2 of their vans every week. Apart from that he has no fixed work. His permission to use his back garden contains the clause that he must do at least twice as many cars as vans. Let x be the number of vans sprayed each week. Let y be the number of cars sprayed each week.
(a) Write down three inequalities which must be satisfied
(b) Draw a graph and use it to list all possible combinations of vehicles which he can spray each week.
12. A travel agent has to fly 1000 people and 35000 kg of baggage from London to Paris. Two types of aircraft are available: A which takes 100 people and 2000 kg of baggage, or B which takes 60 people and 3000 kg of baggage. He can use no more than 16 aircrafts altogether.
(a) What is the smallest number of aircraft he could use?
(b) If the hire charge for each aircraft A is $10000 and for each aircraft B is $12000, find the cheapest option available to him.
(c) If the hire charges are altered so that each A costs $10000 and each B costs $20000, find the cheapest option now available to him.
13. A farmer needs to buy up to 25 cows for a new herd. He can buy either brown cows (x) at $50 each or black cows (y) at $80 each and he can spend a total of no more than $1600. He must have at least 9 of each type. On selling the cows he makes a profit of $50 on each brown cow and $60 on each black cow. How many of each sort should he buy for maximum profit?
14. A transport company has two types of trucks, Type A and Type B. Type A has a refrigerated Capacity of 20 m3 and a non-refrigerated capacity of 40 m3 while Type B has the same overall volume with equal sections for refrigerated and non-refrigerated stock. A grocer needs to hire trucks for the transport of 3000 m3 of refrigerated stock and 4000 m3 of non-refrigerated stock. The cost per kilometer of a Type A is $30, and $40 for Type B. How many trucks of each type should the grocer rent to achieve the minimum total cost?
15. A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers available. The rental cost for a large bus is $800 and $600 for the small bus. Calculate how many buses of each type should be used for the trip for the least possible cost.
16. A store wants to liquidate 200 of its shirts and 100 pairs of pants from last season. They have decided to put together two offers, A and B. Offer A is a package of one shirt and a pair of pants which will sell for $30. Offer B is a package of three shirts and a pair of pants, which will sell for $50. The store does not want to sell less than 20 packages of Offer A and less than 10 of Offer B. How many packages of each do they have to sell to maximize the money generated from the promotion?
