Simplex part i

02/11/13 1FOSTIIMA Business School

STEP BY STEP GUIDE1. Formulate the problem

(i) Pick out important information(ii) Formulate constraints(iii) Formulate objective function

2. Introduce slack variables

3. Form initial tableau

4. Obtain new tableaux(i) Identify pivotal column(ii) Find θ-values(iii) Identify pivotal row(iv) Identify pivot(v) Pivot

5. Get the solution 02/11/13 2FOSTIIMA Business School

THE PROBLEM

• A small factory produces two types of toys: cars and dolls. In the manufacturing process two machines are used: the moulder and the assembler. A doll needs 2 hours on the moulder and 1 hour on the assembler. A car needs 1 hour on the moulder and 1 hour on the assembler. The moulder can be operated for 16 hours a day and the assembler for 9 hours a day. Each doll gives a profit of Rs.16 and each car gives a profit of Rs.14. The profit needs to be maximised.

• How do we formulate this problem?



(i) pick out important information(ii) formulate constraints(iii) formulate objective function



4. Obtain new tableaux

5. Get the solution


PICKING OUT IMPORTANT INFORMATION

• A small factory produces two types of toys: cars and dolls. In the manufacturing process two machines are used: the moulder and the assembler.

• A doll needs 2 hours on the moulder and 1 hour on the assembler. A car needs 1 hour on the moulder and 1 hour on the assembler.

• The moulder can be operated for 16 hours a day and the assembler for 9 hours a day.

• Each doll gives a profit of Rs.16 and each car gives a profit of Rs.14.



(i) pick out important information(ii) formulate constraints(iii) formulate objective function




5. Get the solution




• Using the decision variables

d = number of dolls c = number of cars

make two constraints from this information.02/11/13 11FOSTIIMA Business School

FORMING CONSTRAINT 1THE MOULDER



2d + c ≤ 1602/11/13 12FOSTIIMA Business School

FORMING CONSTRAINT 2The assembler



d + c ≤ 902/11/13 13FOSTIIMA Business School


i) pick out important informationii) formulate constraintsiii) formulate objective function




5. Get the solution


FORMING THE OBJECTIVE FUNCTION


• Let Z be the total profit; formulate the objective function


FORMING THE OBJECTIVE FUNCTION


Z = 16d + 14c


THE LINEAR PROGRAMMING PROBLEM

• MAXIMISE Z = 16d + 14csubject to the constraints:

(i) 2d + c ≤ 16(ii) d + c ≤ 9(iii) c ≥ 0 , d ≥ 0

• VERY IMPORTANT

• DON’T FORGET YOUR NON – NEGATIVITY CONSTRAINTS !






5. Get the solution


INTRODUCING SLACK VARIABLES

To change inequalities (i) and (ii) into equations we add slack variables s and t

This gives:

(i) 2d + c + s = 16

(ii) d + c + t = 9


THE NEW LINEAR PROGRAMMING PROBLEM

• MAXIMISE Z = 16d + 14c + 0s + 0t

subject to the constraints:

2d + c + s + 0t = 16

d + c + 0s + t = 9

c ≥ 0 , d ≥ 0 , s ≥ 0 , t ≥ 0






5. Get the solution


We want to put all the information in the form of a table. This is called the initial tableau.

To form the initial tableau we need tochange the objective function from

Z = 16d + 14c + 0s + 0tto

Z – 16d – 14c – 0s – 0t = 002/11/13 23FOSTIIMA Business School

FORMING THE INITIAL TABLEAU

Label the table with your basic variables, s and t

and with your non – basic variables, d and c.

BASIC VARIABLES

d c s t VALUE

s

t

Z 02/11/13 24FOSTIIMA Business School


2d + 1c + 1s + 0t = 161d + 1c + 0s + 1t = 9

Z – 16d – 14c – 0s – 0t = 0BASIC

VARIABLESd c s t VALUE

s

t

Z

2 16011

1 1 0 1 9

-16 -14 0 0 002/11/13 26FOSTIIMA Business School


This is the objective row

Z

t

s

VALUEtscdBASIC VARIABLES

2 16011

1 1 0 1 9

-16 -14 0 0 0






3. Get the solution


PIVOTAL COLUMN• We now need to find where to pivot and we start by

entering the basis by choosing the column with the most negative entry in the objective row.

Z

t

s


2 16011

1 1 0 1 9

-16 -14 0 0 0

This is the most negative coefficient with corresponding variable d and it’s column is called the pivotal column. d is now called the entering variable.






3. Get the solution


FINDING θ-VALUES

• You are now going to find the pivotal row and the leaving variable.

• You need to find θ-values. 1. Identify positive entries in the pivotal column.2. Divide each entry in value column by the corresponding

positive entry in the pivotal column.

Z

t

s


2 16011

1 1 0 1 9

-16 -14 0 0 0






5. Get the solution


PIVOTAL ROW

• The row with the smallest θ-value is called the pivotal row.

• Here the pivotal row is row (i)

162 8θ = =

91 = 9θ =

• For row (i)

• For row (ii)






5. Get the solution


THE PIVOT

The pivot!

Z

t

s


2 16011

1 1 0 1 9

-16 -14 0 0 0

The pivotal column

The pivotal row






5. Get a solution


PIVOTING

1. Replace the leaving variable with the entering variable.

2. Divide all entries in the pivotal row by the pivot. The pivot becomes 1.

3. Add suitable multiples of the pivotal row to all other rows until all entries, apart from the pivot, in the pivotal column are zero.


Step 1 - Replace the leaving variable with the entering variable.

Z

t

s


2 16011

1 1 0 1 9

-16 -14 0 0 0

d

Z

t

801/21/21s


Step 2 - Divide all entries in the pivotal row by the pivot. The pivot becomes 1.


t

PIVOTING

Step 3 - Add suitable multiples of the pivotal row to all other rows until all

entries, apart from the pivot, in the pivotal column are zero.

row (ii) – ½ row (i)gives

Z

t

s


2 16011

1 1 0 1 9

-16 -14 0 0 0

0 1/2 1 1-1/2


PIVOTING

Z

t

801/21/21d


t 0 1/2 -1/2 1 1


-16 -14 0 0 0

16 8 8 0 128x160 -6 8 0 128

PIVOTING

Z

t

801/21/21d


t 0 1/2 -1/2 1 1


This is our second tableau

Z

t

801/21/21d


t 0 1/2 -1/2 1 1

Z 0 -6 8 0 128


PIVOTING

• Follow the rules for finding a pivot on your second tableau.

• Pivot as before.

• Continue this process until there are no negative entries in the objective row.

• This will be your final tableau. This is called the optimal tableau.


Z

t

801/21/21d


t 0 1/2 -1/2 1 1

Z 0 -6 8 0 128

⇓

⇒

d 1 0 1 -1 7

c 0 1 -1 2 2

Z 0 0 2 12 140

OPTIMAL TABLEAU

• Note there are no negative entries in the objective row.

• Can you see the solution?

BASIC VARIABLES

d c s t VALUE

d 1 0 1 -1 7

c 0 1 -1 2 2

Z 0 0 2 12 140






5. Get the solution


OBTAINING THE SOLUTION

• Remember that since s and t are now non–basic variables they are set to zero.

• This corresponds to the solution:s = 0, t = 0,

d = 7c = 2

Z = 140

BASIC VARIABLES

d c s t VALUE

1 0 1 -1

0 1 -1 2

0 0 2 12

d

c

Z

7

2

140


THE SOLUTION

• Don’t forget to put your solution back into the context of the problem.

Z = 140

d = 7

c = 2

• The maximum profit is Rs.140

• To make this profit the factory should produce 7 dolls and 2 cars.


Z

PIVOTING

Step 3 - Add suitable multiples of the pivotal row to all other rows until all

entries, apart from the pivot, in the pivotal column are zero.

row (iii) + 8 row (i)gives

Z

t

s


2 16011

1 1 0 1 9

-16 -14 0 0 0

0 -6 0 1288


Simplex part i

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