1. Problem Constraints of the Form 
TBF General Mathematics - II Lecture – 11 : Minimization with
Problem Constraints of the Form 
Prof. Dr. Halil İbrahim Karakaş
Başkent Üniversitesi

2. Minimization Problems with Constraints of the Form  .
In our previous lecture we have seen how to solve standard maximization problems with problem constraints of the form . In the present lecture, we will consider minimization problems with problem constraints of the form . We shall see that these two types of problems are closely related.
To solve a minimization problem with problem constraints of the form , we form a maximization problem with problem constraints of the form  which is called the dual problem.
Process of forming the dual problem and getting the solution of the original problem from the dual problem will be explained on an example.

3. Example .
Minimize
subject to
The dual problem is formed as follows:
Using the problem constraints and the objective function, we obtain the following system and the matrix A .

4. Minimize
subject to  
At this point we recall the definition of the “transpose” of a matrix:
The transpose of an m n matris A is the n m matrix AT whose i-j entry is the j-i entry of A .
It follows from this definition that i-th row of AT is the i-th column of A and the j-th column of AT is the j-th row of A.
For example, the transpose of the matrix A above is
Now we asociate with this matrix the system with constraints  above on the right.
We will summarize what we have done so far:

5.     Minimize subject to Original Problem
The dual of the problem in the beginning is the maximization problem with constraints  obtained from the rows of AT or the last system as follows. We use different symbols for the variables of the dual problem, to distinguish between the original problem and the dual.
Maximize
subject to
Dual Problem
Note that new symbols are used for variables to distinguish dual problem from the original problem.

6. Duality Principle. A minimization problem has a solution if and only if its dual has a solution. If a solution exists, then the optimum value of the minimization problem coincides with the optimum value of the dual problem.
Minimize
subject to
Corner C
540
Orijinal
510
528
Maximize
subject to
Corner P
Dual
495
510
480

7. As we see from the above examples
The solution that gives the optimum value for the minimization problem is different from the solution that gives the optimum value for the dual. That is why we use different symbols for the variables of the dual problem.
Let us also note that
If the dual of the minimization problem is in standard form, then it can be solved by simplex method.
If the coefficients of the objective function of the minimization problem are all nonnegative, then the dual problem is in standard form and thus can be solved by simplex method.
In applying simplex method to the dual, we write the initial system by using the variables x1 , x2 of the minimization problem as slack variables. Then the solution of the minimization problem can be read from the final simplex tableou. The entries in the buttom row of the columns of x1 , x2 form the solution of the minimization problem.
The dual of our sample problem is in standard form and thus it can be solved by simplex method. We use x1 , x2 as slack variables and realize the solution in the next slide.

8. Dual Maximize subject to P= 510 is maksimum for y1 = 6, y2 = 15 .
C = 510 is minimum for x1 = 10, x2 = 3 .

9. We summarize what we have obtained so far:
We follow the following steps in solving a minimization prblem with constraints of the form  by applying the simplex method to its dual:
Step 1. Form the matrix A by using the problem constraints and the objective function of the minimization problem. (The coefficients of the objective function are written in the bottom row.)
Step 2. Write down the transpose AT of A.
Step 3. Use the rows of AT to form the dual problem as a maximization problem with constraints of the form  . Use different symbols like y1 , y2 , for the variables of the dual problem.
Step 4. Solve the dual problem( if it is in standard form) by simplex method. In initial system, use the variables x1 , x2 , of the minimization problem as slack variables. In the final simplex tableou, the entries in the buttom row of the columns of x1 , x2 , form the solution of the minimization problem.

10. Example. Minimize C(x1, x2 , x3 )= 120 x1 + 60 x2 + 80 x3
subject to
Solution: Let us write down A and AT. 
The dual is
Maximize
subject to

11. Maximize
subject to
The solution of the minimization problem is obtained from the final tableou.
C = is minimum
for  x1 = 0, x2 = 60 and x3 = 10 .

12. Example. Minimize C(x1, x2 , x3 ) = 13x1 + 10x2 + 16 x3
subject to
Solution: Let us write down A and AT. 
The dual is
Maximize
subject to

13. C = 202 is minimum for x1 = 2 , x2 = 0 and x3 = 11.
Maximize
subject to
We write down initial simplex tableou and proceed
The solution of the minimization problem is obtained from the final tableou:
C = 202 is minimum for x1 = 2 , x2 = 0 and x3 = 11.

14. Example. Let us form the dual of each of the minimization problems below and see if the dual is in standard form.
Minimize
Minimize
subject to
subject to
The duals are
Maximize
Maximize
subject to
subject to
We see that dual of the first minimization problem is in standard form, and the dual of the second one is not. Thus the first problem can be solved by simplex method, but the second one can not be solved by simplex method. Solve the first minimization problem by simplex method and the second one by geometric method.

15. Problem. A mining company operates two mines, each of which produces three qualities of ore: low, medium and high quality. From mine A, 2 tons of low quality, 3 tons of medium quality and 1 ton of high quality ore can be produced per hour of operation and from mine B, 2 tons of low quality, 1 ton of medium quality and 2 tons of high quality ore can be produced per hour of operation. To satisfy the orders, at least 100 tons of low quality, 60 tons of medium quality and 80 tons of high quality ore is to be produced. If it costs 400 TL per hour to operate mine A and 600 TL per hour to oprate mine B, how many hours should each mine be operated to supply the required amounts of ore of each quality and minimize the cost of production? What is the minimum production cost?
Solution. We summarize the data on a table:
Mine
Low quality
Medium quality
High quality
Cost A 2 3 1
400 B
600
Order
100 60 80
Let mine A be operated x1 hours, and mine B be operated x2 hours. Then the total production cost is
C(x1, x2 ) = 400 x x2
TL .

16. Due to orders for each quality of ore, we get the following constraints:
Combining therse with the nonnegative constraints we construct the mathematical model of the problem as a linear programming problem as follows.
Minimize C(x1, x2) = 400x1+ 600x2
Since all coefficients of the objective function are nonnegative, the dual of this problem will be in standard form and we will be able to solve it by simplex method. We form the matrix A and its transpose:
subject to  

17. The dual is obtained fro AT
Maximize P(y1,y2,y3) = 100y1 +60y2 + 80y3
subject to
The initial system and the initial tableou:
y y y x1 x2 P x1 x2 P

18. From the final simplex tableou
y y y x1 x2 P x1 x2 P
y y y x x P
From the final simplex tableou
C= is minimum for x1 = 20 , x2 = 30 .
Thus for minimum production cost, mine A sould be operated 20 hours and mine B should be operated 30 hours. Minimum production cost is TL.

19. 500 1000 Distribution Location A Location B Capacity Factory I 3 2 700
Problem. A shoe company produces sport shoes in two factories and stores them in two locations: Factory I, Factory II; Location A, Location B. At most 700 pairs of shoes can be produced in factory I per month, and at most 900 pairs of shoes can be produced in factorty II per month. Each month at least 500 pairs of shoes should be sent to Location A and at least 1000 pairs to location B. The transportation cost of a pair of shoes from factories to the stores are as follows: From factory I to location A, 3 TL; from factory I to location B, 2 TL; from factory II to location A, 2 TL; from factory II to location B, 4 TL. How many pairs of shoes should be transported from each factory to each location to minimize the transportastion cost? What is the minimum transportation cost?
Solution. Data table:
Distribution
Location A Location B
Capacity
Factory I
700
Factory II
900
Least number

20. 500 1000 Distribution Location A Location B Capacity Factory I 3 2 700
700
Factory II
900
Least number
Assume that the number of pairs of shoes transported from factory I to the locations A and B are x1 and x2; the number of pairs of shoes transported from factory II to the locations A and B are x3 and x4, respectively. Then the transportation cost is C(x1,x2,x3,x4)= 3x1+ 2x2+ 2x3+4x4 TL and the mathematical model of the problem is :
Minimize C(x1,x2,x3,x4)= 3x1+ 2x2+ 2x3+4x4
subject to

21.    Minimize C(x1,x2,x3,x4)= 3x1+ 2x2+ 2x3+4x4 subject to
Dual problem and the solution is on the next slide.

22. Maximize P(y1,y2,y3,y4) = -700y1 – 900y2 + 500y3 + 1000y4
subject to

23. Maximize P(y1,y2,y3,y4) = -700y1 – 900y2 + 500y3 + 1000y4
subject to

25. For minimum transportation cost, 700 pairs of shoes from factory I to location A; 500 pairs of shoes from factory II to location A and 300 pairs of shoes from factory II to location B should be transported.
Minimum transportation cost is TL.