1 Use the simplex algorithm to solve the Giapetto problem (Example 1 in Chapter 3).

Giapetto Woodcarving, Inc., manufactures two types of wooden toys: soldiers and trains. A soldier sells for $27 and uses $10 worth of raw materials. Each soldier that is manufactured increases Giapetto’s variable labor and overhead costs by $14. A train sells for $21 and uses $9 worth of raw materials. Each train built increases Giapetto variable labor and overhead costs by $10. The manufacture of wooden soldiers and trains requires two types of skilled labor: carpentry and finishing. A soldier requires 2 hours of finishing labor and 1 hour of carpentry labor. A train requires 1 hour of finishing and 1 hour of carpentry labor. Each week, Giapetto can obtain all the needed raw material but only 100 finishing hours and 80 carpentry hours. Demand for trains is unlimited, but at most 40 soldiers are bought each week. Giapetto wants to maximize weekly profit (revenues _ costs). Formulate a mathematical model of Giapetto’s situation that can be used to maximize Giapetto’s weekly profit.

Solution:

Objective Function: max z= 3x1+2x22x1+x2<= 100x1+x2<=80x1<=40,x1,x2 >= 0Converting LP to standard form:Z-3x1-2x2=02x1+x2+s1=100x1+x2+s2=80x1+s3=40Canonical Form:

Row Z x1 x2 s1 s2 s3 rhs bfs ratio

0 1 -3 -2 0 0 0 0 Z=0

1 0 2 1 1 0 0 100 S1=100 50

2 0 1 1 0 1 0 80 S2=80 80

3 0 1 0 0 0 1 40 S3=40 40

LP to Standard Form:z-2x1-3x2=0x1+2x2+s1=62x1+x2+s2=8Canonical Form:

3.Use the simplex algorithm to solve the following problem:

LP to Standard Form:z-2x1+x2-x3=03x1+x2+x3+s1=60x1-x2+2x3+s2=10x1+x2-x3+s3=20Canonical Form:

z=25; x1=15, x2=5, s1=10, x3=s2=s3=0

4.

optimal solution to minz = (-)maxz

-max(z)= -2x1+5x2

Converting LP to standard Form:

-Z+2x1-5x2=03x1+8x2+s1=122x1+3x2+s2=6

Canonical Form:

Optimal Solution for –max(z) is (-)Z=7.5,x1=0,x2=1.5,s1=0,s2=1.5But for Optimal Solution for min(z) is Z=-7.5,x1=0, x2= 1.5,s1=0,s2=1.5

5. Suppose that in solving an LP, we obtain the tableau in Table 22. Although x1 can enter the basis, this LP is unbounded. why?

Solution: Considering Maximizing the given LP:

Concept: An unbounded LP for a max problem occurs when a variable with a negative coefficient in row 0 has a nonpositive coefficient in each constraint.

6. Suppose we have obtained the tableau in Table 75 fora maximization problem. State conditions on a1, a2, a3, b,c1, and c2 that are required to make the following statements true:

i) The current solution is optimal, and there are alternative optimal Solutionsii) The current basic solution is not a basic feasible solutioniii) The current basic solution is a degenerate bfs.iv) The current basic solution is feasible, but the LP is unbounded v) The current basic solution is feasible, but the objective function value can be improved by replacing x6 as

a basic variable with x1.

Answer:

i) Case (1): b>=0, c1>=0,c2=0,a1>0,then x2 can be entering variable- Optimal Solution Can be obtained.

Case (2): b>=0,c2>=0,c1=0,then x1 can be entering variable- Optimal Solution Can be obtained. Case (3): b>=0, c1,c2>=0,a2>0then x5 can be entering variable- Optimal Solution Can be obtained.ii) (b<0)iii) (b=0)iv) Case (1) if x2 is entering variable, a1<=0, c2<0 ,b>=0 ; Case (2) if x5 is entering variable, a2<=0 ,b>=0v) c1<0 and 3/a3<=b/4