Question

USING EXCEL!! FOR BOTH PROBLEMS SCREEN SHOTS OF WORK PLEASE

19. Write the following primal problem in canonical form, and find its dual. Max 3x1+1x2+5x3+3x4 s.t. 3x1+1x2+2x3 = 30 2x1+1x2+3x3+1x4 >= 15 2x2+3x4<= 25 x1,x2,x3,x4 >= 0 20. Photo Chemicals produces two types of photograph-developing fluids at a cost of $1.00 per gallon. Let x1 = gallons of product 1 x2 = gallons of product 2 Photo Chemicals management requires that at least 30 gallons of product 1 and at least 20 gallons of product 2 be produced. They also require that at least 80 pounds of a perishable raw material be used in production. A linear programming formulation of the problem is as follows: Min 1x1 + 1x2 s.t. 1x1 >= 30 Minimum product 1 1x2 >= 20 Minimum product 2 1x1 + 2x2 >= 80 Minimum raw material x1, x2 >= 0 a. Write the dual problem. b. Solve the dual problem. Use the dual solution to show that the optimal production plan is x1 = 30 and x2 = 25. c. The third constraint involves a management request that the current 80 pounds of a perishable raw material be used. However, after learning that the optimal solution calls for an excess production of five units of product 2, management is reconsidering the raw material requirement. Specifically, you have been asked to identify the cost effect if this constraint is relaxed. Use the dual variable to indicate the change in the cost if only 79 pounds of raw material have to be used.

Answer 1

1.This is very simple programmed is R which can be seen is forgure 4.The first line override any last explainations of the variable lo1,making for the new situation explaintion.The bad situation ios then explained and in last solved on the last line.
Figure 4.Linear Optimixation*lo1)
lo1 <- list ()
lo1$sense <- "max"
lo1$c <- c(3 ,1 ,5 ,1)
lo1$A <- Matrix (c(3,1,2,0,
2,1,3,1,
0,2,0,3), nrow =3, byrow=TRUE , sparse =TRUE)
blc <- c(30,15,- Inf)
buc <- c(30,Inf ,25); lo1$bc <- rbind(blc , buc );
blx <- c(0 ,0 ,0 ,0)
bux <- c(Inf ,10,Inf ,Inf ); lo1$bx <- rbind(blx , bux );
r <- mosek(lo1)

A.Max 15x1+30x2+20x3
S.T.
x1+x3<4
0.5x1+2x2+x3<3
x1+x2+2x3<6
x1,x2,x3>0
B.From the above values for the extra changing we can see that the optimal answer is x1=15/2,x2=15 and x3=0.
C.The optimal value for the dual is shown in sun part to equal 78.Substututing x1=15/2,x2=15 into the primal objective function we can find the given values.