4x1 + x2 lt 10 x1, x2 ge 0 The optimal solution to this LP is z = 32, x1 = 0,x2

Question

4x1 + x2 lt 10 x1, x2 ge 0 The optimal solution to this LP is z = 32, x1 = 0,x2 = 8, s1 = 0, s2 = 2. Graphically find the range of values of 1 for which the current basis remains optimal. Wivco produces product 1 and product 2 by processing raw material. As much as 90 lb of raw material may be purchased at a cost of $ 10/lb. One pound of raw material can be used to produce either 1 lb of product 1 or 0.33 lb of product 2. Using a pound of raw material to produce a pound of product 1 requires 2 hours of labor or 3 hours to produce 0.33 lb of product 2. A total of 200 hours of labor are available, and at most 40 pounds of product 2 can be sold. Product 1 sells for $ 13/lb, and product 2 sells for $40/lb. ltt RM = pounds of raw material processed PI = pounds of raw material used to produce product 1 P2 = pounds of raw material used to produce product 2 To maximize profit, Wivco should solve the following LP: max z = 13P1 + 40(0.33)P2 - 10RM s.t. RM ge P1 + P2 2P1 + 3P2 lt 200 RM lt 90 0.33P2 lt 40 P1, P2, RM ge 0 Use the LINDO output in Figure 12 to answer the following questions: If only 87 lb of raw material could be purchased, what would be Wivco's profits? If product 2 sold for $39.50/lb, what would be the new optimal solution? What is the most that Wivco should pay for another pound of raw material? What is the most that Wivco should pay for another hour of labor? Suppose that 1 lb of raw material could also be used to produce 0.8 lb of product 3, which sells for $24/lb. Processing 1 lb of raw material into 0.8 lb of product 3 requires 7 hours of labor. Should Wivco produce any of product 3? Consider the following LP and its optimal tableau (Tablt 56): max z = 3x1 + 4x2 + x3 s.t. x1 + x2 + x3 lt 50 2x1 - x2 + x3 ge 15 x1 + x2 = 10 x1, x2, x3 ge 0 Find the dual of this LP and its optimal solution.

Explanation / Answer

Q8)   a) RM reduce from 90lb to 87lb, original obj = 274
Section RightHand Side Ranges Row (4) Allowable increase = 10 Allowable decrease = 23.34. This
means as long as the RM availability is between 66.66 and 100, the current basis is still optimal.
Row (4) dual price is 2.6.
OBJnew = OBJold + ?RM* dual price = 274 + (-3)* 2.6 = 266.2

b) The price of product 2 from $40 to $39.5, decreasing $0.5. Coefficient of P2 decreases 0.5*0.33 =
0.165.
Section Obj Coefficient Ranges Row (P2)Allowable decrease = 0.2, So under 0.165 short, the current
basis is still optimal. OBJnew = OBJold + ?Coef.* P2 = 274 + (-0.165)* 20 = 270.7


c) If 1lb of RM was purchased, from (a) we know that Allowable increase = 10. So the current basis is
still optimal. The dual price is 2.6.
OBJnew = OBJold + ?RM* dual price = 274 + (1)* 2.6 = 276.6. The profit will increase by $2.6. So this is
the most that Wivco should pay for another pound of raw material.

d) If the labor hours increase from 200 to 201, then check Section Obj Coefficient Ranges Row (3)
Allowable increase = 70, So on the condition of 201 labor hours, the current basis is still optimal. Row (3)
dual price is 0.2.
OBJnew = OBJold + ?labor* dual price = 274 + (1)* 0.20 = 274.2
The profit will increase by $0.2. So this is the most that Wivco should pay for additional labor hour.

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