The following questions refer to the Giapetto problem. Giapetto’s LP was
\begin{array}{ll} \max z=3 x_{1}+2 x_{2} \\ \text{s.t.} { 2 x_{1}+x_{2} \leq 100} & {\text { (Finishing constraint) }} \\ {x_{1}+x_{2} \leq 80} & {\text { (Carpentry constraint) }} \\ {x_{1}} { \leq 40} & \text{(Limited demand for soldicrs)} \end{array}
maxz=3x
1
+2x
2
s.t.2x
1
+x
2
≤100
x
1
+x
2
≤80
x
1
≤40
(Finishing constraint)
(Carpentry constraint)
(Limited demand for soldicrs)
(x1 = soldiers and x2 = trains). After adding slack variables s2, and the optimal tableau is as shown in Table 12. Use this optimal tableau to answer the following questions: a. Show that as long long as soldiers (x1) contribute between $2 and$4 to profit, the current basis remains optimal. If soldiers contribute S3.50 to profit, find the new optimal solution to the Giapetto problem. b. Show that as long as trains (x2) contribute between $1.50 and$3.00 to profit, the current basis remains optimal. c. Show that if between 80 and 120 finishing hours are available, the current basis remains optimal. Find the new optimal solution to the Giapetto problem if 90 finishing hours are available. d. Show that as long as the demand for soldiers is at least 20, the current basis remains optimal. e. Giapetto is considering manufacturing toy boats. A toy boat uses 2 carpentry hours and 1 finishing hour. Demand for toy boats is unlimited. If a toy boat contributes $3.50 to profit, should Giapetto manufacture any toy boats? TABLE 12:
\begin{matrix} \text{z} & \text{ x1} & \text{x2} & \text{ s1} & \text{s2} & \text{s3} & \text{rhs} & \text{Basic Variable}\\ \text{1} & \text{0} & \text{0} & \text{1} & \text{1} & \text{0} & \text{180} & \text{z = 180}\\\text{0} & \text{1} & \text{0} & \text{1} & \text{−1} & \text{0} & \text{20} & \text{x1 = 20}\\ \text{0} & \text{0} & \text{1} & \text{−1} & \text{2} & \text{0} & \text{60} & \text{x2 = 60}\\ \text{0} & \text{0} & \text{0} & \text{−1} & \text{1} & \text{1} & \text{20} & \text{s3 = 20}\\ \end{matrix}
z
1
0
0
0
x1
0
1
0
0
x2
0
0
1
0
s1
1
1
−1
−1
s2
1
−1
2
1
s3
0
0
0
1
rhs
180
20
60
20
Basic Variable
z = 180
x1 = 20
x2 = 60
s3 = 20 d. Show that as long as the demand for soldiers is at least 20, the current basis remains optimal. e. Giapetto is considering manufacturing toy boats. A toy boat uses 2 carpentry hours and 1 finishing hour. Demand for toy boats is unlimited. If a toy boat contributes $3.50 to profit, should Giapetto manufacture any toy boats? TABLE 12:
\begin{matrix} \text{z} & \text{ x1} & \text{x2} & \text{ s1} & \text{s2} & \text{s3} & \text{rhs} & \text{Basic Variable}\\ \text{1} & \text{0} & \text{0} & \text{1} & \text{1} & \text{0} & \text{180} & \text{z = 180}\\\text{0} & \text{1} & \text{0} & \text{1} & \text{−1} & \text{0} & \text{20} & \text{x1 = 20}\\ \text{0} & \text{0} & \text{1} & \text{−1} & \text{2} & \text{0} & \text{60} & \text{x2 = 60}\\ \text{0} & \text{0} & \text{0} & \text{−1} & \text{1} & \text{1} & \text{20} & \text{s3 = 20}\\ \end{matrix}
z
1
0
0
0
x1
0
1
0
0
x2
0
0
1
0
s1
1
1
−1
−1
s2
1
−1
2
1
s3
0
0
0
1
rhs
180
20
60
20
Basic Variable
z = 180
x1 = 20
x2 = 60
s3 = 20
