7 The following questions refer to the Giapetto problem (Section 3.1). Giapetto's LP was max z = 3x, + 2x2 2x, + x2 < 100 X, + x2 < 80 < 40 (Finishing constraint) (Carpentry constraint) (Limited demand for soldiers) st. (x, = soldiers and x2 = trains). After adding slack variables S1, 82, and s3, the optimal tableau is as shown in Table 12. Use this optimal tableau to answer the following questions: a Show that as long as soldiers (x1) contribute between $2 and $4 to profit, the current basis remains optimal. If soldiers contribute $3.50 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 fin- ishing 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 Basic Variable rhs 1 1 180 z = 180 1 -1 20 X1 = 20 1 -1 60 X2 = 60 -1 20 S3 = 20

Linear Algebra: A Modern Introduction
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Author:David Poole
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Chapter2: Systems Of Linear Equations
Section2.4: Applications
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7 The following questions refer to the Giapetto problem
(Section 3.1). Giapetto's LP was
max z =
3x, + 2x2
2x, + x2 < 100
X, + x2 < 80
< 40
(Finishing constraint)
(Carpentry constraint)
(Limited demand for soldiers)
st.
(x, = soldiers and x2 = trains). After adding slack variables
S1, 82, and s3, the optimal tableau is as shown in Table 12.
Use this optimal tableau to answer the following questions:
a Show that as long as soldiers (x1) contribute between
$2 and $4 to profit, the current basis remains optimal. If
soldiers contribute $3.50 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 fin-
ishing 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
Basic
Variable
rhs
1
1
180
z = 180
1
-1
20
X1 = 20
1
-1
60
X2 = 60
-1
20
S3 = 20
Transcribed Image Text:7 The following questions refer to the Giapetto problem (Section 3.1). Giapetto's LP was max z = 3x, + 2x2 2x, + x2 < 100 X, + x2 < 80 < 40 (Finishing constraint) (Carpentry constraint) (Limited demand for soldiers) st. (x, = soldiers and x2 = trains). After adding slack variables S1, 82, and s3, the optimal tableau is as shown in Table 12. Use this optimal tableau to answer the following questions: a Show that as long as soldiers (x1) contribute between $2 and $4 to profit, the current basis remains optimal. If soldiers contribute $3.50 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 fin- ishing 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 Basic Variable rhs 1 1 180 z = 180 1 -1 20 X1 = 20 1 -1 60 X2 = 60 -1 20 S3 = 20
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