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Minimize

Graphically solve a profit maximization problem and a job assignment problem in Quantitative Analysis for Management.

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Added on  2023-01-16

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This document discusses the process of minimizing the given objective function subject to constraints using the Solver Engine. It provides the optimal solution and a report of the process.

Minimize

Graphically solve a profit maximization problem and a job assignment problem in Quantitative Analysis for Management.

   Added on 2023-01-16

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3)
Minimize,
10 x1 A + 4 x1 B +11 x1C +12 x2 A + 5 x2 B + 8 x2 C +9 x3 A +7 x3 B +6 x3 C
Subject to:
x1 A + x1 B + x1C 70
x2 A + x2 B + x2C 50
x3 A + x3 B + x3 C 30
x1 A + x2 A + x3 A =40
x1 B + x2 B + x3 B=50
x1 C+ x2 C + x3 C=60
xij 0 for all i j
Report Created: 08-04-2019 15:46:46
Result: Solver has converged to the current solution. All Constraints are satisfied.
Solver Engine
Engine: GRG Nonlinear
Solution Time: 0.578 Seconds.
Iterations: 13 Subproblems: 0
Solver Options
Max Time Unlimited, Iterations Unlimited, Precision 0.000001
Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Central
Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume
NonNegative
Objective Cell (Min)
Cell Name Original Value Final Value
$K$16 unit used 0 1040.000011
Variable Cells
Cell Name Original Value Final Value Integer
$B$4:$J$4
$B$4 x1A 0 38.88888921 Contin
$C$4 x1B 0 31.11111141 Contin
$D$4 x1C 0 0 Contin
$E$4 x2A 0 0 Contin
$F$4 x2B 0 18.88888921 Contin
$G$4 x2C 0 31.11111141 Contin
Minimize_1
$H$4 x3A 0 1.111111252 Contin
$I$4 x3B 0 0 Contin
$J$4 x3C 0 28.88888921 Contin
Constraints
Cell Name Cell Value Formula Status
Slac
k
$K$10 Constraint 4 used 40.00000046 $K$10=$M$10 Binding 0
$K$11 Constraint 5 used 50.00000061 $K$11=$M$11 Binding 0
$K$12 Constraint 6 used 60.00000061 $K$12=$M$12 Binding 0
$K$7 Constraint1 used 70.00000061 $K$7<=$M$7 Binding 0
$K$8 Constraint 2 used 50.00000061 $K$8<=$M$8 Binding 0
$K$9 Constraint 3 used 30.00000046 $K$9<=$M$9 Binding 0
x1A x1B x1C x2A x2B x2C x3A x3B x3C
38.88
889
31.11
111
0 0 18.88
889
31.11
111
1.111
111
0 28.88
889
used Avail
able
Constr
aint1
1 1 1 70 <
=
70
Constr
aint 2
1 1 1 50 <
=
50
Constr
aint 3
1 1 1 30 <
=
30
Constr
aint 4
1 1 1 40 = 40
Constr
aint 5
1 1 1 50 = 50
Constr
aint 6
1 1 1 60 = 60
unit 10 4 11 12 5 8 9 7 6 1040
Final Lagrange
Cell Name Value
Multiplie
r
$K$1 Constraint 4 40.0000004 0
Minimize_2
0 used 6
$K$1
1
Constraint 5
used
50.0000006
1 -6
$K$1
2
Constraint 6
used
60.0000006
1 -3
$K$7
Constraint1
used
70.0000006
1 10
$K$8
Constraint 2
used
50.0000006
1 11
$K$9
Constraint 3
used
30.0000004
6 9
The yellow highlighted portion shows the optimal solution of the variables and the least cost
incurred is 1040 highlighted in Red. (Microsoft, n.d.)
1)
Objective Function: Maximize
z=8 x1 +5 x2
Constraint 1 drawn x1+ x2 10
Consider it as x1+ x2=10
if x1=0
x2=10
if x2=0
x1=10
x1 0 10
x2
1
0 0
Constraint 2 drawn 2 x1 6
Consider it as 2 x1 =6
then, x1=3
Minimize_3

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