ADM 2302 Study Guide - Spring 2018, Comprehensive Midterm Notes - Shadow, Time, Sides Of An Equation
12 Oct 2018
School
Department
Course
Professor
ADM 2302
MIDTERM EXAM
STUDY GUIDE
Fall 2018
Microsoft Excel 14.0 Answer Report
Worksheet: [3. Transhipment_Problem.xls]Alternative Solution
Result: Solver found a solution. All Constraints and optimality conditions are satisfied.
Solver Engine
Engine: Simplex LP
Solution Time: 0.047 Seconds.
Iterations: 17 Subproblems: 0
Solver Options
Max Time 100 sec, Iterations 100, Precision 0.000001
Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 5%, Solve Without Integ
Objective Cell (Min)
Cell Name Original Value Final Value
$P$6 Cost of Transportation Total $275.00 $2,705.00
Variable Cells
Cell Name Original Value Final Value Integer
$B$4 Solution X14 050 Contin
$C$4 Solution X15 0 0 Contin
$D$4 Solution X24 045 Contin
$E$4 Solution X25 0 0 Contin
$F$4 Solution X34 0 0 Contin
$G$4 Solution X35 045 Contin
$H$4 Solution X46 0 5 Contin
$I$4 Solution X47 055 Contin
$J$4 Solution X48 035 Contin
$K$4 Solution X49 0 0 Contin
$L$4 Solution X56 020 Contin
$M$4 Solution X57 0 0 Contin
$N$4 Solution X58 0 0 Contin
$O$4 Solution X59 25 25 Contin
Constraints
Cell Name Cell Value Formula Status Slack
$P$12 Contractor 1 demand LHS 25 $P$12=$R$12 Binding 0
$P$13 Contractor 2 demand LHS 55 $P$13=$R$13 Binding 0
$P$14 Cantractor 3 demand LHS 35 $P$14=$R$14 Binding 0
$P$15 Cantractor 4 demand LHS 25 $P$15=$R$15 Binding 0
$P$16 Transhipment node 4 LHS 0 $P$16=$R$16 Binding 0
$P$17 Transipment node 5 LHS 0 $P$17=$R$17 Binding 0
$P$9 Capacity of plant 1 LHS 50 $P$9<=$R$9 Binding 0
$P$10 Capacity of plant 2 LHS 45 $P$10<=$R$10 Not Binding 10
$P$11 Capacity of plant 3 LHS 45 $P$11<=$R$11 Binding 0
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t Integer Constraints, Assume NonNegative
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Document Summary
Max time 100 sec, iterations 100, precision 0. 000001. Max subproblems unlimited, max integer sols unlimited, integer tolerance 5%, solve without integ. Source (grain elevator) a. chicago b. st. louis. 4x1 + 2x2 8 x1 4 x2 6 x1, x2 0. Special cases in linear programming: redundant constraints: A constraint that does not form a unique boundary of the feasible solution space; its removal would not alter the feasible solution space. The second constraint is redundant: no feasible solution. Occurs in problems where to satisfy one of the constraints, another constraint must be violated. Maximize z = 5x1 + 3x2 subject to: multiple optimal solutions. Problems in which different combinations of values of the decision variables yield the same optimal value. 4x1 + 3x2 120 x1, x2 0. The objective function can be parallel to a constrain line. Where: x1 = number of bowls x2 = number of mugs: unbounded solution. When nothing prevents the solution from becoming infinitely large.
