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# Linear Programming Problem for an Optimal Solution

Simple model formation. Using the attached document, formulate the linear programming model for the school.

A company has three teachers. On a particular day, six classes are scheduled to be completed. A class does not need more than one teacher. The cost for each teacher to do each class is shown in the attached document.

#### Solution Preview

See the attached file. Thanks.

A school has three teachers. On a particular day, six subjects are scheduled to be taught. A subject does not need more than one teacher. The cost for each teacher to do each subject shown below.

Teacher Subject 1 Subject 2 Subject 3 Subject 4 Subject 5 Subject 6
A 3 2 2 6 4 6
B 4 3 7 5 3 3
C 9 9 7 9 7 6
For each of the following questions, formulate a linear mathematical model in a standard format and implement it in Excel to find the optimal answer. Clearly define your decision variables, objective function and constraints in your formulation and provide snapshots of your Excel solutions. You can define the decision variables only once but you need to have a complete formulation for each question.

Questions:
a.) What is the minimum cost assignment when each teacher can do 2 subjects? Provide both the formulation and the Excel solution
Xij is a binary variable which takes a value of 1 if teacher i takes subject j, else it takes a value of 0

Decision Variables
Teacher Subject 1 Subject 2 Subject 3 Subject 4 Subject 5 Subject 6
A 1 1 0 0 0 0
B 0 0 0 1 1 0
C 0 0 1 0 0 1

Objective Function:
Minimize the cost assignment
Cij = Cost if teacher i takes subject j
Min SXij*Cij 26.00

Constraints
Capacity constraints for each teacher (two subjects only) LHS RHS
Teacher 1 X11+X12+X13+X14+X15+x16<=2 2.00 <= 2
Teacher 2 X21+X22+X23+X24+X25+X26<=2 2.00 <= 2
Teacher 3 X31+X32+X33+X34+X35+X36<=2 2.00 <= 2
Each subject should be assigned to at least one teacher
Subject 1 X11+X21+X31>=1 1.00 >= 1
Subject 2 X12+X22+X32>=1 1.00 >= 1
Subject 3 X13+X23+X33>=1 1.00 >= 1
Subject ...

#### Solution Summary

A simple linear model formation are examined for optimal solutions.

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