Operations Research Project

1. GROUP MEMBERS:
OPERATIONS RESEARCH
GROUP SUBMISSION
GROUP - 6
AKSHAY
RATAN
B15068
ANKIT
ANAND
B15071
AVINASH
BHANDARU
B15079
ROHIT
CHALASANI
B15103
AISHWARY
GUPTA
B15066
AAYUSH JAIN
B15063

2. LINEAR PROGRAMMING
Linear programming is a strategy to accomplish the optimal result, in a numerical model whose
constraints are represented by linear relationships. Linear programs have objective functions that are
either to be maximized or minimized, constraints that are of three types (<,>,=) and variables that
usually have an upper and a lower bound. A standard form Linear Programming Problem has the
following characteristics –
 The Objective function needs to be maximized
 All constraints are of the type – less than (<)
 The constraints are bound by non-negative numbers
 All variables are restricted to non-negativity
In algebraic form, these can be represented as-
Objective Function (to be maximized):
Problem Constraints:
Non-Negative Variables:
QUESTION
A corporation plans on building a maximum of 11 new stores in a large city. They will build these stores
in one of three sizes for each location – a convenience store (open 24 hours), standard store, and an
expanded services store. The convenience store requires $4.125 million to build and 30 employees to
operate. The standard store requires $8.25 million to build and 15 employees to operate. The expanded-
services store requires $12.375 million to build and 45 employees to operate. The corporation can
dedicate $82.5 million in construction capital, and 300 employees to staff the stores. On the average,
the convenience store nets $1.2 million annually, the standard store nets $2 million annually, and the
expanded services store nets $2.6 million annually. How many of each should they build to maximize
revenue?
VARIABLES
x1 = number of convenience stores
x2 = number of standard stores
x3 = number of expanded services stores

3. CONSTRAINTS
a. x1 + x2 + x3 ≤ 11
b. 4.125 x1 + 8.25x2 + 12.375x3 ≤ 82.5
c. 30x1 + 15x2 + 45x3 ≤ 300 x1 ≥ 0, x2 ≥ 0
d. x3 ≥ 0
OBJECTIVE FUNCTION
Maximize Y= 1.2x1 + 2x2 + 2.6x3 (in millions)
SOLUTION
Step 1: Entering the data on excel
Step 2: Entering the Objective functions and constraints on the solver

4. Step 3: Solver Solution
Step 4: Interpretation
The organization should build two convenience stores and nine standard stores in order to maximize
revenue.
TRANSPORTATION PROBLEM
We can adopt the simplex technique to solve transportation problems, but easier algorithms have been developed
for solution of such problems. Transportation model as a solution to optimise the distribution cost was originated
in 1941.
Transportation model is a special minimum cost network flow model for which every node is either a pure supply
node (origins) or a pure demand node (destinations). There are no intermediary trans-shipment nodes. All flow
goes from the origin to the destination. The commodity flowing may be people, oil, water, almost anything.
The objective of the transportation problem is to transport various quantities of a single homogenous commodity,
which are initially stored at various origins to various destinations. This must be done in such a way as to maximize
the profit or minimize the cost.
The classic statement of the transportation problem uses a matrix with the rows representing sources and columns
representing destinations. The algorithms for solving the problem are based on this matrix representation.
The transportation problem can be approached in several ways to come up with an initial basic feasible solution.
Some of these ways are:
 North-West Method
 Least Cost Method
 Vogel’s Approximation Method
While any method can be used to come up with a feasible solution, one needs to check for optimality to find the
optimum solution which would reduce the total transportation cost of the problem.
QUESTION
A severe winter ice storm has swept across North Carolina & Virginia, followed by over a foot of snow.
And frigid, single-digit temperatures. These winter conditions have resulted in numerous downed power
lines and power outages in the region causing dangerous conditions for much of the population. Local
utility companies have been overwhelmed and have requested assistance from unaffected utility
companies across the Southeast. The following table shows the number of utility trucks with crews

5. available from five different companies in Georgia, South Carolina, and Florida; the demand for crews in
seven different areas that local companies cannot get to; and the weekly cost ($1,000s) of crews going
to a specific area (based on the visiting company’s normal charges, the distance the crew has to come
and living expenses in an area).
Crew
Area (Cost = $1000s) Crews
AvailableNC-E NC-SW NC-P NC-W VA-SW VA-C VA-T
GA-1 15.2 14.3 13.9 13.5 14.7 16.5 18.7 12
GA-2 12.8 11.3 10.6 12.0 12.7 13.2 15.6 10
SC-1 12.4 10.8 9.4 11.3 13.1 12.8 14.5 14
FL-1 18.2 19.4 18.2 17.9 20.5 20.7 22.7 15
FL-2 19.3 20.2 19.5 20.2 21.2 21.3 23. 12
Crews
Needed 9 7 6 8 10 9 7
Determine the number of crews that should be sent from each utility to each affected area that will
minimize the total costs.
SOLUTION
In the above problem, we need to minimize the total cost associated with sending crews from
unaffected areas to affected areas. This is a transportation problem which can be solved using excel
solver as shown.
Step 1: Formulate the minimization problem. We need to minimize the total cost. Hence Sum product of
transportation matrix with cost matrix mist be minimized.
Step 2: Define the required constraints. We will have three constraints
a. The transportation matrix values are integers since we can have integer number of crews
moving from one location to the other.
b. Corresponding demands at each location is equal in both transportation and cost matrix.
c. Number of crews moving from each location are less than or equal to the maximum crews
available in that location.
Step 3: We create the tables as shown below (Can be seen in the corresponding excel sheet).

6. Formulation
Step 4: Specify the objective and constraints in solver as shown below and solve.
Solver window
Upon solving as per the above steps, we get the following result.

7. Solution
Step 6: Interpretation
As seen from the above solution.
1. From GA-1
a. 2 crews go to NC-W
b. 10 crews go to VA-SW
2. From GA-2
a. 1 crew will go to NC-SW
b. 9 crews will go to VA-C
3. From SC-1
a. 6 will go to NC-SW
b. 6 will go to NC-P
c. 2 will go to VA-T
4. From FL-1
a. 9 will go to NC-E
b. 6 will go to NC-W
5. From FL-2
a. 5 will go to VA-T.
As per the above computation, the minimum cost would be $840500.
ASSIGNMENT PROBLEM
The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in
polynomial time. It was developed by Harold Kuhn in 1955, who gave the name "Hungarian method"
because the algorithm was largely based on the earlier works of two Hungarian mathematicians: Dnes
Knig and Jen Egervry. In 2006, it was discovered that Carl Gustav Jacobi had solved the assignment
problem in the 19th century, and the solution had been published posthumously in 1890 in Latin.
This method was originally invented for the best assignment of a set of persons to a set of jobs. It is a
special case of the transportation problem. The algorithm finds an optimal assignment for a given “n x n”

8. cost matrix. “Assignment problems deal with the question how to assign n items (e.g. jobs) to n machines
(or workers) in the best possible way. Mathematically an assignment is nothing else than a bijective
mapping of a finite set into itself
The Hungarian model can also be used to solve special cases of transportation problem , which asks for
the allocation of a homogeneous commodity from given supplies at a set of sources to satisfy prescribed
demands at a set of destinations, while minimising the total source-destination transportation costs.
Steps to solve assignment problems using Hungarian Model:
Step 1 – Subtract the row minimum from each row.
Step 2 – Subtract the column minimum from each column from the reduced matrix.
Step 3 – Assign one “0” to each row & column.
Step 4 – Tick all unassigned row.
Step 5 – If a row is ticked and has a “0”, then tick the corresponding column (if the column is not yet
ticked).
Step 6 – If a column is ticked and has an assignment, then tick the corresponding row (if the row is not yet
ticked).
Step 7 - Repeat step 5 and 6 till no more ticking is possible.
Step 8 – Draw lines through unticked rows and ticked columns. The number of lines represents the
maximum number of assignments possible.
Step 9 – Find out the smallest number which does not have any line passing through it. We call it Theta.
Subtract theta from all the numbers that do not have any lines passing through them and add theta to all
those numbers that have two lines passing through them. Keep the rest of them the same.
Step 3 – Assign one “0” to each row & column.
The optimal solution is found if there is one assigned “0” for each row and each column.
QUESTION
manager doesnot want to assign china to that Monty department store has six employees available to assign to
four departments in the store—home furnishings, China, Appliances and Jewelry. Cost of assigning each worker to

9. each department is given in the table. Objective is to minimize cost. Employee 3 has not worked in the china
department before so he should not be assigned to china.
SOLUTION
Step 2: Enter data on excel and solver
Step 2: Allocations

10. CONCLUSION
Following allocations bring about minimum cost
1- B
2- D
3- C
4- E
5- F
6- A
It is important to note that 5 and 6 are dummy departments added to balamce the matrix and thus have a cost of
0. So, minimum cost is 1310.
FUZZY MULTI CRITERIA DECISION MAKING
In multiple objective decision problems, all alternatives available to the system are evaluated according
to a number of criteria. Each criterion will induce a particular ordering of the alternatives and a relevant
procedure is then implemented to construct one overall preference ordering. A general formulation of
the multi objective linear programming function looks like –
In a standard goal programming function, goals and constraints are precisely defined. Applying fuzzy set
theory in Multi Objective Decision Making allows us to incorporate the vague aspirations of the Decision
Maker. An optimal decision is the point in the space of alternatives at which the membership of the
fuzzy decision is maximum. Formally, consider that a fuzzy set A in X, with X= {xi} denoting a set of points
xi, is a set of ordered pairs A = {xi,µA (xi)} and xi ∈X, where µ(xi) represents the grade of membership of
xi in A. Assume there is a set of goals G = {g1,g2,.......,gn}, a set of constraints C = {c and that
1,c2,.......,cm} ∩ corresponds to the intersection operator, then the fuzzy decision D is,

11. QUESTION
A farmer has 12 acres of cultivable land and wanted to grow multiple vegetable crops viz., Brinjal, Ladies
Finger, Bitter Guard and Tomato in a season. Out of his experience, he stated that the labour work time
available with him is 220 hours and availability of water is 25 acre-inches. The profit coefficients for
required work time and water have been tabulated below. How many acres does the farmer needs to
allocate for each crop in order to get guaranteed net returns out of volatility among profit coefficients?
Brinjal Ladies
Finger
Bitter Guard Tomato
Profit Coefficients (lakh
rupees) Set-1
0.35 0.32 0.69 0.95
Profit Coefficients (lakh
rupees) Set-2
0.55 0.41 0.78 1.20
Profit Coefficients (lakh
rupees) Set-3
0.65 0.53 1.02 0.68
Profit Coefficients (lakh
rupees) Set-4
0.82 0.62 1.25 0.80
Labour requirement per
acre (‘000 hours)
1.76 1.28 1.6 1.84
Water requirement per
acre (acre-inch)
27.2 17.5 18.2 18
VARIABLES
In the given problem, since the objective is to maximize profit from each acre of the cultivated land
through the various vegetables in their respective sets of combinations,
X1 = Number of acres allocated for cultivating Brinjal
X2 = Number of acres allocated for cultivating Ladies Finger
X3 = Number of acres allocated for cultivating Bitter Guard
X4 = Number of acres allocated for cultivating Tomato
OBJECTIVE FUNCTION
There are 4 objective functions each corresponding to the respective sets of cultivations as detailed in
the table above.

12. Maximize Z1 = 0.35x1 + 0.32X2 + 0.69X3 + 0.95X4
Maximize Z2 = 0.55x1 + 0.41X2 + 0.78X3 + 1.20X4
Maximize Z3 = 0.65x1 + 0.53X2 + 1.02X3 + 0.68X4
Maximize Z4 = 0.82x1 + 0.62X2 + 1.25X3 + 0.80X4
CONSTRAINTS
The problem has 3 constraints and a non-negativity of variables constraints. The 3 constraints pertain to
Land. Labour Hours and Water requirement.
Type of Constraint Constraint
Land Constraint (of 12 acres land) x1 + X2 + X3 + X4 ≤ 12
Labour Constraint (of 25,000 labour hours) 0.55x1 + 0.41X2 + 0.78X3 + 1.20X4 ≤ 25
Water Constraint (of 220 acre inches per acre) 0.55x1 + 0.41X2 + 0.78X3 + 1.20X4 ≤ 220
Non Negativity Constraint X1, X2, X3, X4 ≥ 0
SOLUTION
Step 1: Each of the objective functions have been solved individually to find optimum solutions for each
objective function individually.
Objective Function Optimum Solution (X1, X2, X3, X4) Value of Optimum Solution
Objective Function 1 (0,0,0,12) 11.4
Objective Function 2 (0,0,0,12) 14.4
Objective Function 3 (0,0,11.8919,0) 12.13
Objective Function 4 (0,0,11.8919,0) 14.86
Step 2: Each of the objective functions are then normalized by creating membership functions µ1, µ2, µ3
and µ4. Now, to find a compromised multi criteria solution to the problem, each of these membership
functions are to be greater than equal to α, where α is the level at which each of the objective functions
meet their constraints and the overall solution is compromised.
The new Objective function: Maximize Z = α
4 new constraints are created with the condition of each membership function being greater than α.
Constraint 1: 0.35x1 + 0.32X2 + 0.69X3 + 0.95X4 – 3.196α≤ 8.204
Constraint 2: 0.55x1 + 0.41X2 + 0.78X3 + 1.20X4 – 5.126α≤ 9.274
Constraint 3: 0.65x1 + 0.53X2 + 1.02X3 + 0.68X4 – 3.968α≤ 8.16
Constraint 4: 0.82x1 + 0.62X2 + 1.25X3 + 0.80X4 – 5.263α≤ 9.60
These and the original constraints of land, water and labour hours are used to solve the new LPP to find
X1, X2, X3, X4 and α.

13. SOLUTION
The above LPP and its results are tabulated below. A compromised solution is achieved with α = 0.5107
and (X1, X2, X3, X4) = (0, 0, 5.972, 6.028).
Objective Function Individual Optimum
Solutions
Compromised Multi-Objective Solution
Objective Function 1 11.4 9.84728962
Objective Function 2 14.4 11.89177554
Objective Function 3 12.12973 10.19046742
Objective Function 4 14.86486 12.28738335
CONCLUSION
5.972 acres of Bitter Guard and 6.028 acres of Tomato are to be produced to have a compromised multi
objective solution to the problem.

14. REFERENCES
 Chen, F. (2012). Introduction to operations research. Mayıs, 28, 2013.
 Kuhn, Harold W. "Variants of the Hungarian method for assignment problems." Naval
Research Logistics Quarterly 3.4 (1956): 253-258.
 Gass, Saul I., and Carl M. Harris, eds. Encyclopedia of operations research and
management science. Springer Science & Business Media, 2012.
 https://www.wiwi.uni-kl.de/bisororwiki/Assignment_problem:_Hungarian_method_3
 Russell, Roberta S., and Bernard W. Taylor. Operations management: Quality and
competitiveness in a global environment. Wiley, 2005.