SIT718 Assessment 4: Linear Programming, Game Theory, and R Code

This document presents a comprehensive solution to SIT718 Real World Analytics Assignment 4, tackling various problems in linear programming and game theory. The assignment explores the application of linear programming to optimize resource allocation and minimize costs in scenarios such as feed selection for livestock and production planning. It also delves into game theory, analyzing zero-sum and non-zero-sum games, including the identification of Nash equilibrium and the use of mixed strategies. The solutions incorporate graphical methods, mathematical formulations, and the use of R code to solve complex problems, demonstrating the ability to model and analyze real-world decision-making processes. The document covers problems related to farmer's decision, production planning, and bidding strategies, providing detailed explanations and step-by-step solutions for each question. The document emphasizes the application of operations research techniques and decision-making models to solve complex business-oriented problems, providing a valuable resource for students studying real-world analytics.

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Contents
Question:1 : ....................................................................................................................................... 3
Question:2: ........................................................................................................................................ 6
Question: 3: ....................................................................................................................................... 7
Question: 4: ....................................................................................................................................... 8

Question:1 :-
(Solution)
a. The problem, which was formulated in the question in shown below,
Amount (litres) per 1000 kg of A and B
Sheep Cow Goat Cost ($/kg)
A 30 60 40 5
B 80 40 70 8
Before going forward with Linear Programming problems, we need to focus on its
mother distribution, which is Operations Research or OR. Operations Research is a type of
technique, which is used in organizations to solve complex business-oriented problems. World
war two was the time of inception of such a versatile study process which now a days, is been
utilized to solve numerous types of problem in different domains.
Linear programming falls under OR methods. It works with a single objective function,
e.g. maximization of profit or minimization of cost of the production, with the help of some
decision variables. Why the name linear? The answer is that the objective function of the
problem , the constraints are linear function of the decision variables.
A general and perhaps the simplest form of a Mathematical Programming Problem is
shown below,
min or max f(x1,………,xn)
s.t. gi(x1; : : : ; xn) {= or >= or <=} bi
Where ,
x belongs to X
And Linear programming assumes the linearity of both the functions f and gi.
Looking at the problem we can defiantly conclude that the problem satisfies all the
requirement, to be called and solved as a Linear Programming Problem.
b. Now we need to formulate the Linear Programming Problem,
The objective is to minimize the total cost of production of cheese. The problem is
formulated below,
Minimize,
Z = 5𝑥1 + 8𝑥2
Subject to,
30𝑥1 + 80𝑥2 ≥ 45
60𝑥1 + 40𝑥2 ≥ 50
40𝑥1 + 70𝑥2 ≤ 60

Where, 𝑥1 , 𝑥2 > 0
c. For the solution of the Linear Programming Problem, we need to use
Graphical Method here,
We are taking a choice of Z, let it be 40.
Thus, we get the equation ,
5𝑥1 + 8𝑥2 = 40
Or, 𝑥1 + 𝑥2 = 1
8 5
Which gives the following two co-ordinates, wiz. (8,0) and (0,5) . Now, with the
given three lines, one of them defines the feasible region.
It can be observed, that as we move away from origin (i.e. (0,0)) the value
gradually decreases.
We have to solve the following equations to get the feasible region,
−30𝑥1 − 80𝑥2 + 45 = 0
−60𝑥1 − 40𝑥2 +50 =0
And 40 𝑥1 + 70𝑥2 − 60 = 0
From 1 and 2 we get,
𝑥1 −4000−1800 = 𝑥2 −1500−2700 = 1
−1200−4800
Solving this we get , (x1,x2) = (58/60 , 42/60)
From 1 and 3 we get,
𝑥1
Solving this we get,
4800+3150 = 𝑥2 1800+1800 = 1
−2100−3200
(x1,x2) =(-7950/5300, -3600/5300)
From 2 and 3 we get,
𝑥1
Solving this we get,
2400+3500 = 𝑥2 3600+2000 = 1
−4200−1600
(x1,x2) = (-5900/5800, -5600/5800)
Now from the objective function we are replacing the values of the points
respectively to get the optimal solution,

For A , -5 × 8 - 5×5 = -65
B , -5 × (58/60) - 8×(42/60) = -10.43
C , -5 × (-7950/5300) -8 × (3600/5300) = 2.066
D , -5 × (5900/5800) -8 × (5600/5800) = -12.81.
Looking at the above solutions we say that points (x1,x2) = (-5900/5800, -
5600/5800) is the optimal solution.
Below is the graphical representation .
d. A line, which coordinates with two feasible points but does not appear in any other
feasible set, is considered as a convex feasible set of solution. It appears in various problems
and specially in Linear programming problems. If we come across such a objective function
which is convex in nature and our purpose is to maximize it, it could have multiple solution
points and the Local optimum can be considered as the Global Optimum. But if we have a
contradictory inequality of constraints then there will be no points that could have been used
to get the optimal solutions. And as a result, we will get a null set regarding the feasible
solution. And the solution will be considered as an infeasible. solution. Feasible solutions
need to have both upper and lower bound. Optimal solution may not be found because of
unbounded feasible set. In any solution method of Linear Programming Problem, and
specially in Graphical Method, we put our interest to find the optimal solution through the
vertex of the Graphical representation.

Optimal solution gets changed with the changing value of constraints. Thus,
there is no suitable range for A that will not change the optimal solution.

Question:2:-
Solution:-
a. The Linear programming problem is described below,
Let ,𝑋𝑖𝑗 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑒𝑚𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑗𝑡ℎ 𝑖𝑡𝑒𝑚 𝑎𝑡 𝑖𝑡ℎ 𝑠𝑒𝑎𝑠𝑜𝑛.
Thus the Following matrix shows it all,
Cotton oolW ilS k
ringSp X11 X12 X13
utu nA m X21 X22 X23
interW X31 X32 X33
Assuming demand = Production,
Thus,
P1 = 60 × 4200 – 5 × 4200 – 30X11-45X12-52X23
P2 = 55 × 3200 – 5 × 3200 – 30X21 – 45X22 -50X23
P3 = 60 × 3500 – 5 ×3500 -30X31-45X32-50X33
And therefore the objective function is,
Z = P1+P2+P3 = 586700 -30(X11+X21+X31) – 45(X12+X22+X32) -50(X13+X23+X33)
Subject to,
X11+X12+X13 ≤ 4200
X21+X22 ≤ 3200
X31+X32+X33 ≤3500
And
X11 ≥ 2100
X12 ≥ 1680
X21 ≥ 1920
X22 ≥ 1280
X31 ≥ 1400
X32 ≥ 1750
b. Find the attached file R code 1

This Linear programming problem has nothing to do with Duality, as the constraints are
satisfied firmly. Objective is to maximize the profit and for the same purpose we evaluated the
objective function. And after substituting the values we do consider that the optimizing profit would
be $56 , keeping the condition of the constraints fixed. Where the decision variables will consider
the values 45, 57 and 36 respectivly.

Question: 3:-
a. Zero-sum game is a expansion of Game Theory. The fundamental of the
concept is that a gain of a player is loss to another. Let us take a coin tossing
example, say two players 1 and 2. Tossing an unbiased coin, if tail appears Player 1
is going to lose to Player one and vies-versa. In the problem above there are two
companies 1 and 2. Loss of One signifies gain of another’s bid. Thus, it is clearly a
Zero sum game.
b. The pay-off matrix considering all possible outcomes of bid is given below.
$10 $20 $30 $40 $50
$10
Gain Company
1
Loss Company
1
Loss Company
1
Loss Company
1
Loss Company
1
$20
Gain Company
1
Gain Company
1
Loss Company
1
Loss Company
1
Loss Company
1
$30
Gain Company
1
Gain Company
1
Gain Company
1
Loss Company
1
Loss Company
1
$40
Gain Company
1
Gain Company
1
Gain Company
1
Gain Company
1
Loss Company
1
$50
Gain Company
1
Gain Company
1
Gain Company
1
Gain Company
1
Gain Company
1
c. Saddle point is a point of a certain player, where that player will win for each
time, irrespective of the strategy of another player. Here we are adopting the
MaxMin strategy for Company 1 and MinMax strategy for Company 2. And
calculating the diagonal points for Max gain in bidding for Minimum gain in bidding
for company 1 and the maximum value among them is $ 50 Million, $ 50 Million.
That means if no matter whatever bid the other company produces, Company 1 will
gain if they bid $ 50 Million anytime.
d. For Company 1,
Say, the value of the game is = V.
Thus the objective function becomes maximize Z =1/V = x1+x2+x3+x4+x5
Subject to,
10𝑥1 + 20𝑥2 + 30𝑥3 + 40𝑥4 + 50𝑥5 ≤ 10
10𝑥1 + 20𝑥2 + 30𝑥3 + 40𝑥4 + 50𝑥5 ≤ 20
10𝑥1 + 20𝑥2 + 30𝑥3 + 40𝑥4 + 50𝑥5 ≤ 30
10𝑥1 + 20𝑥2 + 30𝑥3 + 40𝑥4 + 50𝑥5 ≤ 40
10𝑥1 + 20𝑥2 + 30𝑥3 + 40𝑥4 + 50𝑥5 ≤ 50
e. Find the attached R code 2

f. The objective of any Linear programming problem is to identify the optimal
solution. The problem here is to manage the strategy of company 1 , and finding the
feasible region, where they could not get lost . The work was done in a firm and
perhaps the most efficient way possible to find the region, for the solution. When we
do convert an Game theory problem to an Linear Programming Problem we do face
the situation of Duality and which is considered the first hurdle of the Game but the
solution is very simple and straight forward for this problem as we did consider the
profit of Company 1 for all the cases and was interested to find the optimal strategy
for its game itself.

)
Question: 4:-
Solution:-
a. The function to calculate profit is , 𝜋 = 𝑃1𝑄1 − 𝐶𝑄1
𝜋 = 𝑄1(𝑃1 − 𝐶)
And we know,
𝑄1 = 180 − 𝑃1 − (𝑃1 − 𝑃ˉ)
& 𝑄2 = 180 − 𝑃2 − (𝑃2 − 𝑃ˉ)
C = 20 and Pˉ is the average cost.
Now,
𝜋1 (𝑃1 , 𝑃2) = [ 180 − 𝑃1 − (𝑃1 − 𝑃1+𝑃2 ] ( 𝑃1 − 20)
2
𝜋1 (𝑃1 , 𝑃2) = [ 180 − 𝑃2 − (𝑃2 − 𝑃1+𝑃2)] ( 𝑃2 − 20)
2
Replacing the values, for the different firms prices and choices of profits are
obtained in the pay-off matrix for the game of discrete choice for both the firms
are shown below.
P2 = 60 P2 = 70 P2 = 80
P1 =
60 4800 , 4800 4800 , 5500 4800 , 6000
P1 =
70 5500 , 4800 5500, 5500 5500 , 6000
P1 =
80 6000 , 4800 6000 , 5500 6000 , 6000
b. From the payoff matrix above, we conclude that the price for Nash
equilibrium is 70.
And the profits at this equilibrium is (5500, 5500).
c. The pay-off Matrix for C = 30 is shown below,
P2 = 60 P2 = 70 P2 = 80
P1 =
60 3600, 3600 3600 ,4800 3600 ,6000
P1 =
70 4500, 3600 4800 , 4800 4800 , 6000
P1 =
80 6000 ,3600 6000 , 4800 6000 ,6000
And as we can see that the Nash equilibrium does not gets changed with the
change of the value of C.