Real World Analytics Question Answer 2022

Real World Analytics 4
Real World Analytics
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Real World Analytics 4
Real World Analytics
Question 1
(a) The problem can be presented in form of a set of linear equations. There is an
objective in the problem (minimize costs) and the function has a direct linear
relationship with the restrictions on the mount of product A or B than can be used in
making the beverage (Morris 2012). The other condition that the problem satisfy is
the non-negativity assumption of linear programming. Since we cannot use less than
zero units of the ingredients in making the beverage. Therefore, the problem can be
modeled by a linear programming method.
(b) In order to represent the problem in mathematical terms, let the food factory use x =
liters of product A and y liters of product B in producing the beverage. This implies
that the total volume of beverage produced is (x + y) liters.
C = 5x + 7y.
In addition to the cost the restrictions are presented as follows:
Orange Mango Lime
x + 2
3 y ≥ 75, x + 3
2 y ≥ 125, x + 8
3 y ≤2 00
Certainty constraint: x ≥ 0 and y≥ 0.
In mathematical terms, the problem is to:
Min C = 5x + 7y.
Subject to:
x + 2
3 y ≥ 75, x + 3
2 y ≥ 125, x + 8
3 y ≤2 00, x ≥ 0 and y≥ 0.
(c) The equation is plotted in desmos.com as shown in screenshot below

Real World Analytics 4
Using the values at the corner points gives a minimum total cost of 595 at point (35,
60). Therefore, the factory should use 35 liters of product A and 60 liters of product B
to produce 95 liters of the beaverage.
(d) If we increase the liters of A by one unit the new optimal solution is at point (33.33,
62.5). This means that the price should be less than one. After tring values between 5
to 5.9 we realise the values that A can take without chnaging the position of the
optimal solution is between 5 and 5.85 liters of product A.
Question 2
(a) Given xij ≥ 0 is the decision variable that denotes the number of tons of products j for
j ε { 1=Spring;2=Autumn ; 3=Winter } to be produced from materials i for
i ε {C=Cotton ,W =Wool , S=Silk } . Also, the proportion of a particular type of
material in a particular type of product can be calculated as xc1
xc1 + xw 1+ xs 1
proportion
of cotton in Spring. Let profit be the difference between total cost and the total
revenue, given as:
Profit = Cost – Revenue, where Cost = production cost + purcahse cost, Revenue =
sales price x deamnd. Then,
Revenue=60 xc1 +60 xw 1+60 xs 1 +55 xc2 +55 xw 2+55 xs 2 +60 xc3 +60 xw 3+ 60 x s3
Production cost=5 ( xc 1+ xw1 + xs 1 ) +4 ( xc2 + xw 2+ xs 2 )+ 5 ( xc3 + xw 3+ xs 3 )
Purchase cost=30 xc1 + 45 xw 1 +50 xs 1+30 xc 2+ 45 xw 2+50 xs 2 +30 xc3 + 45 xw 3 +50 xs 3
Cost =35 xc 1 +50 xw1 +55 xs 1+ 34 xc2 + 49 xw 2 +54 xs 2 +35 xc3 +50 xw3 +55 xs 3
Profit=25 xc1 +10 xw 1+5 xs 1 +21 xc 2 +6 xw 2+ x s2 +25 xc 3+10 xw 3 +5 xs 3
Therefore, we