Analytics Assignment Question 2022

Analytics Assignment 1
Real World Analytics
by[Name]
Course
Professor’s Name
Institution
Location of Institution
Date

Analytics Assignment 2
Real World Analytics: Assignment 4
Question 1
(a) The problem is described in a way that a system of linear equations can be formed.
There is an objective in the problem (minimize costs). A direct link between the
objective and the constraints can be represented by a linear relationship. Next, the
restrictions on the mount of product A or B than can be used in making the beverage
is provided in the question (Morris 2012). The problem satisfies the non-negativity
assumption of linear programming. From the problem negative units of the product A
and B does not exists.
(b) Suppose 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 x + 2
3 y ≥ 75
Mango x + 3
2 y ≥ 125
Lime x + 8
3 y ≤200 , 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 ≤200, x ≥ 0 and y≥ 0.
(c) The solution on desmos.com is as follows:

Analytics Assignment 3
From 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) Increasing the liters of A by one unit shift optimal solution to (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) Using the hint provided in the question wee have: xij ≥ 0 as the decision variable
interprated as the amount in tons of products j where
j ε { 1=Spring;2=Autumn ; 3=Winter } produced from materials i where
i ε {C=Cotton ,W =Wool , S=Silk } . Further, the proportion of a particular type of
material in a particular type of product is given by the formula ( i.e xc1
xc1 + xw 1+ xs 1
proportion of cotton in Spring).
Now, define profit as the difference between total cost and the total revenue:
Profit = Cost – Revenue, where Cost is the sum of production cost and purcahse
cost. Finally, Revenue is the product of sales price and deamnd.
Using the information provided we have:
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
Thus problem is to
Max 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
Subject to:
( xc 1+ xw 1 + xs 1 ≤ 4500 ) , ( xc 2+ xw 2 + xs 2 ≤ 3000 ), ( xc 3+ xw3 + xs 3 ≤3500 ),
( xc 1−xw 1−xs 1 ≥ 0 ), ( 3 xc1 −7 xw1 +3 xs 1 ≤ 0 ), ( xc 1+ xw 1−4 xs 1 ≤ 0 ),
( xc 2−xw 2−xs 2 ≥ 0 ), ( 2 xc2−3 xw 2 +2 xs 2 ≤ 0 ), ( xc 2+ xw 2−9 x s2 ≤ 0 ),
( 3 xc3 −2 xw 3−2 xs 3 ≥0 ), ( xc 3−xw 3+ xs 3 ≤ 0 ), ( xc 3+ xw3 −9 xs 3 ≤ 0 ),