Linear Programming Assignment: Business Optimization Strategies

This document presents a solved linear programming assignment. Task 1 focuses on maximizing customer ad exposure by optimizing newspaper and radio ad placements, considering budget constraints, and using Excel Solver. Task 2 aims to maximize media exposure to working mothers through different ad packages, analyzing scenarios with varying media exposure levels and budgets. Task 3 addresses minimizing production costs across three facilities, determining optimal production weeks for two products, and evaluating the impact of constraints on the solution. The assignment covers objective functions, constraints, and the use of Excel Solver to find optimal solutions for business optimization problems.

LINEAR PROGRAMMING
STUDENT ID:
[Pick the date]

TASK 1
(a) Let
X is the total number of newspaper ads
Y is the total number of radio ads
Number of customers for Newspaper = 6000
Number of customers for radio ads = 8000
Now,
The aim is to maximize the number of ad exposures to the customers.
Objective function
Max Z=6000 X+8000 Y
Subject to constraints
 For number of newspaper ads
X ≤ 21
 For number of radio ads
Y ≤28
 For total cost of ads
100 X +300 Y ≤6000
X , Y ≥0 ( Non−negativity constraints)
Excel solver model
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(b) It can be concluded that Matt’s Master Photography should give 21 ads in newspaper and
13 ads in radio so as to reach out the maximum number of customers i.e. 230,000.
(c) If the total available cost changes and becomes $8000 then the respective constraints
would also be changed accordingly.
 For total cost of ads
100 X +300 Y ≤8000
Excel solver model
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Hence, it can be concluded that Matt’s Master Photography should give 21 ads in newspaper
and 20 ads in radio so as to reach out the maximum number of customers i.e. 283333.3when
the total cost available is $8000.
(d) From the above computations, it is evident that as there is a change in the constraints, the
optimum mix of ads in various media would also change. One of the key variables
impacting the optimal solution would be the overall cost of advertisements. Another key
aspect is the maximum amount of television and radio ads which are available.
TASK 2
(a) Let
X is the total number of ads package of newspaper
Y is the total number of ads package of radio
Z is the total number of ads package of TV
Now,
The aim is to maximize the exposure of working mother’s
Objective function
Max Z=6 X +12Y +8 Z
Subject to constraints
 For cost of package
X +Y +2 Z ≤ 16
 For available audience
10 X +7 Y +18 Z<¿ 140
 For total cost of ads
100 X +300 Y ≤6000
X , Y , Z ≥ 0 (Non−negativity constraints)
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Excel solver model
Therefore, the maximum exposure of working mother would happen when 16 units of Radio
ads package would be purchased and no newspaper and TV ads package adopted.
Further, the total number of package exposure would be 192.
(b) Media exposure to working mothers has now changed as 6000 to newspaper, 4000 to
radio, 8000 to TV and thus, the new model is
Excel solver model
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(c) Media exposure to working mothers has now changed as 4000 to newspaper, 4000 to
radio, 8000 to TV and thus, the new model is
Excel solver model
d) The key parameter which has been altered in the current problem is the exposure of
different media to working mothers. This clearly impacts the optimal solution as apparent
from comparison of part a, part b and part c. Another important variable to be considered is
the total available advertisement budget that Suzzane has for disposal. The optimal mix
would also be a function of this budget.
TASK 3
(a) Let
A, B and C are the three facilities to produce the two products.
Now,
The aim is to minimize the total cost of the company by determining the optimal number of
weeks for each facility.
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Objective function
MinC =1000 A+ 3000 B+ 4000C
Subject to constraints
 For menu I (Ceylonese)
200 A +200 B+400 C ≥ 2000
 For menu II (Bengali)
100 A +200 B+ 100C ≥1200
 A , B ≥ 0(Non−negativity constraints)
(b) Excel solver model
Therefore, the minimal cost of the company would be $12,000 and the optimal number of
weeks for each facility (A, B and C) would be 12, 0 and 0 respectively.
(c) The optimal solution to the above problem would alter if there are lesser constraints. This
is because in the above solution, it is evident that the optimal distribution of weeks for the
different facilities would be impacted if the constraints are altered. The current solution
indicates the best solution based on the current constraints. It is rational to expect that the
solution can be improved if the constraints are lesser.
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