University Decision Support Systems: Analysis and Applications

This document presents a comprehensive solution to a decision support systems assignment, encompassing various aspects of business development and operations management. It includes detailed answers to questions related to linear programming, focusing on cost minimization, constraint analysis, and sensitivity analysis. The assignment delves into simulation techniques, outlining their advantages and disadvantages compared to other modeling methods, and explores real-world applications in management problem-solving. Furthermore, it covers queuing systems, decision tables, and decision trees, illustrating their use in decision-making processes. The document provides solutions to questions from different years, offering insights into diverse scenarios and problem-solving approaches, making it a valuable resource for students studying decision support systems.

Running Head: DECISION SUPPORT
Decision Support
Name of the Student
Name of the University
Author Note

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Table of Contents
Year 2010.........................................................................................................................................3
Section A.....................................................................................................................................3
Question 2................................................................................................................................3
Question 3................................................................................................................................5
Section B......................................................................................................................................8
Question 4:...............................................................................................................................8
Question 5..............................................................................................................................10
Question 6..............................................................................................................................11
Year 2011.......................................................................................................................................14
Section A...................................................................................................................................14
Answer 2................................................................................................................................14
Section B....................................................................................................................................15
Answer 4................................................................................................................................15
Answer 5................................................................................................................................16
Answer 6................................................................................................................................18
2012...............................................................................................................................................20
Section A...................................................................................................................................20
Answer 2................................................................................................................................20
Section B:...............................................................................................................................21

2
Answer 4:...............................................................................................................................21
Answer 5................................................................................................................................22
Answer 7................................................................................................................................22
2013...............................................................................................................................................25
Section A...................................................................................................................................25
Answer 2................................................................................................................................25
Section B....................................................................................................................................25
Answer 4................................................................................................................................25
Answer 5:...............................................................................................................................26
Answer 6:...............................................................................................................................27

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Year 2010
Section A
Question 2
Part [a]
Here, the objective is to minimize the total cost. Hence, the optimal solution is 740000.
The binding constraints;
Plant capacity constraints
Market deviant constraint
Part [b]
Capacity constraint
140000 desks are to be shipped to W/H1 for optimal cost. However, the allowable increase or
decrease in the number of units to be shipped are infinity and 60000 respectively. Thus, if the
number of desks shipped are more than 80000, there will be no change in the optimal cost.
60,000 desks are shipped to W/H2 for optimal cost. However, the allowable increase or
decrease in the number of units to be shipped are 90000 and 10000 respectively. Thus, if the
number of desks shipped are between 50000 and 150000, the optimal cost will decrease with a
multiple of unity.
Market demand constraint
50,000 desks are received by market 1 for optimal cost. However, the allowable
increase or decrease in the number of units to be received are 90000 and 10000 respectively.
Thus, if the number of desks shipped are between 40000 and 140000, the optimal cost will
increase 3 times.

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100,000 desks are received by market 2 for optimal cost. However, the
allowable increase or decrease in the number of units to be received are 60000 and 90000
respectively. Thus, if the number of desks shipped are between 40000 and 190000, the optimal
cost will increase 4 times.
50,000 desks are received by market 3 for optimal cost. However, the allowable
increase or decrease in the number of units to be received are 10000 and 50000 respectively.
Thus, if the number of desks shipped are between 40000 and 100000, the optimal cost will
increase 4 times.
Part [c]
This will increase the total cost required to ship the desk from plant 1 to W/H1 hence will not
meet the objective.
The new function value will be; = f140000+f120000
To warehouse f260000
Total cost = 260000+150000+360000+10000+50000
= f830000
Part [d]
When the market 2 demand increases the distribution cost increases because from either
warehouse distribution cost is needed. This implies that it has values.
This leads to an increase in the optimal solution to take care of the increase in the
distribution cost.
Part [e]
The shadow price for plant 1 is found to be zero. Thus, there will be no change in cost if the
products are directly supplied to the markets bypassing the warehouses. Hence, no change or
modification in the distribution plan is required as there will be no impact of this change on the
optimal cost.

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Part [f]
Zero reduced cost for shipment between warehouse 1 and market 3 implies that there will be
no change in the objective function if the shipment costs are changed.
Part [g]
Warehouse 2 is currently handling 60,000 desks per. If warehouse 2 cannot handle only 50,000
desks per year, this number lies between the allowable limit for the constraint. Thus, the
optimal solution of the LP will increase by its shadow price, that is, the optimal solution will
increase 3 times.
Question 3
Part [a]
Let x1 be the number of barrels of pruned olives and x2 be the number of barrels of regular olives.
Max Z = 20 x1 + 30 x2
Subject to the constraints:
5 x1 + 2 x2 ≤ 250
x1 + 2 x2 ≤ 150
1.5 x1 ≤ x2
x1, x2 ≥ 0 and takes only integer values.
Part [b]

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Extreme Point Co Ordinates Objective Function Value
O (0, 0) 20 (0) + 30 (0) = 0
A (0, 75) 20 (0) + 30 (75) = 2250
B (25, 62.5) 20 (25) + 30 (62.5) = 2375
C (31.25, 46.875) 20 (31.25) + 30 (46.875) = 2031.25
Hence, the optimal solution is given by x1 = 25, x2 = 62.5 and the maximum profit is £2375
Part [c]
Putting the values of the optimal x1 and x2 in the constraint equations,
5 (25) + 2 (62.5) = 250, thus, this is a binding constraint
(25) + 2 (62.5) = 150, thus this is a binding constraint
1.5 (25) ≤ (62.5), thus this is a non-binding constraint
None of the constraints are redundant
d) The shadow price for the labour constraint has been determined as 1.25, the shadow
price of the land constraint is 13.75 and the shadow price for the third constraint relating to the
number of olives is 0.

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e) If the profit of regular olives increases to £45 per barrel, the new optimal profit becomes
£3375.
Extreme Point Co Ordinates Objective Function Value
O (0, 0) 20 (0) + 45 (0) = 0
A (0, 75) 20 (0) + 45 (75) = 3375
B (25, 62.5) 20 (25) + 30 (62.5) = 3312.5
C (31.25, 46.875) 20 (31.25) + 30 (46.875) =
2734.375
The maximum value of the objective function z = 3375 occurs at the extreme point (0, 75).
Hence, the optimal solution to the given problem is x1 = 0, x2 = 75 and max z = 3375.

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Part [f]
Here, the goals are
Goal 1: x1 + 2 x2 ≤ 120
Goal 2: 20 x1 + 30 x2 ≤ 1800
Limit: 5 x1 + 2 x2 ≤ 250 & 1.5 x1 ≤ x2
Section B
Question 4:
Part [a]
Part [b]
Advantages of simulation compared to other modelling techniques
First, simulation is good in analyzing large and complex problems that cannot be easily
solved by some of the mathematical problems

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Secondly, simulation is always flexible and thus the changes in the variables in the system can
be best made among the best alternatives. Also in simulation process, the experiments are
always carried out with the model without interfering with the policy of the system.
Furthermore decisions on the policy can be made swiftly by knowing in advance the best
options by minimizing the risk of experimenting in the exact system.
Disadvantages of the simulation compared to other modelling techniques
First, simulation does not always generate optimal solutions. A good simulation model
usually takes long period of time. In some cases, simulation tends to be expensive than other
modelling techniques. Finally, the simulation does not always provide the answers by itself and
thus the decision maker must be in a position to provide all the information about the
constraints conditions for examination.
Also in the field of simulation, computers have become very useful tool for the
mathematical modelling since the simulation of the system is always represented as the
running of the system’s model. Computers can be used in simulation to acquire new views and
delve into better advancements and thus estimates the performance of the system to a better
level. Consequently, the magnitude of the events that can be simulated by the computers vary
from one level to another.
Part [c]
The types of management problems that can be solved by modeling techniques are
· Evaluation based on modern business problems like
· Also, they can be applied initially during new operations and procedures, ongoing set of
operations and recommendation for corrective action for the unsatisfactory results.
· Technically on limitations on the availability of capital, workers, supplies, machines etc
· To determine the number of units of a product to sell or produce (i.e. volume) that will
equate total revenue with total cost.

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· Some application areas would be project planning, capital budgeting, inventory analysis,
production planning and scheduling
Some examples of management problems could be
· to rebuild Interstate 10 damaged in the 1994 earthquake in the Los Angeles area - Project
Scheduling Techniques
· the basis for the development of a comprehensive framework for planning environmental
policy in Finland – Decision Analysis
· the overall design plans for Disneyland and Disney World, which lead to the development
of ‘waiting line entertainment’ in order to improve customer satisfaction – Queing Models
· Burger King to find how to best blend cuts of meat to minimize costs – Linear
Programming
Question 5
Part [a]
i. P (4 inch nail) = 451/1719
= 0.26236
ii. P (5 inch nail) = 333/1719
= 0.19371
iii. P (3 inch shorter) = P (1 inch) or P (2 inch) or P (3 inch)
= 651/1719 + 243/1719 +41/1719
= 935/1719
= 0.5439
Part [b]
Decision tables and decision trees are decision support tools and have several merits as
discussed below.

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They are simple to interpret and understand by anyone. Many people tend to
understand the decision tables and decision trees after a short explanation.
They have data values that are not hard. In relation to this therefore, important views
can be generated based on the professionals describing alternatives, probabilities and the
outcomes of the preferences.
Decision tales always allow addition of the new possible scenarios. Also with this, it can
help in the determination of the best, worst values of any expected scenario.
Finally, they can also be combined with other decision techniques in modeling.
Part [c]
From the above figure, it can be said that company XYZ has to decide small system, as it will
give maximum expected monetary value.
Question 6
Part [a]