Case study: Nut supplier Supply Chain assignment 代写

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  • Case study: Nut supplier Supply Chain assignment 代写

    Case study: Nut supplier
    The firm makes four different products from almond nuts grown locally in
    Australia: raw nuts (Raw), roasmixed nuts (Roasted), roasted and salted
    mixed nuts (Salted), and chili-coated mixed nuts (Chili)The firm is barely able to
    keep up with the increasing demand for these products. However, increasing raw
    material prices and foreign competition are forcing the firm to watch its margins
    to ensure it is operating in-the most efficient manner possible. To mee marketing demands for the coming week, the firm needs to produce at leas 1,500 kg of the Raw product, between 600 and 700 kg of the Roasted product, at
    least 500 kg of the Salted product, and no more than 400 kg of the Chili product.
    Each kilogram of the Raw, Roasted, Salted, and Chili product contains,
    respectively, 100%, 70%, 55%, and 40% almond nuts with the remaining weight
    made up of other nuts. The company has 3000 kg of almond nuts and 1000 kg of
    othnuts for use in the next week. The various products are made using four
    different machines that hull the nuts, roast the nuts, coat the nuts in chili (if
    needed), and package the products. The following table summarises the  required by each product on each machine. Each machine has 70 hours of time
    available in the coming week.
    Minutes required per kg
    Machine  Raw  Roasted  Salted  Chili
    Hulling  1.00  1.00  1.00  1.00
    Roasting  0.00  1.00  1.75  1.50
    Coating  0.00  0.00  0.00  1.00
    Packaging  1.50  1.50  1.25  1.00
    The controller recently presented management with the following financial
    summary of the firm’s average weekly operations over the past quarter.
    Product
    Raw  Roasted  Salted  Chili  Total
    Sales Revenue  $5,412  $1,846  $623  $1,165  $9,046
    Variable Costs
    Direct materials  $1,331  $560  $144  $320  $2,355
    Direct labour  $1,092  $400  $96  $130  $1,718
    Manufacturing overhead  $333  $140  $36  $90  $599
    Selling & Administrative  $540  $180  $62  $120  $902
    Allocated Fixed Costs
    Manufacturing overhead  $688  $331  $99  $132  $1,250
    Selling & Administrative  $578  $278  $83  $111  $1,050
    Units Sold  1040  500  150  200  1890
    1. Formulate an LP model for this problem to maximise the total profit and
    briefly describe each equation.
    2Cree a spreadsheet model for this problem and solve it using Solver.
    3. What is the optimal solution? You must report inputs, output, and
    required resources.
    4. Create a sensitivity report for this solution, and answer the following
    questions:
    a. If the firm wanted to decrease the production on any product,
    which one would you recommend and why?
    b. If the firm wanted to increase the production of any product,
    which one would you recommend and why?
    c. Which resources are preventing the firm from making more
    money? If the fm could acquire more of this resource, how much
    should it acquire and how much should it be willing to pay to
    acquire this resource?
    d. If the marketing department wanted to decrease the price of the
    Roasted product by $1, would the optimal solution change and
    why?
     
    Supply Chain Analysis & Design
    Topic 2
    Linear Programming:
    Concepts and Model
    Formulation
    2
    An Introduction to Linear Programming
    • Linear Programming Problem
    • Problem Formulation
    • A Simple Maximization Problem
    • Application of LP models
    • Revenue Management
    2
    3
    Linear Programming
    • Linear programming has nothing to do with computer
    programming.
    • The use of the word “programming” here means
    “choosing a course of action.”
    • Linear programming involves choosing a course of
    action when the mathematical model of the problem
    contains only linear functions.
    3
    4
    Linear Programming (LP) Problem
    • The maximization or minimization of some quantity is
    the objective in all linear programming problems.
    • All LP problems have constraints that limit the degree
    to which the objective can be pursued.
    • A feasible solution satisfies all the problem's
    constraints.
    • An optimal solution is a feasible solution that results in
    the largest possible objective function value when
    maximizing (or smallest when minimizing).
    • A graphical solution method can be used to solve a
    linear program with two variables.
    4
    5
    Linear Programming (LP) Problem
    • If both the objective function and the constraints are
    linear, the problem is referred to as a linear
    programming problem.
    • Linear functions are functions in which each variable
    appears in a separate term raised to the first power and
    is multiplied by a constant (which could be 0).
    • Linear constraints are linear functions that are
    restricted to be "less than or equal to", "equal to", or
    "greater than or equal to" a constant.
    5
    6
    LP Model
    0 ,
    180 5
    100 2 3
    150 5 3
    :
    4 2
    2 1
    2 1
    2 1
    2 1
    2 1
     
     
     
     
    x x
    x x
    x x
    x x
    to Subject
    x x Z Max
    Constraints
    Objective Function
    Non negativity constraints
    6
    7
    Problem Formulation
    • Problem formulation or modeling is the process of
    translating a verbal statement of a problem into a
    mathematical statement.
    • Formulating models is an art that can only be mastered
    with practice and experience.
    • Every LP problems has some unique features, but most
    problems also have common features.
    7
    8
    Formulating LP Models
    • Formulating linear programming models involves the
    following steps:
    1. Understand the problem thoroughly.
    2. Determine the objective function.
    3. Describe each constraint.
    4. Define the decision variables (i.e. controllable
    inputs).
    5. Write the objective function in terms of the decision
    variables.
    6. Write the constraints in terms of the decision
    variables.
    8
    9
    Example 1
    a : Giapetto’s Woodcarving
    • Giapetto’s, Inc., manufactures wooden soldiers and trains.
    1.Each soldier built:
    • Sell for $36 and uses $19 worth of raw materials.
    • Increase Giapetto’s variable labor/overhead costs by $14.
    • Requires 2 hours of finishing labor.
    • Requires 1 hour of carpentry labor.
    2.Each train built:
    • Sell for $21 and used $9 worth of raw materials.
    • Increases Giapetto’s variable labor/overhead costs by $10.
    • Requires 1 hour of finishing labor.
    • Requires 1 hour of carpentry labor.
    a
    example from Operations Research Applications and Algorithms, Wayne L. Winston (2004), 4 th edition, Cengage Learning,
    ISBN- 13: 978-0-534-38058-8
    9
    10
    Ex. 1 - continued
    • Each week Giapetto can obtain:
    • All needed raw material.
    • Only 100 finishing hours.
    • Only 80 carpentry hours.
    • Demand for the trains is unlimited.
    • At most 40 soldiers are bought each week.
    • Giapetto wants to maximize weekly profit (revenues –
    costs).
    • Formulate a mathematical model of Giapetto’s
    situation that can be used maximize weekly profit.
    10
    11
    Example 1: Solution
    • The Giapetto solution model incorporates the
    characteristics shared by all linear programming
    problems.
    • Decision variables should completely describe the
    decisions to be made.
    •x 1 = number of soldiers produced each week
    •x 2 = number of trains produced each week
    • The decision maker wants to maximize (usually
    revenue or profit) or minimize (usually costs) some
    function of the decision variables. This function to
    be maximized or minimized is called the objective
    function.
    •For the Giapetto problem, fixed costs do not
    depend upon the values of x 1 or x 2 .
    11
    12

    Case study: Nut supplier Supply Chain assignment 代写
    Ex. 1 - Solution continued
    • Giapetto’s weekly profit can be expressed in terms of
    the decision variables x 1 and x 2 :
    Weekly profit =
    weekly revenue – weekly raw material costs – the
    weekly variable costs = 3x 1 + 2x 2
    • Thus, Giapetto’s objective is to choose x 1 and x 2 to
    maximize weekly profit. The variable z denotes the
    objective function value of any LP.
    • Giapetto’s objective function is
    Maximize z = 3x 1 + 2x 2
    • The coefficient of an objective function variable is
    called an objective function coefficient.
    12
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    Ex. 1 - Solution continued
    • As x 1 and x 2 increase, Giapetto’s objective function
    grows larger.
    • For Giapetto, the values of x 1 and x 2 are limited by the
    following three restrictions (often called constraints):
    • Each week, no more than 100 hours of finishing time
    may be used. (2 x 1 + x 2 ≤ 100)
    • Each week, no more than 80 hours of carpentry time
    may be used. (x 1 + x 2 ≤ 80)
    • Because of limited demand, at most 40 soldiers
    should be produced (x 1 ≤ 40)
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    14
    • The coefficients of the decision variables in the constraints are
    called the technological coefficients. The number on the right-
    hand side of each constraint is called the constraint’s right-hand
    side (or rhs).
    • To complete the formulation of a linear programming problem,
    the following question must be answered for each decision
    variable.
    • Can the decision variable only assume nonnegative values, or is the
    decision variable allowed to assume both positive and negative
    values?
    • If the decision variable can assume only nonnegative values, the sign
    restriction x i ≥ 0 is added.
    • If the variable can assume both positive and negative values, the
    decision variable x i is unrestricted in sign (often abbreviated urs).
    14
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    Ex. 1 - Solution continued
    • For the Giapetto problem model, combining the sign
    restrictions x 1 ≥0 and x 2 ≥0 with the objective function
    and constraints yields the following optimization
    model:
    Max z = 3x 1 + 2x 2 (objective function)
    Subject to (s.t.)
    2 x 1 + x 2 ≤ 100  (finishing constraint)
    x 1 + x 2 ≤ 80  (carpentry constraint)
    x 1 ≤ 40  (constraint on demand for soldiers)
    x 1 ≥  0  (sign restriction)
    x 2 ≥  0  (sign restriction)
    15
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    Revenue Management
    • Another LP application is revenue management.
    • Revenue management involves managing the short-
    term demand for a fixed perishable inventory in
    order to maximize revenue potential.
    • The methodology was first used to determine how
    many airline seats to sell at an early-reservation
    discount fare and how many to sell at a full fare.
    • Application areas now include hotels, apartment
    rentals, car rentals, cruise lines, and golf courses.
    16
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    Question?
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    Case study: Nut supplier Supply Chain assignment 代写