Course Name: Business Quantitative Analysis QU1 Module: 10 Module Title: Linear Programming

Lectures and handouts by: Paul Jeyakumar, M.Sc., CGA 1

Overview Graphing linear inequalities Linear programming model Graphical sensitivity analysis Applications

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Graphing Linear Inequalities Identify the slope and the y-intercept in the equation for a straight line. Graph an equation for a straight line. Determine the point of intersection of two lines algebraically and graphically. Graph linear inequalities. 3

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Linear Equation: Slope-Intercept Slope-intercept form of the equation for a straight line: y = mx + b where m = slope of the line b = y-intercept

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Linear Equation: Slope-Intercept Example Slope-intercept form: 2x1 + 3x2 = 13 3x2 = - 2x1 + 13 x2 = – 2x1/3 + 13/3 Slope = –2/3

y-intercept = 13/3

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Graphing a Linear Equation 2x1 + 3x2 = 13 At x1 = 0, x2 = 13/3 = 4.333 At x2 = 0, 2x1 + 3(0) = 13, or x1 = 13/2 = 6.5 Plot the points (6.5, 0) and (0, 4.333) on a graph and join them by a straight line. 6

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Graphing Equation - Example

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Solving Two Simultaneous Equations Solve the following two equations using algebra. Graph both equations and verify the answer. 2x1 + 3x2 = 10 …… (1) 3x1 – 4x2 = – 2 …… (2)

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Solving Two Simultaneous Equations (contd.) Multiply the first equation by 3 and the second equation by 2. 6x1 + 9x2 = 30 …… (3) 6x1 – 8x2 = – 4 …… (4) Subtract (4) from (3). 17x2 = 34 or x2 = 2 Substitute x2 = 2 in equation (1). 2x1 + 3(2) = 10 2x1 = 4 or x1 = 2 The lines intersect at (2, 2). 9

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Graphical Solution - Example 2x1 + 3x2 = 10 x1 = 0, x2 = 10/3 = 3.33 x2 = 0, x1 = 10/2 = 5 Points (0, 3.33) and (5, 0) are on the line. 3x1 – 4x2 = – 2 x1 = 0, x2 = 0.5 x2 = 0, x1 = – 2/3 = – 0.667 Points (0, 0.5) and (– 0.667, 0) are on the line. 10

Graphical Solution (contd.)

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Graphing Inequality, 2x1 + 3x2 ≤ 10

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Assumptions Proportionality assumption: No economies of scale or quantity discounts, and for each decision variable, the total amount of each resource input and the associated profit are directly proportional to the value of the variable Example: If one unit of production requires 2 hours of assembly time and the unit profit is $6/unit, then for 10 units we need 20 hours of assembly time and we make a profit of $60. 13

Assumptions (contd.) Divisibility Assumption: Can have fractional values for the decision variables May be true mathematically, but it may not be practical Can we produce 23.6 units? 14

Assumptions (contd.) Additivity Assumption: Given all the values of the decision variables, the total amount of each resource input, and the associated profit, is the same as the sum of the input and profit for each individual process. Example: A unit of Product A takes 2 hours of assembly time, and a unit of Product B takes 3 hours of assembly time. If we produce 10 units of Product A and 15 units of Product B, then the total assembly time is (2 x 10) + (3 x 15) = 65 hours. 15

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Practice Question 1 A company produces round and square goldfish bowls made of glass. The square type has a metal base. Each square bowl requires 4 hours of furnace time while each round bowl requires 2 hours. 2 pounds of glass are required for each type of bowl. Each week the company has 1,600 pounds of glass, 2,000 hours of furnace time, and 400 metal bases. The contribution margins per unit are $8 and $6 for square bowls and round bowls respectively. The company can sell all the bowls it makes. How many of each bowl should be made in a week?

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Practice Question 1 (contd.) Definition of the two decision variables: x1 = number of round bowls to produce/week x2 = number of square bowls to produce/week

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Practice Question 1 (contd.) Objective function: Maximize Z = 6x1 + 8x2

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Practice Question 1 (contd.) Resource constraints: 2x1 + 2x2 ≤ 1,600 (glass amount) 2x1 + 4x2 ≤ 2,000 (furnace time) x2 ≤ 400 (metal bases)

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Practice Question 1 (contd.) Non-negativity constraints: x1 ≥ 0 x2 ≥ 0

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Practice Question 1 (contd.)

Maximize

6x1 + 8x2

Subject to

2x1 + 2x2 ≤ 1,600 2x1 + 4x2 ≤ 2,000 x2 ≤ 400 x1 ≥ 0, x2 ≥ 0

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Practice Question 1 (contd.) To draw the graph, convert the inequality constraints to equality constraints. 2x1 + 2x2 = 1,600 2x1 + 4x2 = 2,000 x2 = 400

(glass amount) (furnace time) (metal bases)

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Practice Question 1 (contd.) For each linear constraint identify 2 points that lie on the line. Plot the 2 points and join the points using a straight line. 2x1 + 2x2 = 1,600 Set x1 to 0. 2(0) + 2x2 = 1,600 or 2x2 = 1,600 or x2 = 800 Set x2 to 0. 2x1 + 2(0) = 1,600 or 2x1 = 1,600 or x1= 800 The points (0,800) and (800,0) are on the line. 23

Practice Question 1 (contd.) 2x1 + 4x2 = 2,000 Set x1 to 0. 2(0) + 4x2 = 2,000 or 4x2 = 2,000 or x2 = 500 Set x2 to 0. 2x1 + 4(0) = 2,000 or 2x1 = 2,000 or x1= 1,000 The points (0,500) and (1,000,0) are on the line.

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Practice Question 1 (contd.) x2 = 400 There is only one set of values (0,400) because x1 does not require any metal bases. 2 non-negativity lines: x1 = 0, and x2 = 0

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Practice Question 1 (contd.)

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Practice Question 1 (contd.) The shaded area in the graph is the set of points that satisfy all the constraints. It is called the solution set or the feasible region. Feasible region has 5 corner points: A, B, C, D, and E

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Practice Question 1 (contd.) Optimal production plan is at one of the extreme points. Compute the contribution margin, 6x1 + 8x2 at each extreme point. Extreme point A: Coordinates: x1 = 0, and x2 = 0 Contribution margin = ($6x0) + ($8x0) = $0 28

Practice Question 1 (contd.) Extreme point B: Coordinates: x1 = 0, and x2 = 400 Contribution margin = ($6x0)+ ($8x400) = $3,200

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Practice Question 1 (contd.) Extreme point E : Coordinates: x1 = 800, and x2 = 0 Contribution margin = ($6x800)+ ($8x0) = $4,800

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Practice Question 1 (contd.) Extreme point D: Point D is at the intersection of the two lines: 2x1 + 4x2 = 2,000 (1) 2x1 + 2x2 = 1,600 (2) With an accurate graph the coordinates could be read as (600, 200). To solve subtract equation (2) from (1): 2x2 = 400, thus x2 = 200 So 2x1 + 4(200) = 2,000. x1 = 1,200/2 = 600 Coordinates: (600, 200) Contribution margin = ($6 × 600) + ($8 × 200) = $5,200

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Practice Question 1 (contd.) Extreme point C: Point C is at the intersection of: 2x1 + 4x2 = 2,000 and x2 = 400 Substitute for x2 = 400 as follows: 2x1 + 4(400) = 2,000, 2x1 + 1,600 = 2,000 2x1 = 400 or x1 = 200 Coordinates: (200, 400) Contribution margin = ($6 × 200) + (8 × 400) = $4,400 32

Practice Question 1 (contd.)

Optimal production plan: 600 round bowls and 200 square bowls

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Sensitivity Analysis How would the solution change if the problem parameters were altered? Shadow price

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Shadow Price Ratio of the increase in the objective function to the increase in resource availability If the resource is non-binding, the shadow price is zero. The shadow price is valid only for a range. 35

Practice Question 1 (contd.) What is the shadow price for metal bases? We have 400 metal bases available. Optimal production plan utilizes only 200. There are 200 unused metal bases. The constraint is non-binding. The shadow price of this resource is zero.

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Practice Question 1 (contd.) Would there be any benefit to increase the furnace time from 2,000 to 2,100 hours? Solve: 2x1 + 2x2 = 1,600 2x1 + 4x2 = 2,100 The solution gives x1 = 550 and x2 = 250, yielding a contribution margin of ($6 × 550) + ($8 × 250) = $5,300. Contribution margin increases from $5,200 to $5,300 by $100. 37

Practice Question 1 (contd.) Shadow price = (increase in objective function) divided by (increase in resource availability) Shadow price = $100/100 hours = $1/hour $1/hour is the maximum additional amount the company should pay for an additional hour of furnace time. The shadow price is valid only for a range. 38

Software Applications We use software applications to solve these linear programming problems dealing with multiple constraints.

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Practice Question 2 Price CD Player $150 Tape Deck $85 Tuner $70

VC $75 $35 $30

CM $75 $50 $40

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Practice Question 2 (contd.) Price VC CM Assembly CD Player $150 $75 $75 3 hours Tape Deck $85 $35 $50 2 hours Tuner $70 $30 $40 1 hour Available assembly hours = 400,000

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Practice Question 2 (contd.) Price VC CM Assembly CD Player $150 $75 $75 3 hours Tape Deck $85 $35 $50 2 hours Tuner $70 $30 $40 1 hour Available assembly hours = 400,000 Sales demand: CD Players Tape Decks Tuners

Maximum 130,000 110,000 90,000

Minimum 50,000 50,000 50,000 42

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Practice Question 2 (contd.) What is our objective? What should we do? Definition of the controllable variables: x1 = number of CD players to make x2 = number of Tape Decks to make x3 = number of Tuners to make

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Practice Question 2 (contd.) CM from producing x1 CD players, x2 Tape decks, and x3 Tuners = 75x1 + 50x2 + 40x3 The objective is to maximize contribution margin. Objective function: Maximize, Z = 75x1 + 50x2 + 40x3

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Practice Question 2 (contd.) Choice of values for x1, x2, and x3 is restricted because of the availability of the assembly time. Time constraint: 3x1 + 2x2 + x3 ≤ 400,000

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Practice Question 2 (contd.) x1 ≥ 50,000 x2 ≥ 50,000 x3 ≥ 50,000 x1 ≤ 130,000 x2 ≤ 110,000 x3 ≤ 90,000

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Practice Question 2 (contd.)

x1 ≥ 0 x2 ≥ 0 x3 ≥ 0

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Practice Question 2 (contd.) Maximize 75x1 + 50x2 + 40x3 Subject to 3x1 + 2x2 + x3 ≤ 400,000 x1 ≥ 50,000 x2 ≥ 50,000 x3 ≥ 50,000 x1 ≤ 130,000 x2 ≤ 110,000 x3 ≤ 90,000 x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 48

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Answer Report – Target Cell Target Cell (Max) Cell Name Original Value $G$3 CM 11350000

Final Value 11350000

The maximum contribution margin as a result of the production plan identified by solver is $11,350,000. 49

Answer Report – Adjustable Cells Cell $B$2 $C$2 $D$2

Name Original Value CD 70000 TD 50000 Tuner 90000

Final Value 70000 50000 90000

The final values indicate the number of units to produce.

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Answer Report – Constraints Name Cell Formula Status Slack Time 400000 $G$5