BONUS QUESTION [5 MARKS]!! Using similar computations, you can verify that the dual price for machine#2 capacity is $2.00/hr and it remains valid for changes (increases or decreases) that move its constraint parallel to itself to any point on the line segment D to E. Please verify?! ques The rate of revenue change resulting from increasing machine#1 capacity by 1 hr (point C to point G): = [28 - zd] - [ Capacity change] = [142 - 128] = [9 - 8] = $14.00/hr The computed rate provides a direct link between the model input (resources) and its output (total revenue) that represents the unit worth of a resource in $/hr) - that is, the change in the optimal objective value per unit change in the availability of the resource (machine capacity). This means that a unit increase (decrease) in machine#1 capacity will increase (decrease) revenue by $14. The dual price of $14.00/hr remains valid for changes (increases or decreases) in machine#1 capacity that move its constraint parallel to itself to any point on the line segment B to F. This means that the range of applicability of the given dual price can be computed as follows: Minimum machine#1 capacity is: (at B = (0, 2.67)] = 2 x 0 + 1 X 2.67 = 2.67 hrs Maximum machine#1 capacity is: [at F = (8,0)] = 2 8 + 1 X0 = 16 hrs We can thus conclude that the dual price of $14.00/hr will remain valid for the range: 2.67 hrs S Machine#1 capacity 5 16 hrs Changes outside this range will produce a different dual price (worth per unit) moodle.ubt.edu.sa JOBCO JOBCO produces two products on two machines. A unit of product#1 requires 2 hours on machine 1 and 1 hour on machine 2. For product#2, a unit requires 1 hour on machine 1 and 3 hours on machine 2. The revenues per unit of products 1 and 2 are $30 and $20, respectively. The total daily processing time available for each machine is 8 hours. Letting X, and x, represent the daily number of units of products #1 and #2, respectively, the LP model is given as: Maximize z = 30x, + 20x2 Subject to 2x, + x2 38 *, + 3x2 5 8 (Machine#1) (Machine#2) Next figure illustrates the change in the optimum solution when changes are made in the capacity of machine#1. If the daily capacity is increased from 8 hours to 9 hours, the new optimum will occur at point G. The rate of change in optimum z resulting from changing machine#1 capacity from 8 hours to 9 hours can be computed as follows: GRAPHICAL SENSITIVITY OF OPTIMAL SOLUTION TO CHANGES IN THE RESOURCES OR RIGHT-HAND- SIDE VALUES The rate of revenue change resulting from increasing machine#1 capacity by 1 hr (point to point G): = 126 -2 = Capacity change) = [142 - 128] = [9 - 8] = $14.00/hr The computed rate provides a direct link between the model input (resources) and its output (total revenue) that represents the unit worth of a resource in $/hr) - that is, the change in the optimal objective value per unit change in the availability of the resource (machine capacity). This means that a unit increase (decrease) in machine#1 capacity will increase (decrease) revenue by $14 The dual price of $14.00/hr remains valid for changes (increases or decreases) in machine#1 capacity that move its constraint parallel to itself to any point on the line segment B to F. This means that the range of applicability of the given dual price can be computed as follows: Minimum machine capacity ise [ot B = 10,2.67)1 = 2 x 0 + 1 x 2.67 = 2.67 hrs Maximum machine capacity is for F = (8,0) = 2x 8+1 X0 = 16 hrs We con thus conclude that the dual price of $14.00/he will remain valid for the range 2.67 hrs S Machine#1 capacity 5 16 hrs Changes outside this range will produce a different dual price (worth per unit). 2:03 1 moodle.ubt.edu.sa = [20-22 : 1 Capacity change] = [142-128) 19-8) = $14.00/hr The computed rate provides a direct link between the model input (resources and its output total revenue) that represents the unit worth of a resource in /hr) - that is, the change in the optimal objective value per unit change in the availability of the resource machine capacity). This means that a unit increase (decrease in machine #1 capacity will increase (decrease) revenue by $14 The dual price of $14.00/hr remains valid for changes increases or decreases) in machine#1 capacity that move its constraint parallel to itself to any point on the line segment B to F. This means that the range of applicability of the given dual price can be computed as follows Minimum machine capacity is for B = (0,2.67)) = 2 x 0 + 1 x 2.67 = 2.67 hrs Maximum machine capacity is fotF (8,0) = 2x 8+1 O 16 hours We con thus conclude that the dual price of $14.00/hr will remain valid for the range 2.67 hrs Machine capacity 5 16 hrs Changes outside this range will produce a different duel price (worth per un BONUS QUESTION 5 MARKS]!! Using similar computations, you can verify that the dual price for machine#2 capacity is $2.00/hr and it remains valid for changes increases or decreases) that move its constraint parallel to itself to any point on the line segment D to E. Please verify?! SS BONUS QUESTION [5 MARKS]!! Using similar computations, you can verify that the dual price for machine#2 capacity is $2.00/hr and it remains valid for changes (increases or decreases) that move its constraint parallel to itself to any point on the line segment D to E. Please verify?! ques The rate of revenue change resulting from increasing machine#1 capacity by 1 hr (point C to point G): = [28 - zd] - [ Capacity change] = [142 - 128] = [9 - 8] = $14.00/hr The computed rate provides a direct link between the model input (resources) and its output (total revenue) that represents the unit worth of a resource in $/hr) - that is, the change in the optimal objective value per unit change in the availability of the resource (machine capacity). This means that a unit increase (decrease) in machine#1 capacity will increase (decrease) revenue by $14. The dual price of $14.00/hr remains valid for changes (increases or decreases) in machine#1 capacity that move its constraint parallel to itself to any point on the line segment B to F. This means that the range of applicability of the given dual price can be computed as follows: Minimum machine#1 capacity is: (at B = (0, 2.67)] = 2 x 0 + 1 X 2.67 = 2.67 hrs Maximum machine#1 capacity is: [at F = (8,0)] = 2 8 + 1 X0 = 16 hrs We can thus conclude that the dual price of $14.00/hr will remain valid for the range: 2.67 hrs S Machine#1 capacity 5 16 hrs Changes outside this range will produce a different dual price (worth per unit) moodle.ubt.edu.sa JOBCO JOBCO produces two products on two machines. A unit of product#1 requires 2 hours on machine 1 and 1 hour on machine 2. For product#2, a unit requires 1 hour on machine 1 and 3 hours on machine 2. The revenues per unit of products 1 and 2 are $30 and $20, respectively. The total daily processing time available for each machine is 8 hours. Letting X, and x, represent the daily number of units of products #1 and #2, respectively, the LP model is given as: Maximize z = 30x, + 20x2 Subject to 2x, + x2 38 *, + 3x2 5 8 (Machine#1) (Machine#2) Next figure illustrates the change in the optimum solution when changes are made in the capacity of machine#1. If the daily capacity is increased from 8 hours to 9 hours, the new optimum will occur at point G. The rate of change in optimum z resulting from changing machine#1 capacity from 8 hours to 9 hours can be computed as follows: GRAPHICAL SENSITIVITY OF OPTIMAL SOLUTION TO CHANGES IN THE RESOURCES OR RIGHT-HAND- SIDE VALUES The rate of revenue change resulting from increasing machine#1 capacity by 1 hr (point to point G): = 126 -2 = Capacity change) = [142 - 128] = [9 - 8] = $14.00/hr The computed rate provides a direct link between the model input (resources) and its output (total revenue) that represents the unit worth of a resource in $/hr) - that is, the change in the optimal objective value per unit change in the availability of the resource (machine capacity). This means that a unit increase (decrease) in machine#1 capacity will increase (decrease) revenue by $14 The dual price of $14.00/hr remains valid for changes (increases or decreases) in machine#1 capacity that move its constraint parallel to itself to any point on the line segment B to F. This means that the range of applicability of the given dual price can be computed as follows: Minimum machine capacity ise [ot B = 10,2.67)1 = 2 x 0 + 1 x 2.67 = 2.67 hrs Maximum machine capacity is for F = (8,0) = 2x 8+1 X0 = 16 hrs We con thus conclude that the dual price of $14.00/he will remain valid for the range 2.67 hrs S Machine#1 capacity 5 16 hrs Changes outside this range will produce a different dual price (worth per unit). 2:03 1 moodle.ubt.edu.sa = [20-22 : 1 Capacity change] = [142-128) 19-8) = $14.00/hr The computed rate provides a direct link between the model input (resources and its output total revenue) that represents the unit worth of a resource in /hr) - that is, the change in the optimal objective value per unit change in the availability of the resource machine capacity). This means that a unit increase (decrease in machine #1 capacity will increase (decrease) revenue by $14 The dual price of $14.00/hr remains valid for changes increases or decreases) in machine#1 capacity that move its constraint parallel to itself to any point on the line segment B to F. This means that the range of applicability of the given dual price can be computed as follows Minimum machine capacity is for B = (0,2.67)) = 2 x 0 + 1 x 2.67 = 2.67 hrs Maximum machine capacity is fotF (8,0) = 2x 8+1 O 16 hours We con thus conclude that the dual price of $14.00/hr will remain valid for the range 2.67 hrs Machine capacity 5 16 hrs Changes outside this range will produce a different duel price (worth per un BONUS QUESTION 5 MARKS]!! Using similar computations, you can verify that the dual price for machine#2 capacity is $2.00/hr and it remains valid for changes increases or decreases) that move its constraint parallel to itself to any point on the line segment D to E. Please verify?! SS