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fertilizer manufacturer has to fulfill supply contracts to its two main customers ​(650 tons to Customer A and 800 tons to Customer​ B). It can meet this demand by shipping existing inventory from any of its three warehouses. Warehouse 1​ (W1) has 400 tons of inventory​ onhand, Warehouse 2​ (W2) has 500 ​tons, and Warehouse 3​ (W3) has 600 tons. The company would like to arrange the shipping for the lowest cost​ possible, where the​ per-ton transit costs are as​ follows:     W 1 W 2 W 3 Customer A $7.50   $6.25   $6.50   Customer B $6.75   $7.00   $8.00     Write the objective function and the constraint in equations. Let Vij​= tons shipped to customer i from warehouse j​,   and so on. For​ example, VA1=tons shipped to customer A from warehouse W1.   This exercise contains only parts​ b, c,​ d, e, and f. Part 2 ​b) The objective function for the LP model​ =

fertilizer manufacturer has to fulfill supply contracts to its two main customers ​(650 tons to Customer A and 800 tons to Customer​ B). It can meet this demand by shipping existing inventory from any of its three warehouses. Warehouse 1​ (W1) has 400 tons of inventory​ onhand, Warehouse 2​ (W2) has 500 ​tons, and Warehouse 3​ (W3) has 600 tons. The company would like to arrange the shipping for the lowest cost​ possible, where the​ per-ton transit costs are as​ follows:     W 1 W 2 W 3 Customer A $7.50   $6.25   $6.50   Customer B $6.75   $7.00   $8.00     Write the objective function and the constraint in equations. Let Vij​= tons shipped to customer i from warehouse j​,   and so on. For​ example, VA1=tons shipped to customer A from warehouse W1.   This exercise contains only parts​ b, c,​ d, e, and f. Part 2 ​b) The objective function for the LP model​ =

Practical Management Science
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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fertilizer manufacturer has to fulfill supply contracts to its two main customers
​(650
tons to Customer A and
800
tons to Customer​ B). It can meet this demand by shipping existing inventory from any of its three warehouses. Warehouse 1​ (W1) has
400
tons of inventory​ onhand, Warehouse 2​ (W2) has
500
​tons, and Warehouse 3​ (W3) has
600
tons. The company would like to arrange the shipping for the lowest cost​ possible, where the​ per-ton transit costs are as​ follows:
 
 
W 1
W 2
W 3
Customer A
$7.50
 
$6.25
 
$6.50
 
Customer B
$6.75
 
$7.00
 
$8.00
 
 
Write the objective function and the constraint in equations. Let
Vij​=
tons shipped to customer i from warehouse
j​,
  and so on. For​ example,
VA1=tons
shipped to customer A from warehouse W1.
 
This exercise contains only parts​ b, c,​ d, e, and f.
Part 2
​b) The objective function for the LP model​ =
 
 
Allowable Decrease Variable Final Value Reduced Cost Objective Coefficient 1.50 Allowable Increase $7.50 1E+30 1.50 VA1 0.00 $6.25 0.25 0.75 100 VA3 $6.50 0.75 0.25 550 0.00 V81 0.00 $6.75 0.50 1E+30 400 0.00 $7.00 0.75 0.50 V82 400 0.75 $8.00 1E+30 0.75 V83 Constraints Constraint RH Side Allowable Increase Allowable Decrease Name Final Value Shadow Price 6.50 650 50 550 650 7.25 800 50 400 C2 800 - 0.50 400 50 400 C3 400 C4 - 0.25 500 550 50 500 C5 550 0.00 600 1E+30 50 (enter your response as an integer). ) Based on the information given in the sensitivity reports, the number of constraints that are binding = ) For variable V43. the range of optimality is from $6.25 to $ (round your response to two decimal places). (round your response to two decimal places). If to customer A, 16 less tons are supplied, how much money might be saved? $
Transcribed Image Text:Allowable Decrease Variable Final Value Reduced Cost Objective Coefficient 1.50 Allowable Increase $7.50 1E+30 1.50 VA1 0.00 $6.25 0.25 0.75 100 VA3 $6.50 0.75 0.25 550 0.00 V81 0.00 $6.75 0.50 1E+30 400 0.00 $7.00 0.75 0.50 V82 400 0.75 $8.00 1E+30 0.75 V83 Constraints Constraint RH Side Allowable Increase Allowable Decrease Name Final Value Shadow Price 6.50 650 50 550 650 7.25 800 50 400 C2 800 - 0.50 400 50 400 C3 400 C4 - 0.25 500 550 50 500 C5 550 0.00 600 1E+30 50 (enter your response as an integer). ) Based on the information given in the sensitivity reports, the number of constraints that are binding = ) For variable V43. the range of optimality is from $6.25 to $ (round your response to two decimal places). (round your response to two decimal places). If to customer A, 16 less tons are supplied, how much money might be saved? $
Minimize Z = $7.50 V+ $6.25 V + $6.50 (shipping cost to customer A) $6.75 V + S7.00 V + $8.00 (shipping cost to customer B) c) Subject to: Customer A's demand Customer B's demand Warehouse 1's supply Warehouse 2's supply Warehouse 3's supply For all Vj 20 non negativity condition Using a software the linear programming problem was solved and the following sensitivity report was obtained: Adjustable Cells Variable Final Value Reduced Cost Objective Coefficient Allowable Increase Allowable Decrease VA1 1.50 $7.50 1E+30 1.50 V42 100 0.00 $6.25 0.25 0.75 VA3 550 0.00 $6.50 0.75 0.25 V81 V82 400 0.00 $6.75 0.50 1E+30 400 0.00 $7.00 0.75 0.50
Transcribed Image Text:Minimize Z = $7.50 V+ $6.25 V + $6.50 (shipping cost to customer A) $6.75 V + S7.00 V + $8.00 (shipping cost to customer B) c) Subject to: Customer A's demand Customer B's demand Warehouse 1's supply Warehouse 2's supply Warehouse 3's supply For all Vj 20 non negativity condition Using a software the linear programming problem was solved and the following sensitivity report was obtained: Adjustable Cells Variable Final Value Reduced Cost Objective Coefficient Allowable Increase Allowable Decrease VA1 1.50 $7.50 1E+30 1.50 V42 100 0.00 $6.25 0.25 0.75 VA3 550 0.00 $6.50 0.75 0.25 V81 V82 400 0.00 $6.75 0.50 1E+30 400 0.00 $7.00 0.75 0.50
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