Microsoft Word Iphwž x bb4bc518e4be336514478adc982a87d839773aecb45a65ab368701722
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Question
Microsoft Word Iphwž x bb4bc518e4be336514478adc982a87d839773aecb45a65ab368701722e1cd6fdd3bfad39/ assets Jypl The Valley Wine Company produces two kinds of wine- Valley Nectar and Valley Red. The wines are produced from 64 tons of grapes the company has acquired this season. A 1 requires 4 tons of grapes and a batch of Red requires 8 tons. However, production is limited by the availability of only 50 cubic yards of storage space for aging and 120 hours of processing time. A batch of each type of wine requires 5 cubic yards of storage space. The processing time for a batch of Nectar is 15 hours and the processing time for a batch of Red is 8 hours. Demand for the two types of wine is limited to 7 batches, each. The profit for a batch of Nectar is $9,000 and the profit for a batch of red is $12,000. The company wants to determine the number of 1,000--gallon batches of Nectar and Red to produce in order to maximize profit. of Nectar a. Formulate a linear programming model for this problem. b. Use POM for windows to obtain the solutions list and results printouts for this problem. c. Answer the questions below from the printouts you obtained in b above and submit both the printout and answers to the questions to Blackboard by the time and date specified. 1. How much is produced of each type of wine to maximize profits and what is the maximum profit? 2. How much of each type of resource is left over? 3. What is the value for an additional unit of each resource? 4. Over what ranges are the values in 3) above valid? 5. What is the profit-per-unit at which the production plan will change for cach type of wine?Explanation / Answer
LPP model is
MAX Z = 9000x1 + 12000x2
subject to
4x1 + 8x2 <= 64
5x1 + 5x2 <= 50
15x1 + 8x2 <= 120
x1 <= 7
x2 <= 7
and x1,x2 >= 0
Solution:
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate
1. As the constraint 1 is of type '?' we should add slack variable S1
2. As the constraint 2 is of type '?' we should add slack variable S2
3. As the constraint 3 is of type '?' we should add slack variable S3
4. As the constraint 4 is of type '?' we should add slack variable S4
5. As the constraint 5 is of type '?' we should add slack variable S5
After introducing slack variables
Negative minimum Zj-Cj is -12000 and its column index is 2. So, the entering variable is x2.
Minimum ratio is 7 and its row index is 5. So, the leaving basis variable is S5.
? The pivot element is 1.
Entering =x2, Departing =S5, Key Element =1
R5(new)=R5(old)
R1(new)=R1(old)-8R5(new)
R2(new)=R2(old)-5R5(new)
R3(new)=R3(old)-8R5(new)
R4(new)=R4(old)
Negative minimum Zj-Cj is -9000 and its column index is 1. So, the entering variable is x1.
Minimum ratio is 2 and its row index is 1. So, the leaving basis variable is S1.
? The pivot element is 4.
Entering =x1, Departing =S1, Key Element =4
R1(new)=R1(old)÷4
R2(new)=R2(old)-5R1(new)
R3(new)=R3(old)-15R1(new)
R4(new)=R4(old)-R1(new)
R5(new)=R5(old)
Negative minimum Zj-Cj is -6000 and its column index is 7. So, the entering variable is S5.
Minimum ratio is 1 and its row index is 2. So, the leaving basis variable is S2.
? The pivot element is 5.
Entering =S5, Departing =S2, Key Element =5
R2(new)=R2(old)÷5
R1(new)=R1(old)+2R2(new)
R3(new)=R3(old)-22R2(new)
R4(new)=R4(old)-2R2(new)
R5(new)=R5(old)-R2(new)
Since all Zj-Cj?0
Hence, optimal solution is arrived with value of variables as :
x1=4,x2=6
Max Z=108000
Answer (1)
Valley Necter x1=4
Valley Red x2=6
Profit Max Z=108000
Answer (2)
Grape = 0 tons, Space =0 cubic yards, Processing Time =12 hours
Answer (3)
For Valley Nector
Grape = 4 tons, Space =5 cubic yards, Processing Time =3 hours?
For Valley Red
Grape = 8 tons, Space =5 cubic yards, Processing Time =8 hours?
Answer (4)
Valley Nector, because minimum resources required.
Answer (5) No changes,
i.e., Valley Nector $9000 and Valley Red $12000
Max Z = 9000 x1 + 12000 x2 + 0 S1 + 0 S2 + 0 S3 + 0 S4 + 0 S5 subject to 4 x1 + 8 x2 + S1 = 64 5 x1 + 5 x2 + S2 = 50 15 x1 + 8 x2 + S3 = 120 x1 + S4 = 7 x2 + S5 = 7 and x1,x2,S1,S2,S3,S4,S5?0