Linear Programming and the Simplex methods : Surpluses
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Consider the following linear programming problem: A workshop of Peter's Potters makes vases and pitchers. Profit on a vase is $3.00; profit on a pitcher is $4.00. Each vase requires ½ hour of labor, each pitcher requires 1 hour of labor. Each item requires 1 unit of time in the kiln. Labor is limited to 4 hours per day and kiln time is limited to 6 units per day. Initial and final tableaux are shown in finding the production plan which will maximize profits: (x = number of vases and y = number of pitchers made per day).
X y u v M x y u v M
½ 1 1 0 0 4 0 1 2 -1 0 2
1 1 0 1 0 6 1 0 -2 2 0 4
-3 -4 0 0 1 0 0 0 2 2 1 20
(initial) ( final)
How much surplus labor is there when the optimal plan is in effect?
a. 0 hours
b. 2 hours
c. 4 hours
d. 1 hour
e. None of the above
If labor were increased by one hour a day, profits could be increased to:
a. $24
b. $22
c. $21
d. Not increased
e. None of the above
If labor were increased by one hour a day, the optimal production plan would require how many vases?
a. 2
b. 3
c. 4
d. 1
e. None of the above
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Linear programming and the simplex methods for analyzed. The surpluses of pitchers are computed.
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Consider the following linear programming problem: A workshop of Peter's Potters makes vases and pitchers. Profit on a vase is $3.00; profit on a pitcher is $4.00. Each vase requires ½ hour of labor, each pitcher requires 1 hour of labor. Each item requires 1 unit of time in the kiln. Labor is limited to 4 hours per day and kiln time is limited to 6 units per day. Initial and final ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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