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yuradex [85]
3 years ago
10

The following is a linear programming formulation of a labor planning problem. There are four overlapping shifts, and management

must decide how many employees to schedule to start work on each shift. The objective is to minimize the total number of employees required while the constraints stipulate how many employees are required at each time of day. The variables X1 - X4 represent the number of employees starting work on each shift (shift 1 through shift 4).
MIN X1 + X2 + X3 + X4

s.t. X1 + X4 ≥ 12 ........(on duty during shift 1)
X1 + X2 ≥ 15 ........(on duty during shift 2)
X2 + X3 ≥ 16 ........(on duty during shift 3)
X3 + X4 ≥ 14 ........(on duty during shift 4)
X1, X2, X3, and X4 ≥ 0

Given the optimal solution: X1* = 13, X2* = 2, X3* = 14, X4* = 0, how many workers would be assigned to shift 2?

a. 13
b. 12
c. 14
d. 15
e. 2
Mathematics
1 answer:
dexar [7]3 years ago
6 0

Answer:

d. 15

Step-by-step explanation:

Putting the values in the shift 2 function

X1 + X2 ≥ 15

where x1=  13, and x2=2

13+12≥ 15

15≥ 15

At least 15 workers must be assigned to the shift 2.

The LP model questions require that the constraints are satisfied.

The constraint for the shift 2 is that the  number of workers must be equal or greater than 15

This can be solved using other constraint functions e.g

Putting  X4= 0 in

X1 + X4 ≥ 12

gives

X1 ≥ 12

Now Putting the value X1 ≥ 12  in shift 2 constraint

X1 + X2 ≥ 15

12+ 2≥ 15

14 ≥ 15

this does not satisfy the condition so this is wrong.

Now from

X2 + X3 ≥ 16

Putting X3= 14

X2 + 14 ≥ 16

gives

X2  ≥ 2

Putting these in the shift 2

X1 + X2 ≥ 15

13+2 ≥ 15

15 ≥ 15

Which gives the same result as above.

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Answer:

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Step-by-step explanation:

There was no figure but the question is clear

Volume of a cylinder is given by the formula \bold{\pi r^2h}\\

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Volume of a cone is given by \bold{\frac{1}{3} \pi r^2 h}

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The radius of the cylinder = \frac{1}{2}(diameter) = \frac{1}{2}(12) = 6cm

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Volume of cylinder V_{cyl} = \pi (6)^2  15 = \pi (36)15 = \bold{540\pi}

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