Can anyone help me with this problem?An oil fired power station burns three grades of oil: grade A, grade B and grade C. The efficiency with which electricity can be produced from the burning of...
Can anyone help me with this problem?
An oil fired power station burns three grades of oil: grade A, grade B and grade C.
The efficiency with which electricity can be produced from the burning of these oils
depends upon the presence of three ingredients: X, Y and Z which are known to be
present in each ton of the three grades as shown in the table below:
Tons of ingredient per ton of oil
X Y Z
Grade A 0.2 0.4 0.3
Grade B 0.1 0.3 0.2
Grade C 0.3 0.2 0.4
Maximum efficiency in the burning process requires the presence of at least 200 tons
of ingredient X, at least 175 tons of ingredient Y, and at least 300 tons of ingredient
Z, in 1 hour of continuous electricity production.
The power station purchases the oil at prices of £60, £80, and £70 per ton for grades
A, B and C respectively, and is required to operate for 12 hours per day.
Formulate the linear program which will minimize the total daily cost of
Can anyone help me with this??
WHAT WE KNOW:
Just to see for element X, we know that we COULD get the (12 hrs/day times 200 tons of element X needed) 2400 tons of element X by buying 10,000 t(ons) of A (Grade A) oil (10,000t · .2t per t of A), plus 1000t of B (1000t · .1t per t of B), plus 1000t of C (1000t · .3t per t of C). This would give us 2000t from A, 100t from B, and 300t from C. It would cost us (10000t · £60/t) + (1000t · £80/t) + (1000t · £70/t) a total of (600,000 + 80,000. + 70,000) = £750,000. We do end up with exactly the 3600t of element Z, but we have 4500t of element Y, over by quite a bit from the 2100t needed.
CHECK FOR REASONABLENESS:
Would all A be best? All B? All C? We can see that B is the more expensive per ton and less productive than A for every element. We would only buy B if A is not available, hence we can eliminate B from consideration.
Let A = tons of Grade A oil, and C = tons of Grade C oil. From the example, we can derive:
Element X: 2400t = .2A + .3C, or 4800t = .4A + .6C, or 7200 = .6A + .9C
Element Y: 2100t = .4A + .2C, or 4200t = .8A + .4C
Element Z: 3600t = .3A + .4C, or 7200 = .6A + .8C
Graphing the above three lines, we could find either one, two, or three points of intersection. There could be none, or an infinite amount as well, which for a production problem such as this would not occur, since we have limits of 0≤A and C, and there is not a multiple of one equation.
Solving for Elements X & Y:
4800t = .4A + .6C
-(2100t = .4A + .2C)
2700t = + .4C, and dividing both sides by .4, we get 6750t needed of Grade C.
Substituting to find A, we get
2100 = .4A + .2·6750
2100 = .4A + 1350
750 = .4A, and 1875 = A
1st possible solution set: (A, C) = 1875t, 6750t
Cost is (1875t ∙ £60/t) + (6750t ∙ £70/t) = £585,000
Solving for Elements X & Z:
7200 = .6A + .9C
-(7200 = .6A + .8C)
0 = .1C, and C=0.
Substituting to find C, we get from Element X:
2400 = .2A + .3(0), or just 2400 = .2A, dividing both sides by .2, we get A = 12000.
2nd possible solution set: (A, C) = 12000t, 0t
Cost is (12000t ∙ £60/t) + (0t ∙ £70/t) = £720,000
Solving for Elements Y & Z:
4200 = .8A + .4C
-(3600 = .3A + .4C)
600 = .5A, and with dividing both sides by .5, we get A = 1200
Substituting to find C, we get from Element Z:
3600 = .3(1200) + .4C
3600 = 360 + .4C, subtract 360 from both sides,
3240 = .4C, and dividing both side by .4, we get C = 8100.
3rd possible solution set: (A, C) = 1200t, 8100t.
Cost is (1200t ∙ £60/t) + (8100t ∙ £70/t) = £574,200
The best solution is the last one. Remember, in a system of linear inequalities, the minimums or maximums occur at the intersections.