Sensitivity Analysis Exeter Mines produces iron ore at four different mines, how

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Sensitivity Analysis Exeter Mines produces iron ore at four different mines, however, the ores extracted at each mine are different in their iron content. Mine 1 produces magnetite ore, which has a 70% iron content; mine 2 produces limonite ore, which has a 60% iron content; mine 3 produces pyrite ore; which has a 50% iron content; and mine 4 produces taconite ore, which has only a 30 % iron content. Exeter has three customers that produce steel of three types known as Armco, Best and Corcom. Armco needs 400 tons of pure (100%) iron, Best requires 250 tons of pure iron, and Corcom requires 290 tons. It costs $37 to extract and process 1 ton of magnetite ore at mine 1, $46 to produce 1 ton of limonite ore at mine 2, $50 per ton of pyrite ore at mine 3, and $42 per ton of taconite ore at mine 4. Exeter can extract 350 tons of ore at mine 1; 530 tons at mine 2; 610 tons at mine 3 and 490 tons at mine 4. The company wants to know how much ore to produce at each mine in order to minimize cost and meet its customers' demand for 100% pure iron. a. Formulate a linear programming model for this problem. b. Solve the model by using the Excel Solver. c. Do any of the mines have slack capacity? If yes, which one(s)? d. If Exeter mines could increase production capacity at any one of its mines, which it should be? Why? e. If Exeter determined that it could increase production capacity at mine 1 from 350 tons to 500 tons, at an increase in production cost to $43 per ton, should it do so?

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A summation is a compact notation for representing the sum of the terms in a sequence over a given range. alower + alower+1 + · · · + aupper = upper X j=lower aj Notational variations are also possible Pupper j=lower aj or P lowerjupper aj . In many cases, the lower bound is 0 or 1. However, this is not required. In the summation Pn j=m aj , the variable j is called the index of summation and can be renamed if desired (so long as it is done consistently and name clashes do not occur). Note that the index of summation need not be a subscript. Xn j=m aj = Xn i=m ai = Xn k=m ak For example, P5 j=1 j 2 = P4 k=0(k + 1)2 P5 j=1 j 2 = 12 + P5 j=2 j 2 = (P4 j=1 j 2 ) + 52 P5 j=1 j 2 = 12 + 22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25 = 55 Double Summation Consider example: X 4 i=1 X 3 j=1 i · j = X 4 i=1 (i + 2i + 3i) = X 4 i=1 6i 6 Summation Over Sets One can also specify a sum over the elements of a set using the following notation: X sS f(s) where f is a function from S to numbers. For instance, X s{0,2,4} s = 0 + 2 + 4. Triangular Numbers and Their Relatives In the definitions below, n is a nonnegative integer. The triangular number Tn is the sum of all integers from 1 to n: Tn = Xn i=1 i = 1 + 2 + · · · + n. For instance, T4 = 1 + 2 + 3 + 4 = 10. By Bn we denote the number of ways to choose two elements out of n. For instance, if we take 5 elements a, b, c, d, e, then there will be 10 ways to choose two: a, b; a, c; a, d; a, e; b, c; b, d; b, e; c, d; c, e; d, e. We conclude that B5 = 10. By Sn we denote the sum of the squares of all integers from 1 to n: Sn = Xn i=1 i 2 = 12 + 22 + · · · + n 2 . For instance, S4 = 12 + 22 + 32 + 42 = 30. By Cn we denote the sum of the cubes of all integers from 1 to n: Cn = Xn i=1 i 3 = 13 + 23 + · · · + n 3 . For instance, C4 = 13 + 23 + 33 + 43 = 100. The harmonic number Hn is defined by the formula Hn = Xn i=1 1 i = 1 1 + 1 2 + · · · + 1 n . For instance, H3 = 1 1 + 1 2 + 1 3 = 11 6 . The factorial of n is the product of all integers from 1 to n: n! = Yn i=1 i = 1 · 2 · · · · · n ( Q stands for product, similarly as P stands for sum). For instance, 4! = 1 · 2 · 3 · 4 = 24. Note that 0! = 1. Triangular numbers can be calculated using the formula Tn = n(n + 1) 2 . To prove this formula, we start with two expressions for Tn: Tn = 1 + 2 + 3 + · · · + (n 2) + (n 1) + n Tn = n + (n 1) + (n 2) + · · · + 3 + 2 + 1. 7 If we add them column by column, we’ll get: 2Tn = (n + 1) + (n + 1) + (n + 1) + · · · + (n + 1) + (n + 1) + (n + 1) = n(n + 1), and it remains to divide both sides by 2. There are also other ways to prove the formula for Tn. Consider these identities: (1 + 1)2 = 12 + 2 · 1 · 1 + 12 , (2 + 1)2 = 22 + 2 · 2 · 1 + 12 , (3 + 1)2 = 32 + 2 · 3 · 1 + 12 , . . . (n + 1)2 = n 2 + 2 · n · 1 + 12 . If we add them column by column, we’ll get: 2 2 + 32 + 42 + · · · + (n + 1)2 = 12 + 22 + 32 + · · · + n 2 + 2Tn + n. Subtract 22 + 32 + · · · + n 2 from both sides: (n + 1)2 = 12 + 2Tn + n. Expand the left-hand side and subtract n + 1 from both sides: n 2 + n = 2Tn. It remains to divide both sides by 2. The relationship between the number Bn and triangular numbers is described by the formula Bn = Tn1 (n 1) It follows that Bn = (n 1)n 2 . There is a simple formula for Cn: Cn = (Tn) 2 = n 2 (n + 1)2 4 . There are no simple precise formulas found for harmonic numbers and factorials

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