Question

Suppose that at Generator A, it costs $20+$1*x dollars per megawatt to produce x megawatts this hour. At Generator B, it costs $30+$1.1*x dollars per megawatt to produce x megawatts this hour. Dollar amounts in this problem are actually somewhat realistic.

i) You need a total of 10 megawatts this hour. What’s the best way use some of A and some of B to get 10 megawatts total? It’s a good idea to do it both “by hand” and by graphing your overall objective function in Desmos and spotting the best point.

ii) Did you notice that Generator B costs more than A no matter what? Did Generator A therefore produce all the power, and B none of the power, in the optimal solution?

Answer 1

20+x1 30+1.1x2 0 10 61 60.9 60.8 60.7 60.6 60.5 60.4 60.3 60.2 60.1 20 21 41 39.9 38.8 37.7 36.6 35.5 34.4 33.3 32.2 31.1 23 4 24 4 26 27 10 0 10 Answer: Suppose that at Generator A, it costs $20+$1*x dollars per megawatt to produce x megawatts this hour. At Generator B, it costs $30+$1.1*x dollars per megawatt to produce x megawatts this hour So let Generator A produces x1 megawatts this hour and Generator B produces x2 megawatts this hour We want X1 + X2-10 And the cost is then (20+ x1) + (30+ 1.1%) = 50+ x1 + 1.1x2 Problem is to minimize z = 50 + x 1.1x Subect to x1 + X210 The following table with the z-values correspondingly, says that minimum occurs when X1-10, X2-0 ¡i) Yes, it is easy to notice that Generator B costs more than A no matter what, because of the fixed cost 30 and the rate per unit 1.1(by generator B) is also higher as compaired to rate per unit (by Generator A) therefore, Generator A produces all the power, and B none of the power, in the optimal solution Clearly, it is better to use generator A as it costs less, so all 10 megawatts by Generator A. That is x1 = 10, x2-0