Production Scheduling In a factory, machine 1 produces...

Production Scheduling In a factory, machine 1 produces 8-inch (in.) pliers at the rate of 60 units per hour (hr) and 6 -in. pliers at the rate of $70 \mathrm{units} / \mathrm{hr}$. Machine 2 produces 8 -in. pliers at the rate of 40 units/hr and 6-in. pliers at the rate of 20 units/hr. It costs $$\$ 50 / \mathrm{hr}$$ to operate machine 1 , and machine 2 costs $$\$ 30 / \mathrm{hr}$$ to operate. The production schedule requires that at least 240 units of 8 -in. pliers and at least 140 units of 6-in. pliers be produced during each 10 -hr day. Which combination of machines will cost the least money to operate?

Key Concepts

Machine Scheduling

Machine scheduling involves determining the operational timing and workload distribution among available machines. It requires analyzing production rates, operating costs, and production targets to identify the most cost-effective schedule that meets all constraints.

Linear Programming

Linear programming is a mathematical method used for optimizing a linear objective function, subject to linear equality and inequality constraints. In production scheduling, it provides a structured framework to allocate scarce resources among competing activities to minimize costs or maximize profits.

Production Rates

Production rates refer to the number of units produced by a machine per unit time. They are crucial in scheduling problems as different machines may operate at different speeds, and these rates determine how long a machine needs to run to meet production targets.

Objective Function and Cost Minimization

The objective function represents the goal of the optimization model, which, in this context, is to minimize the total operating cost of the machines. It combines the operating costs of different machines based on the hours they are used, reflecting how production decisions directly impact overall costs.

Constraint Satisfaction

Constraints in production scheduling ensure that the production targets are met and that the operational limits of machines are not exceeded. They include the minimum production requirements for different product types and any potential upper limits on machine operating hours.

Transcript

00:01 So in this problem, we want to find the minimum cost necessary to fulfill a production quota given two different types of machines.

00:10 So i'm going to start by labeling my variables.

00:13 I'm going to call x the number of hours machine one works.

00:23 Y will be the number of hours machine two works.

00:30 And z will equal our total cost.

00:35 And in this situation, we want to minimize this.

00:40 We're going to write our objective function now, which starts with z equals, and the cost for machine 1 per hour is 50.

00:48 So i do 50x plus 30y.

00:54 So now that i have my objective function, i can go ahead and write down my constraints.

00:58 So my constraints are going to start with x and 1.

01:02 Both being greater than or equal to zero because you can't have a negative amount of machines.

01:09 And then i'm going to move on to the units needed for our pliers, our 8 inch pliers.

01:16 So 60x plus 40y must be greater than or equal to 240.

01:24 Likewise, for our 6 inch pliers, 70x plus 20y must be greater than or equal to 140.

01:34 Since we're told that we have a 10 -hour workday, we know that x plus y must be less than or equal to 10.

01:44 So now that we have our constraints, we're going to graph this.

01:48 We know that x and y are both greater than are equal to zero, meaning that this will take place in our first quadrant.

01:55 And here i can start by rearranging this equation.

01:58 By isolating the y, y would be greater than or equal to negative 3 over 2 x plus 6.

02:13 So i know i have a y intercept of 6, and since i have a slope of negative 3 over 2, i know that my x intercept will be at 4.

02:25 My line will look like this.

02:28 Next, i can rewrite this equation to be 20y is greater than or equal to negative 70x plus 140.

02:38 I isolate the y by dividing both sides by 20 and i end up with y is less than is greater than or equal to negative 7 over 2x plus 7.

02:52 So i know my y intercept will be at 7 this time and since i have a slope of negative 7 halves my x intercept will be at 2.

03:04 And my line will look like this.

03:07 And now i have y is lessen or equal to negative x.

03:14 Excuse me, that's a negative x plus 10.

03:17 Meaning that my y intercept will be at 10 and my x intercept will also be at 10...

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