Dyes Making Company Case Study
A dyes making company decide how to allocate next week’s time on making dyes. The company takes various chemical as input, and can produce two type of dyes used for making clothes. The company’s two products come off at different rates: Tons per hour: Rit :200 Direct: 140 and they also have different profit:
Profit per ton Rit: $25
Direct: $30
To further complicate matters, the following weekly production amounts are the most that can be justified in light of the currently booked orders:
Maximum tons: Rit: 6,000
Direct: 4,000
The question facing the company is as follows: If 40 hours of production time are available this week, how many tons of bands and how many tons of coils should be produced to bring in the greatest total …show more content…
This number cannot exceed the 40 hours available. Since hours per ton is the reciprocal of the tons per hour given above, we have a constraint on the variables:
(1/200) XB + (1/140) XC ≤ 40. There are also production limits: 0 ≤ XB ≤ 6000 0 ≤ XC ≤ 4000
In the statement of the problem above, the upper limits were specified, but the lower limits were assumed — it was obvious that a negative production of bands or coils would be meaningless. Dealing with a computer, however, it is necessary to be quite explicit. By analogy with the formula for total hours, the total profit must be
(profit per ton of Rit) × XB + (profit per ton of Direct) × XC
That is, our objective is to maximize 25 XB + 30 XC.
Putting this all together, we have the following linear program: Maximize : 25 XB + 30 XC
Subject to: (1/200) XB + (1/140) XC ≤ 40 0 ≤ XB ≤ 6000 0 ≤ XC ≤ 4000
This is a very simple linear program, so we’ll solve it by hand in a couple of ways, and then check the answer with AMPL.
First, by multiplying profit per ton times tons per hour, we can determine the profit per hour of mill time for each product:
Profit per hour: Rit $5,000 Direct
Profit per ton Rit: $25
Direct: $30
To further complicate matters, the following weekly production amounts are the most that can be justified in light of the currently booked orders:
Maximum tons: Rit: 6,000
Direct: 4,000
The question facing the company is as follows: If 40 hours of production time are available this week, how many tons of bands and how many tons of coils should be produced to bring in the greatest total …show more content…
This number cannot exceed the 40 hours available. Since hours per ton is the reciprocal of the tons per hour given above, we have a constraint on the variables:
(1/200) XB + (1/140) XC ≤ 40. There are also production limits: 0 ≤ XB ≤ 6000 0 ≤ XC ≤ 4000
In the statement of the problem above, the upper limits were specified, but the lower limits were assumed — it was obvious that a negative production of bands or coils would be meaningless. Dealing with a computer, however, it is necessary to be quite explicit. By analogy with the formula for total hours, the total profit must be
(profit per ton of Rit) × XB + (profit per ton of Direct) × XC
That is, our objective is to maximize 25 XB + 30 XC.
Putting this all together, we have the following linear program: Maximize : 25 XB + 30 XC
Subject to: (1/200) XB + (1/140) XC ≤ 40 0 ≤ XB ≤ 6000 0 ≤ XC ≤ 4000
This is a very simple linear program, so we’ll solve it by hand in a couple of ways, and then check the answer with AMPL.
First, by multiplying profit per ton times tons per hour, we can determine the profit per hour of mill time for each product:
Profit per hour: Rit $5,000 Direct