Question

3. A company makes two products (X and Y) using two machines (A and B). Each unit of X that is produced requires 50 minutes processing time on machine A and 30 minutes processing time on machine B. Each unit of Y that is produced requires 24 minutes processing time on machine A and 33 minutes processing time on machine B. At the start of the current week there are 30 units of X and 90 units of Y in stock. Available processing time on machine A is forecast to be 40 hours and on machine B is forecast to be 35 hours. The demand for X in the current week is forecast to be 75 units and for Y is forecast to be 95 units. Company policy is to maximise the combined sum of the units of X and the units of Y in stock at the end of the week. Required: (i) Formulate the problem of deciding how much of each product to make in the current week as a linear program. (i) Solve this linear program graphically

Answer 1

Step-by-step explanation:

OR-Notes are a series of introductory notes on topics that fall under the broad heading of the field of operations research (OR). They were originally used by me in an introductory OR course I give at Imperial College. They are now available for use by any students and teachers interested in OR subject to the following conditions.

A full list of the topics available in OR-Notes can be found here.

Linear programming solution examples

Linear programming example 1997 UG exam

A company makes two products (X and Y) using two machines (A and B). Each unit of X that is produced requires 50 minutes processing time on machine A and 30 minutes processing time on machine B. Each unit of Y that is produced requires 24 minutes processing time on machine A and 33 minutes processing time on machine B.

At the start of the current week there are 30 units of X and 90 units of Y in stock. Available processing time on machine A is forecast to be 40 hours and on machine B is forecast to be 35 hours.

The demand for X in the current week is forecast to be 75 units and for Y is forecast to be 95 units. Company policy is to maximise the combined sum of the units of X and the units of Y in stock at the end of the week.

Formulate the problem of deciding how much of each product to make in the current week as a linear program.

Solve this linear program graphically.

Solution

Let

x be the number of units of X produced in the current week

y be the number of units of Y produced in the current week

then the constraints are:

50x + 24y <= 40(60) machine A time

30x + 33y <= 35(60) machine B time

x >= 75 - 30

i.e. x >= 45 so production of X >= demand (75) - initial stock (30), which ensures we meet demand

y >= 95 - 90

i.e. y >= 5 so production of Y >= demand (95) - initial stock (90), which ensures we meet demand

The objective is: maximise (x+30-75) + (y+90-95) = (x+y-50)

i.e. to maximise the number of units left in stock at the end of the week

It is plain from the diagram below that the maximum occurs at the intersection of x=45 and 50x + 24y = 2400

Solving simultaneously, rather than by reading values off the graph, we have that x=45 and y=6.25 with the value of the objective function being 1.25

Linear programming example 1995 UG exam

The demand for two products in each of the last four weeks is shown below.

Week

1 2 3 4

Demand - product 1 23 27 34 40

Demand - product 2 11 13 15 14

Apply exponential smoothing with a smoothing constant of 0.7 to generate a forecast for the demand for these products in week 5.

These products are produced using two machines, X and Y. Each unit of product 1 that is produced requires 15 minutes processing on machine X and 25 minutes processing on machine Y. Each unit of product 2 that is produced requires 7 minutes processing on machine X and 45 minutes processing on machine Y. The available time on machine X in week 5 is forecast to be 20 hours and on machine Y in week 5 is forecast to be 15 hours. Each unit of product 1 sold in week 5 gives a contribution to profit of £10 and each unit of product 2 sold in week 5 gives a contribution to profit of £4.

Answer 2

Given:

Processing time of X on machine A = 50 minutes

Processing time of X on machine B = 30 minutes

Processing time of Y on machine A = 24 minutes

Processing time of Y on machine B = 33 minutes

No. of units of X in stock for the current week = 30

No. of units of Y in stock for the current week = 90

Available processing time on machine A = 40 hours

Available processing time on machine B = 35 hours

Demand for X in the current week = 75 units

Demand for Y in the current week = 95 units

To find:

No. of each product by formulating as a linear program.

Solution:

Let $x$ be the no. of product X and $y$ be the no. of product Y.

The no. of product X and the no. of product Y are our decision variables.

The constraints are the available processing time on the machines. The available processing time on machine A is forecast to be 40 hours. Since it is in hours whereas it is given in minutes for processing X and Y, we need to convert hours into minutes.

$1hour=60min$

$40hours=40*60=2400min$

Therefore,

$50x+24y\leq 2400$

Similarly, the available processing time on machine B is forecast to be 35 hours.

$35hours=35*60=2100min$

Therefore,

$30x+33y\leq 2100$

The total demand for products X and Y in the current week is forecasted as 75 units and 95 units respectively. To maximise the no. of units of X and no. of units of Y left in stock at the end of the week,

$S=(x+30-75)+(y+90-95)$

$S=x-45+y-5$

$S=x+y-50$

is the objective function.

The demand for X and Y are 75 units and 95 units respectively whereas initially, they are in stock of 30 units of X and 90 units of Y.

The non-negativity constraints,

$x\geq 75-30$

$x\geq 45$

$y\geq 95-90$

$y\geq 5$

Hence, on formulating the linear program,

Maximize, $S=x+y-50$

subject to, $50x+24y\leq 2400$

$30x+33y\leq 2100$

$x\geq 45,y\geq 5$

To graphically solve this linear program, we plot these inequalities on a graph sheet as shown below:

Here, we obtain the corner points as A(45, 6.25), B(45.6, 5), and C(45, 5). Test these corner points on $S=x+y-50$ and check which point gives the maximum sum of products.

For A(45, 6.25),

$S=45+6.25-50=1.25$

For B(45.6, 5)

$S=45.6+5-50=0.6$

For C(45,5),

$S=45+5-50=0$

Here, the maximum sum obtained is for point A(45, 6.25). Hence, no. of units of X should be $45$ and no. of units of Y should be $6.25$ to obtain the maximum sum of products in the current week.

No. of units of X should be 45 and no. of units of Y should be 6.25 to obtain the maximum sum of products in the current week.