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Linear programming program​

Answer 1

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

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.

It may not be possible to produce enough to meet your forecast demand for these products in week 5 and each unit of unsatisfied demand for product 1 costs £3, each unit of unsatisfied demand for product 2 costs £1.

Formulate the problem of deciding how much of each product to make in week 5 as a linear program.

Solve this linear program graphically.

Solution

Note that the first part of the question is a forecasting question so it is solved below.

For product 1 applying exponential smoothing with a smoothing constant of 0.7 we get:

M1 = Y1 = 23

M2 = 0.7Y2 + 0.3M1 = 0.7(27) + 0.3(23) = 25.80 

M3 = 0.7Y3 + 0.3M2 = 0.7(34) + 0.3(25.80) = 31.54 

M4 = 0.7Y4 + 0.3M3 = 0.7(40) + 0.3(31.54) = 37.46

The forecast for week five is just the average for week 4 = M4 = 37.46 = 31 (as we cannot have fractional demand).

For product 2 applying exponential smoothing with a smoothing constant of 0.7 we get:

M1 = Y1 = 11 

M2 = 0.7Y2 + 0.3M1 = 0.7(13) + 0.3(11) = 12.40 

M3 = 0.7Y3 + 0.3M2 = 0.7(15) + 0.3(12.40) = 14.22 

M4 = 0.7Y4 + 0.3M3 = 0.7(14) + 0.3(14.22) = 14.07

The forecast for week five is just the average for week 4 = M4 = 14.07 = 14 (as we cannot have fractional demand).

We can now formulate the LP for week 5 using the two demand figures (37 for product 1 and 14 for product 2) derived above.

Let

x1 be the number of units of product 1 produced

x2 be the number of units of product 2 produced

where x1, x2>=0

The constraints are:

15x1 + 7x2 <= 20(60) machine X

25x1 + 45x2 <= 15(60) machine Y

x1 <= 37 demand for product 1

x2 <= 14 demand for product 2

The objective is to maximise profit, i.e.

maximise 10x1 + 4x2 - 3(37- x1) - 1(14-x2)

i.e. maximise 13x1 + 5x2 - 125

The graph is shown below, from the graph we have that the solution occurs on the horizontal axis (x2=0) at x1=36 at which point the maximum profit is 13(36) + 5(0) - 125 = £343

Answer 2

Answer:

$\huge\cal\blue{Your Answer}$

Step-by-step explanation:

A = 1R + 2G + 3Bl

B = 2R + 3G + 4Bl

C = 3R + 4G + 1Bl

Stock R = 14

G = 20

Bl = 14

Let say X unit of A & Y unit of B then 14-(X + Y) units of C

14R = XR + 2YR + (14 - X - Y)3R

=> 14 = X + 2Y + 42 - 3X - 3Y

=> 2X + Y = 28

G

20 = 2X + 3Y + 4(14 - X - Y)

=> 2X + Y = 36

28 ≠ 36

Hence there is mistake in data

Bl

14 = 3X + 4Y + (14 - X - Y)

=> 2X + 3Y = 0

Sum of two units produced can not be Zero

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