Mathematical Problem (Linear Programming)

Problem:

Zenith Inc. manufactures two types of kitchen utensils:

1.      Knives.

2.      Forks.

Both must be pressed and polished. The shop manager estimates that there will be a maximum 70 hours available next week in pressing machine center and 100 hours in polishing machine center. However each case of knives requires an estimated 12 minutes (.20 hour) of pressing and 30 minutes (.50 hour) of polishing while in case of forks requires 24 minutes (.40hour) of pressing and 15 minutes (.25 hour) of polishing. The company can sell as many knives as it produces at the prevailing market price of Tk 12 per case. Forks can be sold for Tk 9 per case. Cost of production per case knives is Tk 4 and forks Tk 3.

Zenith wants to determine how many cases of knives and forks the company should produce to maximize profit.

Solution (Equation type expression):

Zenith’s problem is to determine the quantity of knives and forks that will maximize profit and selected quantities cannot be used more than available pressing and polishing time.

Objective: To maximize profit.

Zenith’s total profit = The contribution from knives + the contribution from forks.

Since,

Each case of knives can be sold @ Tk 12

Each case of knives’ production cost@ Tk 4

So, contribution from each case of knives = Tk (12-4) = Tk 8

Since,

Each case of forks can be sold @ Tk 9

Each case of fork’s production cost@ Tk 3

So, contribution from each case of forks = Tk (9-3) = Tk 6

This Tk 8 and Tk 6 per case knives and per case forks contribution multiplied by number of cases gives total profit from knives and forks respectively.

By letting,

X= Number of knives zenith will produce in next week.

Y= Number of forks zenith will produce in next week.

Z= Zenith’s want of total profit.

Now we can represent Zenith’s total profit, 
                     Z= 8X + 6Y ------------- (I)

The company wants to choose the level of decision variables (X and Y) that maximizes total profit (Z). The objective can be expressed as,
                       Maximize Z = 8X + 6Y ---------(II)

Restrictions:

Available pressing and polishing capacity will limit knives and forks Zenith can produce. Since each case of knives uses .20 hour of pressing time i.e., .20X is the total time required to press X cases knives, similarly .40Y is the total time required to press Y cases of forks.

Consequently, .20X + .40Y gives the total time required to press X cases of knives and Y cases of forks.

Zenith can select any product combination doesn’t require more than 70 hours available pressing time. The mathematical representation of this condition will be-
                     .20X + .40Y ≤ 70 (hour) ----------- (III)

Another system constraint deals with polishing operation management of the business/ company knows that each case of knives uses .50 hour and each case of forks uses .25 hour polishing hour.

Since, there are only 100 hours of polishing time. So we can represent the above description as following:
                           .50X + .20Y ≤ 100 (hour) ------------ (IV)

It is physically impossible for zenith to produce negative number of knives and forks. Therefore, management must ensure that decision variables X and Y have values greater than or equal to 0 (zero).

                           Symbolically, X ≥ 0 and Y ≥ 0
                           In abbreviated form X, Y ≥ 0 ------------ (V)

Complete Formulation:

By collecting objective function (II), system constraints (III) and (IV) and non-negativity condition (V), zenith’s management can be represented the machine shop problem with the following mathematical equation/ function:

Maximize, Z = 8X + 6Y
Subject to,
.20X + .40Y ≤ 70 (System Constraints)
.50X + .25Y ≤ 100 (System Constraints)
 X, Y ≥ 0 (Non-negative function)

Solution (Numerical Expression):

Let us, first consider the inequality into equation we have, 
.20X + .40Y = 70 --------- (I)

And .50X + .25Y = 100 -------- (II)

For the equation number (I), when X=0 then

.20 x 0 + .40Y = 70

Or, Y = 175             [X, Y= 0, 175]

When Y=0 then

.20X   + .40 x 0 = 70

Or, X = 350            [X, Y= 350, 0]

For the equation number (II), when X=0 then

.50 x 0 + .25Y = 100

Or, Y = 400             [X, Y= 0, 400]

When Y=0 then

.50X   + .25 x 0 = 70

Or, X = 200            [X, Y= 200, 0]

Now equation number (I) multiplied by 5 and equation number (II) multiplied by 2. Then deduct (II) from (I), we get-

X + 2Y = 350

X + .50Y = 200

-------------------

1.50Y = 150

So, Y = 100

Putting the value of Y, in equation (I) we get,

.20X + .40 x 100= 70

Or, X = 150

So, X, Y = 150, 100

Now all the values of X and y are ---

For equation Number (I)

When X = 0,   then    X, Y = 0, 175

When Y = 0    then X, Y = 350, 0

For equation Number (II)

When X = 0,   then    X, Y = 0, 400

When Y = 0    then X, Y = 200, 0

And X, Y = 150, 100

 

Putting the above values on graph, we get---

Click on the picture to see the original size.

 

By putting values of variables on X and Y axis and using shadow, we get the crossing point of two lines at (150, 100). So it would be the target point where zenith may get highest profit or lowest profit.

Now, Z = 8X + 6Y

 = 8 x 1500 + 6x 100

 = 1800

Competitive analysis:

Corner point	Total profit
0, 175	Z = (8 x 0) + (6 x 175)= 1050
0, 0	Z = (8 x 0) + (6 x 0)= 0
200, 0	Z = (8 x 200) + (6 x 0)= 1600
150, 100	Z = (8 x 150) + (6 x 100)= 1800

Comment:

Under the above solution, we suggest the production manager to produce 150 cases of knives and 100 cases of forks to maximize profit or minimize costs.