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)
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)
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)
.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)
.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)
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)
.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.
