A manufacturing company produces small and large industrial microwave ovens.
Components for both the products need to machined then assembled.
It takes 4 hours to machine the components for the small ovens
and 3 hours to machine the components for the large ones.
A total of 100 machine hours are available per day.
It takes 2 hours to assemble the small ovens
and 3 hours to assemble the large ones.
There are 48 hours of assembly time available per day.
The market potential for the small ovens is 15 per day.
The market potential for the large ovens is 20 per day.
The marketing department
wants the company to produce at least 10 products per day,
in any combination.
Each small oven contributes $2000 to profit and
each large oven contributes $2500 to profit.
How many small and large ovens should be produced per day to maximize profit?
| Product | Machine | Assembly | Market | Profit |
|---|---|---|---|---|
| _______ | Time | Time | Potential | ______ |
| Small | 4 | 2 | <= 15 | 2000 |
| Large | 3 | 3 | <= 20 | 2500 |
| Total | <= 100 | <= 48 | >= 10 |
Maximize 2000 X1 + 2500 X2
subject to
4 X1 + 3 X2 <= 100
2 X1 + 3 X2 <= 48
X1 <= 15
X2 <= 20
And:
X1 + X2 >= 10
Which in standard form is:
-X1 - X2 <= -10
X1, X2 >= 0
Considering the units:
small= number of small ovens large= number of large ovens mach= number of hours of machine time asmb= number of hours of assembly time spot= pot for small ovens lpot= pot for large ovens mreq = minimum number of required ovens Maximize (profit/small)(small) + (profit/large)(large) (mach/small)(small)+ (mach/large)(large) <= mach (asmb/small)(small)+ (asmb/large)(large) <= asmb (spot/small)(small)+ (spot/large)(large) <= spot (lpot/small)(small)+ (lpot/large)(large) <= lpot (mreq/small)(small)+ (mreq/large)(large) >= mreq
The dual problem:
Minimize 100 Y1 + 48 Y2 + 15 Y3 + 20 Y4 -10 Y5 subject to 4 Y1 + 2 Y2 + Y3 -Y5 >= 2000 3 Y1 + 3 Y2 + Y4 -Y5 >= 2500 Y1, Y2, Y3, Y4, Y5 >= 0The units for the dual:
Y1: profit/mach Y2: profit/asmb Y3: profit/spot Y4: profit/lpot Y5: profit/(mreq decreasing)
Solving the primal problem:
The initial dictionary:
X3 = 100 - 4 X1 - 3 X2 X4 = 48 - 2 X1 - 3 X2 X5 = 15 - 1 X1 + 0 X2 X6 = 20 + 0 X1 - 1 X2 X7 = -10 + 1 X1 + 1 X2 ---------------------------- z = 0 +2000 X1 +2500 X2
Initial tableau:
X1 X2 X3 X4 X5 X6 X7
[ 4 3 1 0 0 0 0 ] [X1] [ 100 ]
[ 2 3 0 1 0 0 0 ] [X2] [ 48 ]
[ 1 0 0 0 1 0 0 ] [X3] = [ 15 ]
[ 0 1 0 0 0 1 0 ] [X4] [ 20 ]
[ -1 -1 0 0 0 0 1 ] [X5] [ -10 ]
[X6]
[X7]
The final dictionary:
X3 = 22 + 1 X4 + 2 X5 X2 = 6 - 0.33 X4 + 0.67 X5 X7 = 11 - 0.33 X4 - 0.33 X5 X6 = 14 + 0.33 X4 - 0.67 X5 X1 = 15 + 0 X4 - 1 X5 ------------------------------------------------- z = 45000 -833.33 X4 -333.33 X5 The optimal solution: 45000 The dual solution: Y1= 0, Y2= 833.33, Y3= 333.33, Y4=0, Y5=0
Thus the optimal strategy is to produce 15 small machines and 6 large machines.
Recall the units: Y2: profit/assembly Y3: profit/spotential
From this, we can see that increasing the assembly hours by one unit increases profit by $833.33 and if there was potential to increase the number of small machines that could be sold by one unit, it would increase the profit by $333.33.
The corresponding tableau:
[ 0 0 1 -1.00 -2.00 0 0 ] [X1] [ 22]
[ 0 1 0 0.33 -0.67 0 0 ] [X2] [ 6]
[ 0 0 0 0.33 0.33 0 1 ] [X3] = [ 11]
[ 0 0 0 -0.33 0.67 1 0 ] [X4] [ 14]
[ 1 0 0 0.00 1.00 0 0 ] [X5] [ 15]
[X6]
[X7]
How does the optimal strategy change if t2 extra assembly hours are available and the potential for selling the small machines increases by t3?
I will also consider the addition of t1 extra machine hours, the potential for selling t4 extra large machines, and decrease of t5 to the minimum number of machines required.
Initial tableau:
X1 X2 X3 X4 X5 X6 X7
[ 4 3 1 0 0 0 0 ] [X1] [ 100 ]
[ 2 3 0 1 0 0 0 ] [X2] [ 48 ]
[ 1 0 0 0 1 0 0 ] [X3] = [ 15 ]
[ 0 1 0 0 0 1 0 ] [X4] [ 20 ]
[ -1 -1 0 0 0 0 1 ] [X5] [ -10 ]
[X6]
[X7]
The basis matrix B:
X3 X2 X7 X6 X1 [ 1 3 0 0 4 ] [X3] [ 100 ] [ 0 3 0 0 2 ] [X2] [ 48 ] [ 0 0 0 0 1 ] [X7] = [ 15 ] [ 0 1 0 1 0 ] [X6] [ 20 ] [ 0 -1 1 0 -1 ] [X1] [ -10 ]
I computed the inverse of B using the following Matlab commands:
---------------------------------
B= [1 3 0 0 4 ;
0 3 0 0 2 ;
0 0 0 0 1 ;
0 1 0 1 0 ;
0 -1 1 0 -1]
invB= inv(B)
---------------------------------
The inverse is:
invB = 1 -1.00 -2.00 0 0 0 0.33 -0.67 0 0 0 0.33 0.33 0 1 0 -0.33 0.67 1 0 0 0.00 1.00 0 0
If the update to b is [ t1 t2 t3 t4 t5]T and the ti's are not too big, the change to the optimal solution is:
[ 1 -1.00 -2.00 0 0 ] [t1 ] [ 0 0.33 -0.67 0 0 ] [t2 ] [ 0 0.33 0.33 0 1 ] [t3 ] [ 0 -0.33 0.67 1 0 ] [t4 ] [ 0 0.00 1.00 0 0 ] [t5 ]
So the new solution would be:
X3= 22 + t1 - 1.00 t2 - 2.00 t3 X2= 6 + 0.33 t2 - 0.67 t3 X7= 20 + 0.33 t2 + 0.33 t3 + + t5 X6= 14 - 0.33 t2 + 0.67 t3 + t4 X1= 15 + 0.00 t2 + 1.00 t3
In terms of the original variables:
X1= 15 + 0.00 t2 + 1.00 t3 X2= 6 + 0.33 t2 - 0.67 t3
X1= 15, X2= 7,
profit increases by 833.33 * 3 = 2500
(2) Increase number of small units that can be sold by 3 (t3=3):
X1= 18, X2= 4,
profit increases by 333.33 * 3 = 1000
(3) Increase both by 3 (t2=3, and t3=3):
X1= 18, X2= 5,
profit increases by 833.33*3 + 333.33*3
= 3500
If there is too much of an increase to t2 (assembly hours), there is no longer any benefit:
There are two equations that impose a feasibility constraint on t2:
(1) X3= 22 + 1 t1 - 1.00 t2 - 2.00 t3 + 0 t4 + 0 t5 (2) X6= 14 + 0 t1 - 0.33 t2 + 0.67 t3 + 1 t4 + 0 t5
The constraints are:
(1) t2 <= 22 (tightest)
(2) t2 <= 42
Consider (1):
X3= 22 + 1 t1 - 1.00 t2 - 2.00 t3 + 0 t4 + 0 t5
If t2 > 22, then constraint 1 is violated:
X1= 15, X2 > 6 + 22/3
Constraint 1 is: 4 X1 + 3 X2 <= 100
And with this assignment of variables:
4(15) + 3(6+22/3) > 100 (not enough mach. hrs.)
If there is too much of an increase to t3 (potential for small machines), there is no longer any benefit:
There are two equations that impose a feasibility constraint on t3:
(1) X2= 6 + 0 t1 + 0.33 t2 - 0.67 t3 + 0 t4 + 0 t5 (2) X3= 22 + 1 t1 - 1.00 t2 - 2.00 t3 + 0 t4 + 0 t5
The constraints are:
(1) t3 <= 9 (tightest)
(2) t3 <= 11
If t3 > 9, then the value of X2 would go negative.
This means that the basis is no longer valid.
X2 would exit the basis, and no large machines would be made.