Question 1
Solve the following using graphical method.
i). Maximize Z = 3x1 + 4x2 soln:
Subject to: x1
≤ 350 when
x1=0, x2 ≤ 400
x2
<=250
x1
+ x2 <=400
x1,
x2>=0
When plotting these inequalities on graph paper, we
obtained the following solutions. As the
objective function was drawn and moved away from the origin, it approached
x1=150 and x2=250. Therefore, the optimal solution is (150,250).
so the maximum value is:
Z = 3*150 + 4*250 = 1450
ii). Maximize Z
= 4x1 + x2` 1) let x1
= 0, x2 <=2.5
subject to : x1
+ 2x2 <= 5 x2 = 0, x1 <=
5
3x1
+ 2x2 <= 9 2). Let x1 =0, x2 <=
4.5
x1,x2
>= 0 x2 = 0,
x1 <= 3
By plotting the above values on the graph and drawing
objective function and by moving it away from the origin, the optimal solution
that this linear curve approached is: (3,0)
X1 = 3, X2 = 0.
So the maximum value is, Z = 4*3 + 0 = 12
iii). Minimize Z = x1 - x2 Working:
Subject to : x1 - 2x2 ≤ 3 1) Let x1 = 0, x2 >= -1.5
x1
- 4x2 <= 5 x2 = 0, x1 <= 3
x1,x2
>= 0 2). Let
x1 = 0, x2 >= - 1.25
x2
= 0, x1 <= 5
These inequalities were plotted and as the objective
linear curve was drawn and moved towards the origin, the optimal point it
approached was (0,0). Actually this is the origin. Therefore, the minimum value
is: Z = 0
iv). Maximize
Z = 400x1 + 200x2 working:
Subject to: x1 + x2 = 30 1). Let x1 = 0, x2 = 30
2x1
+ 8x2 >=80 x2 = 0, x1
= 30
x1 <= 20 2).
Let x1 = 0 , x2 >=10
x1,x2
>= 0 x2=0,
x1 >=40
Upon plotting the inequalities on the graph paper,
there were two possible solutions to this problem but the most acceptable
optimal solution is: (20,10).
x1 = 20, x2
= 10
Thus, the maximum value is; Z = 400*20 + 200*10 = 10
000
v). Minimize Z = 400x1 + 200x2 working:
Subject to: x1 + x2 = 30 1). Let x1 = 0, x2 = 0
2x1
+ 8x2 >=80 x2 = 0, x1
= 30
x1 <= 20 2).
Let x1 = 0 , x2 >=10
X1,
x2 >= 0 x2=0,
x1 >=40
From the outer space and drawing objective linear
curve towards the origin, the point which the curve reached was (0,10),
thus x1 = 0 & x2 = 10
Thus, the minimum value is:
Z = 200*10 = 2000
Question 2
Product
|
BoP
|
HE
|
Constraints
|
Labor(hrs)
|
4
|
3
|
30
|
Timber(m3)
|
2
|
8
|
80
|
1. Objective function:
profit
from Bop + profit from HE
Z = K400/BoP + K200/HE Let
x1 = BoP, x2 = HE
Z
= 400x1 + 200x2
2. The key resources are time and special timber.
a). Time: 4hrs/BoP
+ 3hrs/HE <= (restricted to 30 hrs/month)
i.e. 4x1 +
3x2 <= 30hrs
b). Timber: 2m3/BoP
+ 8m3/HE <= (restricted to 80m3/month)
i.e. 2x1 + 8x2
<= 80
LP Model: working
Maximize Z =
400x1 + 200x2 1). Let x1 = 0 , x2 <= 10
Subject to: 4x1
+ 3x2 <= 30 x2=0, x1
<= 7.5
2x1
+ 8x2
<= 80 2).
Let x1 =0, x2 <= 10
x1 <= 20 x2=0,
x1 <= 40
x1,x2
>= 0
In order to satisfy the above conditions to reach the
optimum solution, there was a linear objective curve drawn and moved away from
the origin. As such, the curve approached (7.5, 0). So this is the
optimum solution, x1 = 7.5 & x2 = 0
Hence, the maximum profit is: Z = 400*7.5
= K3000
Using
Excel Solver.
Question 2.
Question 2.
BoP x1
|
HE x2
|
Profit
|
Limit
|
||
Decision variables
|
7.5
|
0
|
|||
Objective function
|
400
|
200
|
3000
|
||
A
|
4
|
3
|
30
|
<=
|
30
|
B
|
2
|
8
|
15
|
<=
|
80
|
Sensitivity Report
Adjustable Cells
|
||||||
Final
|
Reduced
|
Objective
|
Allowable
|
Allowable
|
||
Cell
|
Name
|
Value
|
Cost
|
Coefficient
|
Increase
|
Decrease
|
$B$2
|
Decision variables x1
|
7.5
|
0
|
400
|
1E+30
|
133.3333333
|
$C$2
|
Decision variables x2
|
0
|
-100
|
200
|
100
|
1E+30
|
Constraints
|
||||||
Final
|
Shadow
|
Constraint
|
Allowable
|
Allowable
|
||
Cell
|
Name
|
Value
|
Price
|
R.H. Side
|
Increase
|
Decrease
|
$D$4
|
A Profit
|
30
|
100
|
30
|
130
|
30
|
$D$5
|
B Profit
|
15
|
0
|
80
|
1E+30
|
65
|
Answer Report
Target Cell (Max)
|
|||||
Cell
|
Name
|
Original Value
|
Final Value
|
||
$D$3
|
Objective function Profit
|
0
|
3000
|
||
Adjustable Cells
|
|||||
Cell
|
Name
|
Original Value
|
Final Value
|
||
$B$2
|
Decision variables x1
|
0
|
7.5
|
||
$C$2
|
Decision variables x2
|
0
|
0
|
||
Constraints
|
|||||
Cell
|
Name
|
Cell Value
|
Formula
|
Status
|
Slack
|
$D$4
|
A Profit
|
30
|
$D$4<=$F$4
|
Binding
|
0
|
$D$5
|
B Profit
|
15
|
$D$5<=$F$5
|
Not Binding
|
65
|
There
are three parts to the answer report. (1). Target cell, 2). Adjustable cell and
constraint
cell. The optimal value is given in the final
value column of the target cell.
The
amount of slack for each constraints which means there is a flexibility
existing when
removing
one labor will affect the production. the sensitivity report shown above contain
relevant
information concerning the effect of changes to objective function coefficient.
adjustable
cells section include the reduced cost and ranges of optimality for the
objective
function
coefficient. The constraint section gives the shadow prices and ranges of
feasibility
for
right hand side values(allowable increases or decreases)
Question3
Product
|
Copper
|
Gold
|
Constraints
|
Labor (hrs)
|
3
|
4
|
240
|
Ore (tons)
|
2
|
1
|
1000*T
|
1. objective function:
|
||||
profit from copper +
profit from gold
|
||||
Z = $1000/ton Cu + $1500/
ton Au
|
Let x1 = copper
|
|||
Z = $1000x1 +
$1500x2
|
x2=gold
|
|||
2. Key resources are
tonnes of ore and time.
|
|||
a). Time:
|
3hrs/Cu + 4hrs/Au
<= (restricted to 240hrs/week)
|
||
i.e. 2x1 + 4x2 <=240 hrs
|
|||
b). Ore:
|
2ton/Cu +
1ton/Au <= (restricted to 1000*T tons/week)
|
||
i.e. 2x1 + x2 <=1000*T tons
|
|||
3. Market condition, production at two given conditions.
|
||||
a). X1 + x2 <= 700
|
||||
b). X1 - x2
< = 350
|
||||
LP Models
|
|||||
Maximize Z =
1000x1+1500x2
|
Working.
|
||||
subject to:
|
3x1 + 4x2
<= 240
|
1) x1 <=
80, x2<=60
|
|||
2x1 + x2
<= 1000*T
|
2) x1
<=500*T, x2 <= 1000*T
|
||||
x1 + x2 = 700
|
3)x1
<=700, x2 <= 700
|
||||
x1 - x2
<= 350
|
4) x1 <=
350, x2 >= -350
|
||||
x1,x2>=0
|
|||||
By plotting these inequalities on the graph
paper and drawing the linear objective curve and moving
towards
the origin, the optimal solution that this curve approached was such that it
satisfy the above
given
conditions. So the optimal solution is: (80, 0).
x1
= 80 & x2 = 0
Thus,
the maximum profit is: Z = 1000*80 = $80
000.
Using the Excel
Solver:
copper x1
|
gold x2
|
Profit
|
Limit
|
||
Decision variables
|
80
|
0
|
|||
Objective function
|
1,000
|
1500
|
80000
|
||
Labor (hrs)
|
3
|
4
|
240
|
<=
|
240
|
Ore (tonnes)
|
2
|
1
|
160
|
<=
|
1,000,000
|
total production
|
1
|
1
|
80
|
<=
|
700
|
Mix production
|
1
|
-1
|
80
|
<=
|
350
|
Answer Report
Solver Options
Max
Time 100 sec, Iterations 100, Precision
0.000001
Max
Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%.
Objective Cell (Max)
|
||||||||
Cell
|
Name
|
Original Value
|
Final Value
|
|||||
$D$3
|
0
|
240
|
||||||
Variable Cells
|
||||||||
Cell
|
Name
|
Original Value
|
Final Value
|
Integer
|
||||
$B$1
|
0
|
80
|
Contin
|
|||||
$C$1
|
0
|
0
|
Contin
|
|||||
Constraints
|
||||||||
Cell
|
Name
|
Cell Value
|
Formula
|
Status
|
Slack
|
|||
$D$3
|
240
|
$D$3<=$F$3
|
Binding
|
0
|
||||
$D$4
|
160
|
$D$4<=$F$4
|
Not Binding
|
840
|
||||
$D$5
|
80
|
$D$5<=$F$5
|
Not Binding
|
620
|
||||
$D$6
|
80
|
$D$6<=$F$6
|
Not Binding
|
270
|
||||
Sensitivity Report
Variable Cells
|
||||||
Final
|
Reduced
|
Objective
|
Allowable
|
Allowable
|
||
Cell
|
Name
|
Value
|
Cost
|
Coefficient
|
Increase
|
Decrease
|
$B$1
|
80
|
0
|
3
|
0
|
3
|
|
$C$1
|
0
|
0
|
4
|
1E+30
|
0
|
Constraints
|
||||||
Final
|
Shadow
|
Constraint
|
Allowable
|
Allowable
|
||
Cell
|
Name
|
Value
|
Price
|
R.H. Side
|
Increase
|
Decrease
|
$D$3
|
240
|
1
|
240
|
810
|
1E+30
|
|
$D$4
|
160
|
0
|
1000
|
1E+30
|
840
|
|
$D$5
|
80
|
0
|
700
|
1E+30
|
620
|
|
$D$6
|
80
|
0
|
350
|
1E+30
|
270
|
The optimal value is given in the final
value column of the target cell.
The amount of slack for each constraint
which means there is a flexibility existing for example,
Removing 1 labor will affect the
production. The sensitivity report shown above contain
Relevant information concerning the
effect of changes to objective function coefficient.
Adjustable cells sections include the
reduced cost and ranges of optimality for the objective
function coefficient. The constraint
section gives the shadow prices and ranges of feasibility
for right hand side values(allowable
increase or decrease)
2. answer the following questions:
i). What are the decision variables?
The decision variables are: x1 = 80
&, x2 = 0.
ii). What is the maximum profit? Z =
$80000
iii). If the objective function curve
equals the curve for production function of copper and gold., what other
decision variables are available?
If this situation arises, there the will be
multiple solutions to this problem. As such,
we would have an additional optimal
solutions, and they are:
x1 = 0, & x2=60
This gives us a maximum value of: Z = 60*1500 = $90 000
iv). Find optimal solutions
if the objective function changes by:
|
||||||
a). Maximize Z = 1000x1
+ 4500x2
|
solutions:
|
|||||
b). Maximize Z = 1000x1
- 4500x2
|
x1 = 80,
& x2 = 0
|
Z = $80000
|
||||
c). Maximize Z = 1000x1
+ 500x2
|
x1 =
234.29, & x2 =-115.71
|
Z = $755000
|
||||
x1 =
234.29,& x2 =-115.71
|
Z = $176428.6
|
|||||
a).
Copper x1
|
Gold x2
|
Profit
|
Limit
|
||
Decision variables
|
80.00
|
0.00
|
|||
Objective function
|
1,000
|
4500
|
80000
|
||
Labor (hrs)
|
3
|
4
|
240
|
<=
|
240
|
Ore (tonnes)
|
2
|
1
|
160
|
<=
|
1,000,000
|
Total production
|
1
|
1
|
80
|
<=
|
700
|
Mix production
|
1
|
-1
|
80
|
<=
|
350
|
b).
Copper x1
|
Gold x2
|
Profit
|
Limit
|
||
Decision variables
|
234.29
|
-115.71
|
|||
Objective function
|
1,000
|
-4500
|
755000
|
||
Labor (hrs)
|
3
|
4
|
240
|
<=
|
240
|
Ore (tonnes)
|
2
|
1
|
352.8571
|
<=
|
1,000,000
|
Total production
|
1
|
1
|
118.5714
|
<=
|
700
|
Mix production
|
1
|
-1
|
350
|
<=
|
350
|
c).
Copper x1
|
Gold x2
|
Profit
|
Limit
|
||
Decision variables
|
234.29
|
-115.71
|
|||
Objective function
|
1,000
|
500
|
176428.6
|
||
Labor (hrs)
|
3
|
4
|
240
|
<=
|
240
|
Ore (tonnes)
|
2
|
1
|
352.8571
|
<=
|
1,000,000
|
Total production
|
1
|
1
|
118.5714
|
<=
|
700
|
Mix production
|
1
|
-1
|
350
|
<=
|
350
|
There were no separate graphs drawn for
the above objective functions because the inequalities are the same ones and
only the objective functions changed. But they were solve using excel solver
only. As such there were some negative
value generated by the excel solver. These were resulted when in the options
section of solver, the “assume non-negative” option was not clicked. If it was
click then it would give zero as a substitute of negative values. After all,
the maximum values or maximum profits were given at the last row of objective
function under the profit column for the above tables.
It was seen that, for different objective functions there were different decision variables being generated and consequently various maximum values were also generated. However, when solved graphically, the solution obtained cannot be changed even with the various objective functions because it must satisfy the given conditions, i.e. inequalities or the constraints. Therefore, no matter what, the graphical solution remains constant so only one graph was drawn. In order to satisfy this condition; x1, x2 ≥ 0, the negative values can be ignored and only the values from zero to positive values can be accepted.
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