SIMPLEX Sample Problem
24
Ex: Anton, Inc has P100,000 available for tment and must decide how much of this be invested in each of the three avail natives: stock A, stock B, and bond X Stock A is priced at P50 per share and ted to return P10 per share in annual ends. Although this return is high, st so considered risky. Stock B is priced er share and returns P3 per share. Bon quoted at P200 with an annual return of cent.
description
MBA quantitative analysis, simplex sample problem
Transcript of SIMPLEX Sample Problem
Ex:
Anton, Inc has P100,000 available forinvestment and must decide how much of thisis to be invested in each of the three availablealternatives: stock A, stock B, and bond X
Stock A is priced at P50 per share and isexpected to return P10 per share in annualdividends. Although this return is high, stock Ais also considered risky. Stock B is priced atP25 per share and returns P3 per share. BondX is quoted at P200 with an annual return of9 percent.
Anton has decided to limit its investment in the riskier stock A to a maximum of one-fourth of its total investment. For every shareof stock A purchased, Anton will purchase atleast three shares of stock B. There are a maximum of 250 bonds available to Anton atthe stated price. Anton also stipulates thatthe total investment in bonds must be at leastas great as half of its investment in stocks.
How do you recommend that Anton invest its funds?
Suggested definition of the decision variables:
Let x1 = number of shares of stock A to purchase
x2 = number of shares of stock B to purchase
x3 = number of bonds X to purchase
Max: Z = 10x1 + 3x2 + 18x3
Subject to:
1. 50x1 + 25x2 + 200x3 100,000
2. 50x1 0.25(50x1 + 25x2 + 200x3)
3. x2 3x1
4. 200x3 0.5(50x1 + 25x2)
5. x1, x2, x3 0
Constraint (1) {50x1 + 25x2 + 200x3 100,000}
This is the budget constraint as discussed above.
Constraint (2) {50x1 0.25(50x1 + 25x2 + 200x3)}
This constraint limits the total investment in stock A (which is 50x1 pesos to one- fourth (0.25) the total investment [(50x1 + 25x2 + 200x3) pesos ].
Constraint (3) {x2 3x1}
This constraint states that the total number of shares of stock B (x2) must be at least as great as three times of the total number of shares of stock A (x1).
Constraint (4) {200x3 0.5(50x1 + 25x2)}
This constraint expresses the fact that, since there are only 250 bonds available to Anton, the number of bonds it decides to purchase cannot exceed that amount.
Constraint (4) {200x3 0.5(50x1 + 25x2)}
This constraint states that the total investment in bonds (200x2 pesos) must be at least as great as one-half (0.50) its total investment in stocks, which is (50x1 + 25x2) pesos for stocks A and B combined.
SIMPLEX METHOD
Usage of the following:
Slack variable, Si - fills the gap between left and right members of the equation.
- unused quantity, therefore , it does contribute to any profit, hence, the coefficient in the objective is zero
Artificial Variable, Ai - contributes the biggest amount to costs, therefore in the objective it has a coefficient greater than any of the coefficients of solution variables.
For convenience, use a quantity in powers of 10.
Ex: 10, 100, 1000, 10,000
SIMPLEX MAXIMIZATION METHOD
1. Add a slack variable (S) to the left member of the constraint containing = & symbols.
2. For a symbol of the constraint, convert it first to a symbol by multiplying the constraint by -1, then add the slack variable.
Example:
Maximize: Z = 10x1 + 8x2
Subject to: 30x1 + 20x2 120
2x1 + 2x2 9
4x1 + 6x2 24
x1, x2 0
Maximize: Z =
Subject to: 30x1 + 20x2 + S1 = 120
2x1 + 2x2 + S2 = 9
4x1 + 6x2 + S3 = 24
x1, x2, S1, S2, S3 0
New Program
10x1 + 8x2 + 0S1 + 0S2 + 0S3
Cj
SolutionVariables
SolutionQuantityx1 x2 S1 S2 S3
10 8 0 0 0
S1
S2
S3
30 20 1 0 0
2 2 0 1 0
4 6 0 0 1
120
9
24
0
0
0
Zj 0 0 0 0 0
Cj - Zj 10 8 0 0 0
optimum column
pivotal row
contribution to profit
outgoingvariable
enteringvariable
pivot
T1
0
Cj
SolutionVariables
SolutionQuantityx1 x2 S1 S2 S3
10 8 0 0 0
x1
S2
S3
1 2/3 1/30 0 0
0 2/3 -2/30 1 0
0 10/3 -4/30 0 1
4
1
8
10
Zj 10 20/3 10/30 0 0
Cj - Zj 0 4/3 -10/30 0 0
0
0
40
T2
Cj
SolutionVariables
SolutionQuantityx1 x2 S1 S2 S3
10 8 0 0 0
x1
x2
S3
1 0 1/10 -1 0
0 1 -1/10 3/2 0
0 0 6/30 -5 1
3
3/2
3
10
Zj 10 8 2/10 2 0
Cj - Zj 0 0 -2/10 -2 0
8
0
42
T3
Decision:
X1 = 3
X2 = 3/2
Max Value = 42
SIMPLEX MINIMIZATION METHOD
1. Add an artificial variable (A) if the symbol of the constraint is =.
2. Add a slack variable (S) if the symbol is .
3. Subtract a slack variable and add artificial variable if .
Example:
Minimize: Z = 6x + 16y
Subject to: x + y = 400
x 150
y 200
x, y 0
New Program
Minimize: Z =
Subject to: x + y + A1 = 400
x + S1 = 150
y - S2 + A2 = 200
x, y, S1, S2 0
6x + 16y + 0S1 + 0S2 + 100A1 + 100A2
Cj
SolutionVariables
SolutionQuantityx y S1 S2 A1 A2
6 16 0 0 100 100
A1
S1
A2
1 1 0 0 1 0
1 0 1 0 0 0
0 1 0 -1 0 1
400
150
200
100
Zj 100 200 0 -100 100 100
Cj - Zj -94 -184 0 100 0 0
0
100
60,000
T1
Cj
SolutionVariables
SolutionQuantityx y S1 S2 A1 A2
6 16 0 0 100 100
A1
S1
y
1 0 0 1 1 -1
1 0 1 0 0 0
0 1 0 -1 0 1
200
150
200
100
Zj 100 16 0 84 100 -84
Cj - Zj -94 0 0 -84 0 184
0
16
23,200
T2
Cj
SolutionVariables
SolutionQuantityx y S1 S2 A1 A2
6 16 0 0 100 100
A1
x
y
0 0 -1 1 1 -1
1 0 1 0 0 0
0 1 0 -1 0 1
50
150
200
100
Zj 6 16 -94 84 100 -84
Cj - Zj 0 0 94 -84 0 184
6
16
9,100
T3
Cj
SolutionVariables
SolutionQuantityx y S1 S2 A1 A2
6 16 0 0 100 100
S2
x
y
0 0 -1 1 1 -1
1 0 1 0 0 0
0 1 -1 0 1 0
50
150
250
0
Zj 6 16 -10 0 16 0
Cj - Zj 0 0 10 0 84 100
6
16
4,900
T4
Decision:
X = 150
y = 250
Z = 4,900 (minimum value)
