The document presents a linear programming problem involving a company that manufactures three products (X, Y, Z) using a single raw material. It formulates the problem as a linear program with constraints on minimum sales volumes and maximum production capacities. The optimal product mix is determined both manually and using a solver. The dual price of the raw material constraint is computed from the optimal solution and used to analyze the effect of changes in raw material availability and product demand.

1. SMQ 3043
LINEAR PROGRAMMING
ASSIGMENT
LECTURER:
PN NORSIDA BINTI HASAN
GROUP:
D
NURSUHADA BINTI ARSHAD D20111048843
NORIZAN BINTI NORDIN D20111048874
NOR ATIQAH FATIHAH BINTI ABDULLAH D20111048872
FARAH WAHEEDA BINTI AHMAD SOBRI D20111048879

2. Question:
A company manufactures three products, X, Y, and Z. The sales volume for X is at least 50%
of the total sales of all three products. However, the company cannot sell more than 80 units
of X per day. The three products use one raw material, of which the maximum daily
availability is 240 lb. The usage rates of the raw material are 2 lb per unit of X, 4 lb per unit
of Y, and 3 lb per unit of Z. The unit prices for X, Y and Z are RM 20, RM 50 and RM 35
respectively:
a) Formulate a mathematical model for the problem.
s.t

3. b) Determine the optimal product mix for the company:
i) by hand (manually) AND
ANSWER:
By using Big M method to solve the problem.
Standard form:
Max t = 20x + 50y + 35w - Ma1
s.t X - Y - Z - e1 + a1 = 0
X + s2 = 80
2X + 4Y + 3Z + s3 = 240
X, Y, Z , e1, a1, s2,s 0
BV ROW t X Y Z e1 a1 s2 s3 RHS
t 0 1 -20 -50 -35 0 M 0 0 0
a1 1 0 1 -1 -1 -1 1 0 0 0
s2 2 0 1 0 0 0 0 1 0 80
s3 3 0 2 4 3 0 0 0 1 240
*RO = R0 – MR1
Initial Tableau
BV ROW t X Y Z e1 a1 s2 s3 RHS RATIO
t 0 1 -20-M -50+M -35+M M 0 0 0 0 -
a1 1 0 1 -1 -1 -1 1 0 0 0 0
s2 2 0 1 0 0 0 0 1 0 80 80
s3 3 0 2 4 3 0 0 0 1 240 120
*R0 = R0 – (-20 - M) R1 *R2 = R2 – R1 *R3 = R3 – 2R1

4. Tableau 1
BV ROW t X Y Z e1 a1 s2 s3 RHS RATIO
t 0 1 0 -70 -55 -20 20 + M 0 0 0 -
X 1 0 1 -1 -1 -1 1 0 0 0 -
s2 2 0 0 1 1 1 -1 1 0 80 80
s3 3 0 0 6 6 2 -2 0 1 240 40
*R3 = *R0 = R0 + 70R3 *R1 = R1 + R3 *R2 = R2 – R3
Tableau 2
BV ROW t X Y Z e1 a1 s2 s3 RHS
t 0 1 0 0 M- 0 2800
X 1 0 1 0 0 40
s2 2 0 0 0 1 40
Y 3 0 0 1 0 40
Since row 0 of Tableau 2 consist of nonnegative it has become an optimal tableau.
Therefore: t = 2800
X = 40
Y = 40
S2 = 40
Z = 0

5. ii) using a solver (e.g: LINDO, LINGO, Microsoft Excel and etc.)
ANSWER:
By using a solver LINDO to solve the problem.

6. Result

7. c) Determine the dual price of the raw material resource and its allowable range. If
available raw material is increased by 120 lb, determine the optimal solution and the
change in total revenue using the dual price.
s.t
Initial tableau
BV X Y Z RHS
T -20 -50 -35 0 M 0 0 0 M 0 0
1 -1 -1 -1 1 0 0 0 1 0 0
1 0 0 0 0 1 0 80 0 1 0
2 4 3 0 0 0 1 240 0 0 1
The column , , and , are identical.
Optimal tableau
BV X Y Z RHS
T 0 0 -35 M 2800 M 0
X 1 0 -1 40 0
0 0 0 40 1
0 1 3 40 0
For constraint ≥, the same idea remains applicable except that the dual price will assume the
opposite sign of that associated with the ≤ constraint.
From the optimal tableau, we get the new optimal solution.

8. t = 2800 + M +
X = 40 + +
= 40 + +
Y = 40 +
Dual Price
t = 2800 + M +
Resource Optimal t- equation
coefficient of
Dual price
Constraint 1 M RM
Constraint 2 0 RM
Constraint 3
RM
A unit change in operation 1 capacity change t by M unit.
The zero dual price in constraint 2 means that there is no economic advantage in allocating
more production time to this operation.
A unit change in operation 3 capacity change t by unit.
Feasible Range
The solution remains feasible when all basic variables remains non-negative.
The optimal tableau remains.
t = 2800 + M +

9. X = 40 + +
= 40 + +
Y = 40 +
Range for
40 + 40 + 40
Range for
40 +
Range for
40 + 40
When the raw material increased by 120 lb,
t = 2800 + M +
Raw material will only effect the third constraint so , the new optimal solution is
X = 40 + +
= 40 + +

10. Y = 40 +
t = 2800 +
= 4200
Therefore, when we increased raw material by 120 lb, the total revenue price increased to
RM4200.
The change in revenue:
= The new revenue – The old revenue
= RM 4200 – RM 2800
= RM 1400

11. d) Use the dual price to determine the effect of changing the maximum demand for product X
by ± 10 units.
We choose because the change of ± 10 is still in the range of
When the demand for product X change by -10 units:
X = 40 + +
= 40 +
Y = 40 +
When the demand for product X change by +10 units:
X = 40 + +
= 40 + +
Y = 40 +
Hence, we can conclude that the optimal solution remain unchanged since the changing of
maximum demand for product X by ± 10 units still in the range of D2.

12. REFERENCES
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