Q1.An Animal Feed Company Must Produce at Least 200 Kgs of A Mixture Consisting of Ingredients

 

Q1.An animal feed company must produce at least 200 kgs of a mixture consisting of ingredients

 and  daily.  costs Birr 3 per kg and  Birr 8 per kg. No more than 80 kg of  can be used and

at least 60 kg of  must be used. a. Formulate a mathematical linear programming model to the problem.  b. Solve minimization by using graphical approach. approach. Minimize (total cost)

 

Z=3x1+8x2 Subject to constraints X1+X2=200 X1=60 X1>=0 X1>=0 SOLUTION

Where

 X 

1

 = no. of kgs of ingredient

 X 1 +  X 2 + X 3 =200

,

 X 

 and

1

 X 1 +  X 4 = 80

,

 X 

2

 = no. kgs of ingredient

 X 2− X 5 =60

The first constraint X1 + X2 =200 can be represented as follows. We set X1 + X2 = 200 When X1 = 0 in the above constraint, we get,  0 + X2 = 200 X2 = 200 Similarly when X2 = 0 in the above constraint, we get, X1 + 0 = 200 X1 = 200 The second constraint X1 ≤ 80 can be represented as follows, We set X1 = 80 The third constraint X2 >= 60 can be represented as follows,

 X 

2

 

We set X2 = 60 Cost = 3x1 + 8x2 as Cost of x is less hence we must use maximum a but Not more than 80 kg. Of a can be used Hence a = 80 kg X2 = 200 - 80 = 120 kg Cost = 3 * 80 + 8 * 120= 240 + 960= 1200 Cost of mixture = Rs 1200  X 1 =80

,

 X 2 = 120

,

S =0

Point

X1

X2

Z = 3X1 +8X2

0

0

0

0

A

0

200

Z = 3 x 0 + 8 x 200 = 1,600

B

200

0

Z = 3 x 200 + 8 x 0 = 600=minimum

C

80

120

Z = 3 x 80 + 8 x 120 = 1200

Q2. A nutritionist advises an individual who is suffering from iron and vitamin v itamin B deficiency to

B1, and 1500 mg of vitamin B2 take at least 2400 milligrams (mg) of iron, iron, 2100 mg of vitamin B1, over a period. Two vitamin pills are suitable, brand-A suitable,  brand-A and and brand-B  brand-B.. Each brand-A pill costs 6 cents and contains 40 mg of iron, 10 mg of vitamin B1, and 5 mg of vitamin B2. Each brand-B

 

costs 8  8 cents and contains 10 mg of iron and 15 mg each of vitamins B1 and B2.  pill costs a. Formulate a mathematical linear programming model to the problem  b. What combination of pills should the individual purchase in order to meet the minimum iron and vitamin requirements at the lowest cost? Use graph method Solution  

first tabulate the given information:

  Co C ost/Pill   Iron   Vitamin B1   Vitamin B2

Brand-A 6 40 mg 10 mg 5mg

Brand-B 8 10 mg 15 mg 15 mg

Minimum Requirement 2400 mg 2100 mg 1500 mg

 

Let x be the number of brand-A pills and  y the number of brand-B pills to be b e purchased.



The cost C  (in  (in cents) is given by and is the objective function to be minimized.   C= 6X+8Y



The amount of iron contained in x brand-A pills and y brand-B pills is given by 40 x + 10 y mg, and this must be greater than or equal to 2400 mg. This translates into the inequality   40X+10Y>=2400  The amount of vitamin B1 contained in x brand-A pills and y brand-B pills is given by 10 x + 15 y mg, and this must be greater or equal to 2100 mg. This translates into the inequality   10X+15Y>=2100  The amount of vitamin B2 contained in x brand-A pills and y brand-B pills is given by 5 x  + 15 y mg, and this must be greater or equal to 1500 mg. This translates into the inequality   5X+15Y>=1500  In short, we want to minimize the objective function C= 6X+8Y 40X+10Y>=2400 10X+15Y>=2100 5X+15Y>=1500 X>=0 Y>=0

 



Graphical Solutions of Linear Programming Problems



Which asked us to determine the optimal combination of pills to be purchased in order to meet the minimum iron and vitamin requirements at the lowest cost.



We restated the problem as a linear programming p rogramming problem in which we wanted to minimize the objective function C= 6X+8Y Subject to the system of inequalities 40X+10Y>=2400 10X+15Y>=2100 5X+15Y>=1500 X, Y>=0



We can now solve the problem graphically. We first graph the feasible set S  for  for the  problem. Graph the solution for the inequality 40X+10Y>=2400 considering only positive values for  x and y:

 



We first graph the feasible set S  for  for the problem. 

Graph the solution for the inequality 10X+15Y>=2100 considering only positive values for  x and y:



We first graph the feasible set S  for  for the problem. 

Graph the solution for the inequality 5X+15Y>=1500 considering only positive values for  x and y:



We first graph the feasible set S  for  for the problem.



Graph the intersection of the solutions to the inequalities, yielding the feasible set S . (Note that the feasible set S  is  is unbounded)

 

 



 Next, find the vertices of the feasible set S . The vertices are A(0, 240), B(30, 120), C(120, 60), and D(300, 0).

 Now, find the values of C  at  at the vertices and tabulate them: Vertex

C = 6x + 8y

A(0, 240)

1920

B(30, 120)

1140

 C(  C(120, 60)

1200

D(300, 0)

1800

 

 

Finally, identify the vertex with the lowest value for C :

We can see that C   is is minimized at the vertex B (30, 120) and has a value of 1140. Vertex  A(0, 240)

1920

 B(  B(30, 120)

1140

 C(  C(120, 60)

1200

D(300, 0)



C = 6x + 8y

1800

Finally, identify the vertex with the lowest value for C :



We can see that C  is  is minimized at the vertex  B (30, 120) and has a value of 1140.



Recalling what the symbols x, y, and C  represent;  represent; we conclude that the individual should  purchase 30 brand-A pills and 120 brand-B pills at a minimum cost of $11.40.

 



The Simplex Method: Standard Maximization Problems

x

y

u

v

p

constant

1

0

3/ 5

–1/5

0

48

0

1

–1/5

2/5

0

84

0

0

9/25

7/25

1

148 4/5

Q3. A company is

involved in the

Produce X and Y are twofold, namely machine time for automatic processing and craftsman time For hand finishing. The table below gives the number of minutes required for each item: Machine time Craftsman time Mach Ma chin inee Time Time Cra Craft ftsm sman an Tim Timee Item X Item Y

13 19

Mach Machin inee Tim Timee Craf Crafts tsma man n Time Time 20 29

The company has 40 hours of machine time available in the next working week but only 35 Hours of craftsman time. Machine time is costed at Birr 10 per hour worked and craftsman time Is costed at Birr 2 per hour worked. Both machine and craftsman idle times incur no costs. The Revenue received for each item produced (all production is sold) is Birr 20 for X and Birr 30 for Y. The company has a specific contract to produce 10 items of X per week for a particular customer. a. Formulate the problem of deciding how much to produce per week as a linear program.  b. Find optimization solution  Solution Let  be the number of items of  ,  be the number of items of . Maximize 20 + 30 − 10(h  ) − 2(  ) Subject to: 13 + 19≤ 40(60) h  20 + 29≤ 35(60)   ≥ 10 contract ,≥ 0

 

 So that the objective function becomes Maximize 20 + 30 −10 (13 + 19) 60 −2 (20  + 29) 60 =17.1667  + 25.8667  Subject to: 13 + 19≤ 2400  20 + 29≤ 2100 ≥ 10 ,≥ 0

It is plain from the diagram below that the maximum occurs at the intersection of =10 and 20 + 29≤ 2100. Solving simultaneously, rather than by reading values off the graph, we have that =10 and =65.52 with the value of the objective function being 1866.5.

 

Q4. Consider the following problem. Max 3x1 + 2x2 Subject Function 3x1 + x2 ≤ 12 X1 + x2 ≤ 6 5x1 + 3x2 ≤ 27 X1, x2 ≥ 0. a. Solve the problem by the original simplex method (in tabular form).  b. Solve the dual of this problem manually by the dual simplex method.

 

Step 3- obtaining the initial simplex tableau BV S1 S2 S3

Cj CB 0 0

3 X1 3 1

2 X2 1 1

0 S1 1 0

0 S2 0 1

0 S3 0 0

bj

0 Zj Cj-Zj

5 0 3

3 0 2

0 0 0

0 0 0

1 0 0

27 0

bj/aj

12 6

 

Step 4: Test if the current solution is optimum or not:

Since our solution is not optimal (because Cj-Zj≥ 0). Step 5: Identify the Entering and Leaving Variables BV S1 S2 S3

Cj CB 0 0 0 Zj Cj-Zj

Incoming Variable

3 X1 3 1 5 0 3

2 X2 1 1 3 0 2

0 S1 1 0 0 0 0

0 S2 0 1 0 0 0

key Element

0 S3 0 0 1 0 0

bj

bj/aj

12 6 27

12/3=4 12/3=4 6/1=6 27/5=5.4

key row

Step 6. Evaluate the new solution by constructing a second simplex tableau. 

Entering variable value =old value /key element 3/3=1, 1/3=1/3 3/3=1 1/3=1/3,, 1/3=1/3 1/3=1/3,, 0/3=0 0/3=0, 0/3=0 0/3=0, 12/3=4 12/3=4

 



 New row element element = old row elem element-(key ent-(key column element*new key row or replacement row value). S21-1*1= 1-1*1=0 1-1*1/3=0. 1-1*1/3= 0-1*0=0 0-1*0= 0, 0. 0 1-1*1/3=0, 1-1*1/3=0

1-1*0=1 1-1*0= 1,

6-1*4=2 6-1*4= 2,

S35-5*1= 5-5*1=0 0,

0-5*1/3=-5. 0-5*1/3=-5.

1-5*0=1 1-5*0=1

3-5*1/3=-2,, 3-5*1/3=-2

0-5*0=0 0-5*0= 0,

27-5*4=7 27-5*4= 7,

 

BV X1 S2 S3



Cj CB 3 0 0 Zj Cj-Zj

3 X1 1 0 0 3 0

2 X2 1/3 0 -2 3 -1

0 S1 1/3 0 -5 3 -3

0 S2 0 1 0 0 0

0 S3 0 0 1 0 0

bj

bj/aj

4 2 7 12

4/1=4 2/0=0 7/0=0

Since our solution is satisfied optimality condition (Cj-Zj≤ 0), therefore the optimal. Max Z 3x1 + 2x2 X1=4 X2=0 3*4+2*0=12=12 3*4+2*0=12=12 Z=12

Q5. If j=Cj-Zj is equ equal al to 24 – 7M for for x1 x1;28-7M ;28-7M for x2;45-4M for x 3 and 2 2882M in respect of x4 in a minimization problem, problem, then the what is entering variable? 

  X1-24-7M Let m=1

24-7*1 =24-7=17

 

X2-28-7M 28-7*1 =28-7=21 X3-45-4M =45-4*1 

 

=45-4=41 X4--28-2M =28-2*1 28-2=26

Therefore from the above equitation Therefore equitation X1-24-7M the small positive number and the most negative numbers for this reason the entering variable is X1-24-7M