A manufacturer has 14700 hours capacity in its assembly department and 6300 hour

Question

A manufacturer has 14700 hours capacity in its assembly department and 6300 hours available in its packaging department. Assembly and packaging times for each of products A;B and C are given in the table below, along with the profit per unit for each product:

A B C

Assembly time (hr/unit) 0:8 0:3 0:4

Packaging time (hr/unit) 0:2 0:6 0:1

Profit per unit (€.000) 3 4 1

The manufacturer wishes to determine the production levels so maximise profit.

(a) Express this maximisation problem in its standard form - i.e., formulate the equations that describe the problem outlined above.

(b) Use the simplex method to show that the optimal table is given below:

Basic A B C S1 S2 Solution

z 0 0 1/2 5/21 23/42 139/2

A 1 0 1/2 1/7 -1/14 33/2

B 0 1 0 -1/21 4/21 5

(c) In what units are the decision variables (A; B; C) expressed? What is the maximum level of profit?

(d) The solution above is optimal. How is this recognised from the table? (e) What are the production levels required to maximise profit?

(f) The following constraint A+B+2C 20; related to administration time, must now be taken into consideration also. Using the optimal table solution from part (b), set up the new augmented matrix and outline (but do not solve) the method that would be used to find the new optimal solution.

part f in particular

Explanation / Answer

(a) The formulation of the problem is given below.

The objective function is

Maximize Z = 3A + 4B +C

The constraints are

1) Assembly time (hrs/unit)

8A + 3B + 4 C <= 147

2) Packaging time

2A + 6B + C <= 63

Where A,B and C represend number of units of products A,B,C respectively.

Hence the non negativity restriction, A>=0,B>=0,C>=0.

Therefore the standard form of the problem is

Maximize Z = 3A + 4B +C + 0s1 +0s2

subject to the constraints

8A + 3B + 4 C + s1 = 147

2A + 6B + C+ s2 = 63

A,B.C, s1,s2 >=0

where s1 and s2 are slack variables.

b)The solution of the problem using simplex method is given below

Table #1
==========================================================
Basis A B C s1 s2 solution
==========================================================
Z -3 -4 -1 0 0 0

s1 8 3 4 1 0 147
s2 2 6 1 0 1 63   
  

Table #2
==========================================================
Basis A B C s1 s2 solution
=========================================================

Z -5/3 0 -1/3 0 2/3 42   

s1 7 0 7/2 1 -1/2 231/2
B 1/3 1 1/6 0 1/6 21/2

Table #3( Optimal Table)

==========================================================
Basis A B C s1 s2 solution   
==========================================================
Z 0 0 1/2 5/21 23/42 139/2

A 1 0 1/2 1/7 -1/14 33/2   
B 0 1 0 -1/21 4/21 5

c) The decision variables A,B,C are expressed in number (quanity)of items of products A,B and C.

Maximum level of profit is Z = 139/2.

d) The optimality of the solution can be identified from the Z row of each table. If all tvalues in the Z row are > or = to zero, then the solution is optimal.

In this problem, from table3 , all Z values are >= zero. Hence the optimality of the solution.

The product levels that maximizes the profit are A = 33/2; B =5; C=0

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