Please solve the below question using “Solver” function in excel. Also help me understand the steps to use it on my own.
ACME makes two types of printed products Black and White (A) and Glossy covers (B) for brochures each requiring time on each of two machines. These requirements, along with the limitations on the available labour-hours are given in the following table:
Product Type | Machine 1 | Machine 2 | Unit Profit |
A | 0.2 | 0.53 | $600 |
B | 0.4 | 0.3 | $500 |
Max labour availability
–hours per week |
1000 | 2400 |
A. Determine how many units of Product A and Product B must be produced each week to maximize the total profit. (fractional units allowed)
B. Suppose that a single order is received for 1600 units of Product A per week, and if it is decided that this order must be filled, determine the new values below (fractional units allowed):-
C. In addition to the above, the company estimates that at most 1900 units of Product can be handled by its distribution network each week. Determine the new values below (fractional units allowed):-
Expert Answer
1. In the question posted, there is no minimum requirement as to how much Product A or B should be manufactured. Here, unit profit rate of 1 unit of A and B is given which is $600 and $500 respectively and therefore it is benificial to produce A only, since no minimum obligation information is provided by qtn.
Each product goes thrrough Mach. A as well as B. We have been given the maximim machine hours available and in such case we shall first calculate the maximum units which can be produced of A.
Machine 1 | Machine 2 | |
Time/unit | 0.20 hours | 0.53 hours |
Availabe hours | 1000 | 2400 |
MAx units can be produced | 5000 | 4528.30 |
Thus 4528.30 rounded to 4529 units can be produced of Product A to maximise total profit which would be 4529 x 600 = $ 2717400
Answer to Part (b)
Units to be produced will be same as they are already producing A at its maximum capacity.
Answer to Part (c)
If only 1900 units of Product A is manufactured than in this case hours left can be used to produce B as follows. Accordingly, for Machine 1; 1000 – (1900 x 0.2) = 620 hours and for Machine 2; 2400-(1900 x 0.53) = 1393 hours are left from which maximum of 1550 units (620/.4) can be produced and total profit in this case would be 1900 x 600+1550 x 500 = $1915000