A. Explain the value of the reduced cost (?0.3) for the number of plugs to produ

Question

A. Explain the value of the reduced cost (?0.3) for the number of plugs to produce.

B. If the gross margin for rails is decreased to $1.05, can you predict what the optimal solution and profit will be?

C. Suppose that the gross margin for rivets is increased to $0.85. Can you predict what the optimal solution and profit will be?

D. If the gross margin for clips is reduced to $1.10, can you predict what the optimal solution and profit will be? What if the gross margin is reduced to $1.00?

E. Suppose that an additional 500 minutes of machine capacity is available. How will the optimal solution and profit change? What if planned maintenance reduces capacity by 300 minutes?

Example 13.31:

Explanation / Answer

A. Explain the value of the reduced cost (0.3) for the number of plugs to produce.

The reduce cost value in sensitivity report for a particular non-basis variable explains that if the profit margin of this variable is improved by that amount, the variable would be included in the optimal solution.

For the non-basis variable X1 the reduced cost of -0.3 explains that if the objective coefficient of the plugs is increased by $0.3, the variable would be included in the optimal solution. If the Gross margin/unit of the Plug is improved to 0.3-(-0.3) = $0.6 per unit, then it would be optimal to produce Plugs.

B. If the gross margin for rails is decreased to $1.05, can you predict what the optimal solution and profit will be?

The variable X2 is non-basis variable, that is , it is not optimal to produce any units of Rails due to its lower gross margin per unit. The reduced cost for the Rails is –0.2. To be included in the optimal solution, the gross margin per unit has to improve or increase by $0.2.

Thus, decreasing the gross margin of Rails will not have effect on the optimal solution, the optimal mix and profit will remain unchaged.

C. Suppose that the gross margin for rivets is increased to $0.85. Can you predict what the optimal solution and profit will be?

The variable X3 is non-basis variable that is; it is not optimal to produce any units of Rivets due to its lower gross margin per unit. The reduced cost for the Rails is –0.15. If the gross margin of Rivet is improved to or more than 0.75 – (-0.15) = $0.90, then it would be optimal to produce rivets.

As suggested gross margin of rivets is increased to $0.85 which is less than the required increase, it is not optimal to produce any units of Rivets. Thus, the optimal solution and profit will remain unchanged.

D. If the gross margin for clips is reduced to $1.10, can you predict what the optimal solution and profit will be? What if the gross margin is reduced to $1.00?

The variable X4 is basis variable and the allowable decrease in the profit margin is 0.16. If the gross margin of the clips is reduced by 0.16 the optimal solution will change, beyond this change it is not optimal to produce any unit of Clips.

Thus, suggested change in gross margin of Clips is 1.2-1.10 = 0.10, as the decrease in the margin is less than the allowable decrease, it would be still optimal to produce Clips. The Optimal Solution will remain same but optimal profit will reduce to $16800 – $0.10(140,000) = $t2800.

If the gross margin of Clips is reduced to $1.00, the suggested reduction is of $1.2 - $1.10 = 0.2 which is more than the allowable decrease, thus it would be not optimal to produce only Clips, the optimal solution will change.

E. Suppose that an additional 500 minutes of machine capacity is available. How will the optimal solution and profit change? What if planned maintenance reduces capacity by 300 minutes?

For present optimal solution, capacity constraint is binding constraint as complete capacity is utilized. The shadow price of the capacity constraint is 0.6, which means if additional unit of the binding constraint is made available or reduced the profit will change by $0.6. If the additional machine hours of 500 mins. are made available, the optimal profit will increase by 0.6*500 = $300 with out changing optimal solution.

If the machine hours is reduced by 300 mins., the profit will reduce by 0.6*300 = $180.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.