Report Created: 2/24/2013 10:20:01 AMe Variable Cells the op Final Reduced Objec
ID: 357357 • Letter: R
Question
Report Created: 2/24/2013 10:20:01 AMe Variable Cells the op Final Reduced Objective AllowableAllowable lue Cost Coefficient IncreaseDecrease Cell Name SB$4 Product 1 0-10.2 0 1210.2 1E+30 18 SC$4 Product 2 SDS4 Product 3 480 2) 7.285714286 1.5 15 21 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease SE$6 Machine 160 ?wit? 10 56 40 12 160 288 4.2 17.5 0 0.6 $E$9 Product 2 4 0 SES7 Labor SE$8 Materials 200 1E+30 200 13.33333333 1E+30 232 16 Zerb axe a. What are the optimal production quantities and the optimal profit? 742 >+eL are lop b. which variables are basc and which are non-basic? A new competitor entered the market providing product 2 with a competitive price, the manager decided to decrease the profit contribution of product 2 by 8. What is the effect of the manager decision? they need the optinal Chornc ad tc d. If the manager wants to produce product 1, what change should be made to the efe objective function? e. Calculate the range of optimality of product 3 and interpret the result. f. Which constraints have slack/surplus? What are their values? g. Which one of the three products will have the highest increase in profit for each dollar increase in profit contribution?Explanation / Answer
a) Optimal solution is Final value of variables
Product 1 = 0
Product 2 = 4
Product 3 = 48
Objective Functions coffecient are
Product 1 = 12
Product 2 = 18
Product 3 = 15
Objective Function = 12* quantity of Product 1 + 18*quantity of Product 2 + 15* quantity of Product 3
At optimal solution
Optimal Profit = 12*0 + 18*4 + 15*48 = 792
b) Basic variables - The variables with optimal solution of non zero values
Non Basic variables - Variables with zero final value in optimal solution
In this example ,
Basic Variable - Product 2 & Product 3
Non Basic variables - Product 1
c) Profit contribution = Objective coefficient values of Variables
Currently for Product 2, Profit contribution = 18
Allowable decrease for Objective coefficient i.e Profit contribution is 7.285714
That means if we decrease in profit contibution of product 2 less than 7.285714 , Optimal solution will be same
Product 1 = 0
Product 2 = 4
Product 3 = 48
Any decrease by value above 7.285714 will change the optimal solution
As 8 > 7.285714 so optimal solution will change
d) Allowable increase of objective coefficient of product 1 is 10.2
so increase of profit contribution for product 1 less than will not change the existing optimal solution and product 1 will be not produced. To include the product 1 in optimal solution , profit contribution of product 1 should be equal or greater than 10.2
Increase of profit coefficent for product 1 >= 10.2
e ) Range of optimality of Product 3
Allowable increase = 21
Allowable decrease = 1.5
Current value of objective coefficeint of product 3 = 15
So range (15+21,15-2.1) excluding the two value
(12.9 < Objective coefficent for product 3 <36) will not chnage the optimal solution
f) Slack/Surplus =Constraint RH Side - Final value of Constraint
Machine = 160-160 = 0
Labor = 288-232 =56
Materials = 200- 200
Product 2= 16-4 =12
So constraint with non zero value has slack/surplus
g) With in Allowable increase for all the product , optimal solution will remain same. so Product 3 will be highest number = 48
so with 1 unit increase in profit co-efficient profit will increase by 48
For Product 2, 1 unit increase in profit co-efficient profit will increase by 4
For Product 1, 1 unit increase in profit co-efficient profit will increase by 0
Answer is Product 3 if increase is within allowable increase