Question: Cell Name Final Value Reduced Cost Obj Coefficient Allowable Increase Allowable Decrease $B$10 Bowls= 24 0 40 26.67 15 $B$11 Mugs= 8 0 50 30
| Cell | Name | Final Value | Reduced Cost | Obj Coefficient | Allowable Increase | Allowable Decrease |
| $B$10 | Bowls= | 24 | 0 | 40 | 26.67 | 15 |
| $B$11 | Mugs= | 8 | 0 | 50 | 30 | 20 |
| Constraints | ||||||
| $E$6 | Labor(hr/unit)Usage | 40 | 16 | 40 | 40 | 10 |
| $E$7 | Clay(lb/unit)Usage | 120 | 6 | 120 | 40 | 60 |
I was not given the excel or the objective function
1. Assuming all other parameters remain unchanged, if the objective function coefficient associated with "Bowls" decreases by $8, what will be the change in the objective function? A. Increases by $200 B. Decreases by $200 C. Increases by $192 D. Decreases by $192
2. Assuming all other parameters remain unchanged, if the objective function coefficient associated with "Mugs" increases by $15, what will be the change in the objective function value? A. Increase by $185 B. Decrease by $200 C. Increases by $120 D. Decrease by $210
3. Assuming all other parameters remain unchanged, if the objective function coefficient associated with "Bowls" increases by $24, what will be the change in the optimal solution? A. The optimal number of "Bowls" decreases B. The optimal number of "Bowls" remains the same C. The optimal number of "Bowls" increases D. The optimal number of "Bowls" can either decrease or increases
4. Assuming all other parameters remain unchanged, what is the range of the objective function coefficient "Mugs" associated with for which the current optimal solution still remains optimal? A. Between 20 and 25 B. Between 15 and 40 C. Between 30 and 80 D. Between 40 and plus infinity
5. If the "Labor" constraint's right-hand side were 70, what would have been optimal profit of the Personal Mini Warehouses? A. $1,360-16*30=$1,360-$480=$880 B. $1,360+16*30=$1,360+$480=$1,840 C. $1,200+16*30-$1,680 D. $1,350-6*30=$870
6. If the "Clay" available wer only $80, wht would Personal Mini Warehouses' new optimal profit be? A. $3,800 $1,360-$6*40=$1,360-$240=$1,120 B. $3,700 $1,360+$6*40=$1,360+$240=$1,600 C. $3,500 $1,360-$16*40=$1,360-$640=$720 D. $3,600 $1,360+$16*40=$1,360+$640=$2,000
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