Julia’s Food Booth Case Problem
Julia’s Food Booth Case Problem
MAT 540- Quantitative Methods
February 23, 2013
(A) Formulate and solve an L.P. model for this case.
The following variables were be used: X1 = Slices of Pizza X2 = Hot Dogs X3 = BBQ Sandwiches
The objective is to maximize profit. maximize Z= 0 .75X1+1.05X2+1.35X3
Subject to:
0.75X1+1.05X2+1.35X3≤1,500 (Budget)
24X1+16X2+25X3≤55,296in2 (Oven Space)
X1≥X2+X3
X2X3≥2.0
X1, X2, X3≥0 (B) Evaluate the prospect of borrowing money before the first game.
Following table shows the sensitivity report from excel solver | | | | | | | | | | | | | | | | Food items: | | Pizza | Hot Dogs | Barbecue | | | | | Profit per item: | 0.75 | 1.05 | 1.35 | | | | | Constraints: | | | | Available | Usage | Left over | | Budget ($) | 0.75 | 0.45 | 0.90 | 1,500 | 1,500.00 | 0 | | Oven space (sq. in.) | 24 | 16 | 25 | 55,296 | 50,000.00 | 5296 | | Demand | 1 | -1 | -1 | 0 | - | 0 | | Demand | 0 | 1 | -2 | 0 | 1,250.00 | -1250 | | | | | | | | | | | Stock | | | | | | | | | Pizza= | 1250 | slices | | | | | | | Hot Dogs= | 1250 | hot dogs | | | | | | | Barbecue= | 0 | sandwiches | | | | | | | Profit= | 2,250.00 | | | | | | | | | | | | | | | | |
| | | | | | | | Adjustable Cells | | | | | | | | | Final | Reduced | Objective | Allowable | Allowable | | Cell | Name | Value | Cost | Coefficient | Increase | Decrease | | $B$12 | Pizza= | 1250 | 0 | 0.75 | 1 | 1.00 | | $B$13 | Hot Dogs= | 1250 | 0 | 1.05 | 1E+30 | 0.27 | | $B$14 | Barbecue= | 0 | 0 | 1.35 | 0.375000011 | 1E+30 | | | | | | | | | Constraints | | | | | | | | | Final | Shadow | Constraint | Allowable | Allowable | | Cell | Name | Value | Price | R.H. Side | Increase | Decrease | | $G$9 | Demand Usage | 1,250.00 | - | 0 | 1250 | 1E+30 | |
MAT 540- Quantitative Methods
February 23, 2013
(A) Formulate and solve an L.P. model for this case.
The following variables were be used: X1 = Slices of Pizza X2 = Hot Dogs X3 = BBQ Sandwiches
The objective is to maximize profit. maximize Z= 0 .75X1+1.05X2+1.35X3
Subject to:
0.75X1+1.05X2+1.35X3≤1,500 (Budget)
24X1+16X2+25X3≤55,296in2 (Oven Space)
X1≥X2+X3
X2X3≥2.0
X1, X2, X3≥0 (B) Evaluate the prospect of borrowing money before the first game.
Following table shows the sensitivity report from excel solver | | | | | | | | | | | | | | | | Food items: | | Pizza | Hot Dogs | Barbecue | | | | | Profit per item: | 0.75 | 1.05 | 1.35 | | | | | Constraints: | | | | Available | Usage | Left over | | Budget ($) | 0.75 | 0.45 | 0.90 | 1,500 | 1,500.00 | 0 | | Oven space (sq. in.) | 24 | 16 | 25 | 55,296 | 50,000.00 | 5296 | | Demand | 1 | -1 | -1 | 0 | - | 0 | | Demand | 0 | 1 | -2 | 0 | 1,250.00 | -1250 | | | | | | | | | | | Stock | | | | | | | | | Pizza= | 1250 | slices | | | | | | | Hot Dogs= | 1250 | hot dogs | | | | | | | Barbecue= | 0 | sandwiches | | | | | | | Profit= | 2,250.00 | | | | | | | | | | | | | | | | |
| | | | | | | | Adjustable Cells | | | | | | | | | Final | Reduced | Objective | Allowable | Allowable | | Cell | Name | Value | Cost | Coefficient | Increase | Decrease | | $B$12 | Pizza= | 1250 | 0 | 0.75 | 1 | 1.00 | | $B$13 | Hot Dogs= | 1250 | 0 | 1.05 | 1E+30 | 0.27 | | $B$14 | Barbecue= | 0 | 0 | 1.35 | 0.375000011 | 1E+30 | | | | | | | | | Constraints | | | | | | | | | Final | Shadow | Constraint | Allowable | Allowable | | Cell | Name | Value | Price | R.H. Side | Increase | Decrease | | $G$9 | Demand Usage | 1,250.00 | - | 0 | 1250 | 1E+30 | |
References: Taylor, B. M. (2010). Introduction to management science (11th ed.). Upper Saddle River, NJ: Pearson/Prentice Hall.