5320 Assignment

Task 1: Scottsville Textile Mill Scheduling (1.1) Formulate the problem Defining the decision variables: M1D =Number of Fabric 1 make by Dobbie loom M2D = Number of Fabric 2 make by Dobbie loom M3D = Number of Fabric 3 make by Dobbie loom M3R = Number of Fabric 3 make by Regular loom M4D = Number of Fabric 4 make by Dobbie loom M4R = Number of Fabric 4 make by Regular loom M5D = Number of Fabric 5 make by Dobbie loom M5R = Number of Fabric 5 make by Regular loom B1 = Number of Fabric 1 purchased B2 = Number of Fabric 2 purchased B3= Number of Fabric 3 purchased B4= Number of Fabric 4 purchased B5= Number of Fabric 5 purchased (1.2)&(1.3) optimal production schedule Profits from make: Fabric Selling price Variable cost ($/yard) ($/yard) 1 0.99 (0.66) 2 0.86 (0.55) 3 1.10 (0.49) 4 1.24 (0.51) 5 0.7 (0.5) Profits from buy: Fabric Selling price Purchase price ($/yard) ($/yard)

Profit ($/yard) 0.33 0.31 0.61 0.73 0.2 Profit ($/yard)

1 0.99 0.8 0.19 2 0.86 0.7 0.16 3 1.10 0.6 0.50 4 1.24 0.7 0.54 5 0.70 0.7 0 Production times in hours per yard: Fabric Dobbie Regular (hours/yard) (hours/yard) 1 1/4.63=0.215983 2 1/4.63=0.215983 3 1/5.23=0.191205 1/5.23=0.191205 4 1/5.23=0.191205 1/5.23=0.191205 5 1/4.17=0.239808 1/4.17=0.239808 1

Defining the constraints Demand constraints M1D+ B1=16500 M2D + B2=22000 M3D +M3R +B3=62000 M4D +M4R +B4=7500 M5D +M5R +B5=62000 Loom constraints Dobbie: 0.215983M1D + 0.215983M2D + 0.191205M3D + 0.191205M4D + 0.239808M5D ≤5760 (Dobbie hours available: 8Loom *30 days *24 hour/day =5760) Regular: 0.191205M3R + 0.191205M4R + 0.239808M5R ≤ 21600 (Regular hours available: 30Loom* 30days* 24hours/day =21600) Nonnegativity conditions M1D ,M2D ,M3D ,M3R ,M4D ,M4R ,M5D ,M5R ,B1 ,B2 ,B3 ,B4 ,B5 ≥ 0 (1.3) after entering all appropriate parameters and choosing any necessary options for model, the solver determined that the optimal amount for the final production schedule is: Fabric Make by

1

2

3

4

5

Dobbie

4668.8

22000

0

0

0

Regular

0

0

27707.808

7500

62000

11831.2

0

34292.192

0

0

Fabric purchase

Fabric1 purchase 11831.2 yards Fabric1 Dobbie 4668.8 yards Fabric2 purchase 0 Fabric2 Dobbie 22000 yards Fabric3 purchase 34292.192 yards Fabric3 Dobbie 0 Fabric3 regular 27707.808 Fabric4 purchase 0 Fabric4 Dobbie 0 ; Fabric4 regular 7500 yards Fabric5 purchase 0 Fabric5 Dobbie 0 Fabric5 regular 62000 yards assignment for each fabric  loom assignment for fabric 1: 2

Dobbie: 4668.8/4.63= 1008.3801  loom assignment for fabric 2: Dobbie: 22000/4.63= 4751.6199  loom assignment for fabric 3: Regular: 27707.808/5.23=5297.8601  loom assignment for fabric 4: regular: 7500/5.23= 1434.0344  loom assignment for fabric 5: regular: 62000/4.17= 14868.10552 (1.4)Maximize profit objective function: =0.33M1D+0.31M2D+0.61M3D+0.61M3R+0.73M4D+0.73M4R+0.2M5D+0.2M5R+0.19B1+0 .16B2+0.50 B3+0.54B4+0B5 the objective in an optimization problem is represented mathematically by an objective function, the objective function in Scottsville Textile Mill Scheduling problem expresses the mathematical relationship between the decision variable(such as the amount of the each fabric should be produced) in the model to determined the maximize profit of the production. As the solve determined that the maximize profit is $ 62531.49, and the profit from make is equal to $43137.47, profit from buy is equal to $ 19394.02. (1.5) Alternative way - Cost Minimizing Costs from make: Costs from buy: Fabric Variable cost ($/yard) 1 (0.66) 2 (0.55) 3 (0.49) 4 (0.51) 5 (0.5) Fabric Purchase price ($/yard) 1 0.8 2 0.7 3 0.6 4 0.7 5 0.7 Minimize cost objective function: =0.66M1D+0.55M2D+0.49M3D+0.49M3R+0.51M4D+0.51M4R+0.5M5D+0.5M5R+0.8B1+0. 7B2+ 0.6B3+0.7B4+0.7B5 The minimizing cost value is 93623.5. Above solution shows that production quantity for all fabric types are same in Profit 3

maximization and Cost minimization model. therefore, both the profit maximization and cost minimization models will provide the same optimal solution. However, the coefficients of the objective function are different for the two model. In the cost minimization model, the coefficients are costs per unit, therefore, the solutions allows company to fill the order at a minimum cost of $ 93623.5. on the other hands, in the profit maximization model, the coefficients are profit contributions per unit from make or buy. Task 2:　 The MasterDebt Lockbox (2.1) Formulate the problem: First we must calculate the losses due to lost interest for each possible assignment. Defining the decision variables: If the central sends to Sacramento, then on average there will be 4* 45= 180 (in $1,000s) in process on any given day, interest rate of 15%, this corresponds to a yearly loss of 27 (in $1,000s) then We can calculate the losses for the other possibilities in a similar fashion to get follow table. Annual Interest Loss($1,000s) Sacramento Denver Chicago Dallas New York Atlanta Central 27.00 13.50 13.50 13.50 20.25 20.25 Mid-Atlantic 58.50 39.00 29.25 39.00 19.50 19.50 Midwest 22.50 15.00 22.50 15.00 37.50 30.00 Northeast 81.00 54.00 27.00 67.50 27.00 40.50 Northwest 21.00 31.50 52.50 42.00 63.00 73.50 Southeast 84.00 48.00 36.00 24.00 48.00 24.00 Southwest 18.00 27.00 54.00 18.00 63.00 54.00 Let Xij=be the decision to select or unselect check sent from the region i to lockbox j, where i=1,2,3,4,5,6,7;j=1,2,3,4,5,6.

Let Yi= be the decision to select or unselected lockbox j , where i= Sacramento(S), Denver (D), Chicago(C), Dallas(Da), New York(NY), Atlanta(A).

Minimize total yearly loss objective function: =27X11+13.5X12+13.5X13+13.5X14+20.25X15+20.25X16…….. +25YS+30YD+35YC+35YDA+30YNY+35YA MIN

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where RI CI (2.2)&(2.3) Optimal solution Defining the constraints: Binary constraints: All Xij must be binary All YI must be binary All Xij must be integer Nonnegative condition: X11+X12+X13+X14+X15+X16 =1 X21+ X22+X23+X24+X25+X26 =1 X31+X32+X33+X34+X35+X36=1 X41+X42+X43+X44+X45+X46=1 X51+X52+X53+X54+X55+X56=1 X61+X62+X63+X64+X65+X66=1 Or, Linking constrain( with “big M”) X11+X21+X31+X41+X51+X61+X71≤7YS OR X11+X21+X31+X41+X51+X61+X71-7Ys≤0 X12+X22+X32+X42+X52+X62+X72≤7YD OR X12+X22+X32+X42+X52+X62+X72-7YD≤0 X13+X23+X33+X43+X53+X63+X73≤7YC OR X13+X23+X33+X43+X53+X63+X73-7YC≤0 X14+X24+X34+X44+X54+X64+X74≤7YDA OR X14+X24+X34+X44+X54+X64+X74-7YDA≤0 X15+X25+X35+X45+X55+X65+X75≤7YNY OR X15+X25+X35+X45+X55+X65+X75-7YNY≤0 X16+X26+X36+X46+X56+X66+X76≤7YA OR X16+X26+X36+X46+X56+X66+X76-7YA≤0

(2.4) The total costs are 224,000 and MasterDebt should use Dallas for Central, Midwest, Northwest, Southeast and Southwest while New York for Mid-Atlantic and Northeast. (2.5) The total costs are 227,500 when a maximum of four regions could be assigned to any lockbox location and MaterDebt should use Sacramento for Midwest, Northwest and Southwest while Atlanta for Central, Northeast, Southeast, MidAtlantic. Task 3: Brief Report The objectives of the assignment are to complete two problems with three tasks. We have established a schedule since week 3 and made regular meeting time at three times a week which was right after the class on Monday, Tuesday and the other one was held during weekend with two hours each time discussions. Based on the exercises and knowledge learnt from lectures and tutorials, we solved the problems 5

with Excel together during the meeting and discussed how to put them into the Word. In conclusion, our group has been effective at collaborating on this assignment and successful at developing a plan while each member has been following through with their responsibilities.

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