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Math Help - Tricky LP assignment!!!

  1. #1
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    Tricky LP assignment!!!

    Hi all,

    Im a final year student needing a bit of help with the following lp assignment. Im a bit rusty at linear programming and am find it a bit confusing developing the layout and constraints for the following problem below. Any help would be most appreciated. Thanks!

    The problem:

    Katherine Rally is meeting with Ted Lawson, the production manager, to decide upon next monthís production plan for the Autumn line of clothes. Specifically, she must determine the quantity of each clothing item to produce given the plantís capacity, limited resources, and demand forecasts. Accurate planning for next monthís production is critical to Autumn sales since the items produced next month will appear in stores during September and women generally buy the majority of the Autumn fashions when they appear in September.

    She studies the clothing patterns and material requirements. Her Autumn lines consist of both professional and casual fashions. In her analysis of costs she has established the following table of information concerning the Autumn professional fashions

    Clothing Item / Material Requirements / Price(£) / Labour and Machine Costs (£)


    Tailored wool slacks / 3 yards of wool, 2 yards of acetate / 300 / 160
    Cashmere sweater / 1.5 yards of cashmere / 450 / 150
    Silk blouse / 1.5 yards of silk / 180 / 100
    Silk camisole / 0.5 yards of silk / 120 / 60
    Tailored skirt / 2 yards of rayon, 1.5 yards of acetate / 270 / 120
    Wool blazer / 2.5 yards of wool, 1.5 yards of acetate / 320 / 140


    The Autumn casual fashions include:

    Velvet pants / 3 yards of velvet, 2 yards of acetate / 350 / 175
    Cotton sweater / 1.5 yards of cotton / 130 / 60
    Cotton mini-skirt / 0.5 yards of cotton / 75 / 40
    Velvet shirt / 1.5 yards of velvet / 200 / 160
    Button-down blouse / 1.5 yards of rayon / 120 / 90

    She knows that for the next month she has ordered 45,000 yards of wool, 28,000 yards of acetate, 9,000 yards of cashmere, 18,000 yards of silk, 30,000 yards of rayon, 20,000 yards of velvet and 30,000 yards of cotton for production. The prices of the materials are listed below.


    Material / Price per yard (£)


    Wool / 9.00
    Acetate / 1.50
    Cashmere / 60.00
    Silk / 13.00
    Rayon / 2.25
    Velvet / 35.00
    Cotton / 2.50


    Any material not used in production can be sent back to the textile wholesaler for a full refund, although scrap material cannot be sent back to the wholesaler.

    She knows that the production of the silk blouse and cotton sweater results in scrap material. Specifically, for the production of one silk blouse or one cotton sweater, 2 yards of silk and cotton, respectively are needed. From these 2 yards, in both cases 1.5 yards is used in the clothe item and 0.5 yards is scrap material. She does not want to waste the material, so she plans to use the rectangular scrap of silk or cotton to produce a silk camisole or cotton mini-skirt respectively. Thus whenever a silk blouse is produced a silk camisole is produced and whenever a cotton sweater is produced, a cotton mini-skirt is produced. Note that it is possible to produce a silk camisole without producing a silk blouse and a cotton mini-skirt without producing a cotton sweater.

    Some items have limited demand. A maximum of 5,500 pairs of velvet pants and 6,000 velvet shirts can be sold. The company does not want to produce more than the demand since the items go out of fashion and cannot be sold next year. The cashmere sweater also has limited demand and it is estimated that no more than 4,000 cashmere sweaters can be sold. Similarly, there is limited demand for the silk blouse and camisoles. Specifically, at most 12,000 silk blouses and 15,000 silk camisoles can be sold.

    The demand forecasts also indicate that the wool slacks, tailored skirts and wool blazers have a great demand. There is demand for up to 7,000 pairs of wool slacks and 5,000 wool blazers. Katherine wants to meet at least 60% of demand for these two items to maintain her loyal customer base. The demand for tailored skirts is estimated to be 2,800.

    Formulate and solve a linear programming problem maximizing profit given the production, resource and demand constraints.
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  2. #2
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    i am studying OR in this course too, my exam is soon , so I tried to solve this one
    but I am not sure 100 %
    Formulate and solve a linear programming problem maximizing profit given the production, resource and demand constraints.
    First we need to formulate the objective function which needs to calculate the profit for example:
    In order to calculate the profit for Tailored wool slacks = 300 – cost
    Cost = labor and Machine Costs+( Material Requirements *price per yard)
    Cost for Tailored wool slacks=160+27+3=190

    Profit for Tailored wool slacks = 300 – 190=110

    …. and we do the same thing with all clothing items
    (p.s. price per yard for velvet = 12 not 35, right?)

    the model
    X1 =number of items produced of Tailored wool slacks,
    X2= Cashmere sweater, X3= Silk blouse
    X4=Silk camisole, X5=Tailored skirt, X6=Wool blazer
    X7=Velvet pants , X8= Cotton sweater X9=Cotton mini-skirt
    X10=Velvet shirt X11=Button-down blouse


    Objective Function =
    Max (110 X1+291 X2+60.5 X3+53.5 X4+143.25 X5+155.25 X6+67 X7+66.25 X8+33.75 X9+22 X10+26.63 X11)

    Subject To
    resource constraints
    3 X1+2.5 X6 <=45000 (wool)
    2 X1+1.5 X5+1.5 X6+2 X7 <= 28000 (acetate)
    1.5 X <= 9000 (cashmere)
    1.5 X3+0.5 X4 <= 18000 (silk)
    2 X5+1.5 X11 <= 30000 (rayon)
    3 X7+1.5 X10 <= 20000 (velvet)
    1.5 X5+0.5 X9 <= 30000 (cotton)

    Demand constraint
    X7<=5500 (velvet pants)
    X10<=6000 (velvet shirts)
    X2<=4000 (cashmere sweater)
    X3<=12000 (silk blouse)
    X4<=15000 (silk camisole)
    X1>=4200 (meet at least 60% of demand for to 7,000 pairs of wool slacks)
    X7>=3000 (meet at least 60% of demand for to 5,000 wool blazers)
    X5<=2800 (Tailored skirt)

    <br /> <br />
xi \geq 0    <br />
\ , i=1..11<br />

    Thus whenever a silk blouse is produced a silk camisole is produced and whenever a cotton sweater is produced, a cotton mini-skirt is produced. Note that it is possible to produce a silk camisole without producing a silk blouse and a cotton mini-skirt without producing a cotton sweater.
    I could not formulate this constraint
    I hope someone can help us here and also if my solution is correct

    This is only try , I hope it will help you
    Last edited by Logical; January 14th 2009 at 12:42 PM. Reason: adding non negative constraint+trying latex :S
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  3. #3
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    Hi mate, i've taken a look at the way you've gone about solving this problem, but I have to solve it using a software called Ampl. Where did u get this problem from???

    Regards

    ASA
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  4. #4
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    Thus whenever a silk blouse is produced a silk camisole is produced and whenever a cotton sweater is produced, a cotton mini-skirt is produced. Note that it is possible to produce a silk camisole without producing a silk blouse and a cotton mini-skirt without producing a cotton sweater.
    I thought of this
    I think constraint should be
    X4>=X3
    X9>=X8

    Where did u get this problem from???
    the question you mean ? from here
    the solution is just try from me

    I use WINQSB to solve it, it is easy .
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  5. #5
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    Hi Logical,

    This is my final solution for the problem maximising the profit i get

    IP OPTIMAL SOLUTION, Objective = 54829034.24999999


    DECISION.VARIABLES

    Variable / Activity / U bound
    __________________________________________________ ____________________________
    1 "prodamount['tailored_wool_slacks']" / 4484 / 7000
    2 "prodamount['cash_sweater']" / 4000 / 4000
    3 "prodamount['silk_blouse']" / 7000 / 9000
    4 "prodamount['silk_camisole']" / 8000 / 15000
    5 "prodamount['tailored_skirt']" / 2800 / 2800
    6 "prodamount['wool_blazer']" / 5000 / 5000
    7 "prodamount['velvet_pants']" / 3666 / 5500
    9 "prodamount['cotton_mini_skirt']" / 60000 / 60000
    10 "prodamount['velvet_shirt']" / 6000 / 6000
    11 "prodamount['button_down_blouse']" / 16266 / 20000
    12 "prodamount['free_camisole']" / 7000 / 9000
    ___________________

    The number relating to the activity is how much of each product should be produced and the number corresponding to the ubound is the "upper bound" which is the max amount that could be produced. Therefore in order to maximise the profits, you would produce the number stated as the activity. As you can see my max profit taking all parameters into account is the figure above.

    Did you by anychance get something around this number???

    Thanks

    ASA
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  6. #6
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    did you use same model I suggest ?
    no my objective function was different than urs
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  7. #7
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    yes i did it just like yours. A few people doing this problem have the same result that i got, but I just wanted to confirm with you.
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  8. #8
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    Quote Originally Posted by asa16 View Post
    yes i did it just like yours. A few people doing this problem have the same result that i got, but I just wanted to confirm with you.
    hi asa,i'm also doing this assignment! but i got different results.
    can you help me with it?
    Last edited by mr fantastic; January 21st 2009 at 11:21 PM. Reason: Deleted email address.
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  9. #9
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    Quote Originally Posted by asa16 View Post
    Hi all,

    Im a final year student needing a bit of help with the following lp assignment. Im a bit rusty at linear programming and am find it a bit confusing developing the layout and constraints for the following problem below. Any help would be most appreciated. Thanks!

    The problem:

    Katherine Rally is meeting with Ted Lawson, the production manager, to decide upon next monthís production plan for the Autumn line of clothes. Specifically, she must determine the quantity of each clothing item to produce given the plantís capacity, limited resources, and demand forecasts. Accurate planning for next monthís production is critical to Autumn sales since the items produced next month will appear in stores during September and women generally buy the majority of the Autumn fashions when they appear in September.

    She studies the clothing patterns and material requirements. Her Autumn lines consist of both professional and casual fashions. In her analysis of costs she has established the following table of information concerning the Autumn professional fashions

    Clothing Item / Material Requirements / Price(£) / Labour and Machine Costs (£)


    Tailored wool slacks / 3 yards of wool, 2 yards of acetate / 300 / 160
    Cashmere sweater / 1.5 yards of cashmere / 450 / 150
    Silk blouse / 1.5 yards of silk / 180 / 100
    Silk camisole / 0.5 yards of silk / 120 / 60
    Tailored skirt / 2 yards of rayon, 1.5 yards of acetate / 270 / 120
    Wool blazer / 2.5 yards of wool, 1.5 yards of acetate / 320 / 140


    The Autumn casual fashions include:

    Velvet pants / 3 yards of velvet, 2 yards of acetate / 350 / 175
    Cotton sweater / 1.5 yards of cotton / 130 / 60
    Cotton mini-skirt / 0.5 yards of cotton / 75 / 40
    Velvet shirt / 1.5 yards of velvet / 200 / 160
    Button-down blouse / 1.5 yards of rayon / 120 / 90

    She knows that for the next month she has ordered 45,000 yards of wool, 28,000 yards of acetate, 9,000 yards of cashmere, 18,000 yards of silk, 30,000 yards of rayon, 20,000 yards of velvet and 30,000 yards of cotton for production. The prices of the materials are listed below.


    Material / Price per yard (£)


    Wool / 9.00
    Acetate / 1.50
    Cashmere / 60.00
    Silk / 13.00
    Rayon / 2.25
    Velvet / 35.00
    Cotton / 2.50


    Any material not used in production can be sent back to the textile wholesaler for a full refund, although scrap material cannot be sent back to the wholesaler.

    She knows that the production of the silk blouse and cotton sweater results in scrap material. Specifically, for the production of one silk blouse or one cotton sweater, 2 yards of silk and cotton, respectively are needed. From these 2 yards, in both cases 1.5 yards is used in the clothe item and 0.5 yards is scrap material. She does not want to waste the material, so she plans to use the rectangular scrap of silk or cotton to produce a silk camisole or cotton mini-skirt respectively. Thus whenever a silk blouse is produced a silk camisole is produced and whenever a cotton sweater is produced, a cotton mini-skirt is produced. Note that it is possible to produce a silk camisole without producing a silk blouse and a cotton mini-skirt without producing a cotton sweater.

    Some items have limited demand. A maximum of 5,500 pairs of velvet pants and 6,000 velvet shirts can be sold. The company does not want to produce more than the demand since the items go out of fashion and cannot be sold next year. The cashmere sweater also has limited demand and it is estimated that no more than 4,000 cashmere sweaters can be sold. Similarly, there is limited demand for the silk blouse and camisoles. Specifically, at most 12,000 silk blouses and 15,000 silk camisoles can be sold.

    The demand forecasts also indicate that the wool slacks, tailored skirts and wool blazers have a great demand. There is demand for up to 7,000 pairs of wool slacks and 5,000 wool blazers. Katherine wants to meet at least 60% of demand for these two items to maintain her loyal customer base. The demand for tailored skirts is estimated to be 2,800.

    Formulate and solve a linear programming problem maximizing profit given the production, resource and demand constraints.
    Judging from the posts in this thread, this looks like a graded assignment. MHF does not provide assistance in such circumstances.

    Thread closed.
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