

Subject:
Linear Programing / Decision Making Question
Category: Science > Math Asked by: thanksmatega List Price: $100.00 
Posted:
18 Oct 2003 14:03 PDT
Expires: 17 Nov 2003 13:03 PST Question ID: 267510 
The "Mill" produces five different fabrics. Each fabric can be woven on one or more of the mill's 38 looms. The sales department has forecast demand for the next month. The demand data are shown in Table 1.0, along with data on the selling price per yard, variable cost per yard, and purchase price per yard. The mill operates 24 hours a day and is scheduled for 30 days during the coming month. The Mill has two types of looms: dobbie and regular. The dobbie looms are more versatile and can be used for all five fabrics. The regular looms can produce only three of the fabrics. The Mill has a total of 38 looms: 8 are dobbie and 30 are regular. The rate of production for each fabric on each type of loom is given in Table 1.1. The time to change over from producing one fabric to another is negligible and does not have to be considered. The Mill satisfies all demand with either its own fabric or fabric purchased from another mill. That is, fabrics that cannot be woven at The Mill because of limited loom capacity will be purchased from another mill. The purchase price of each fabric is also shown in Table 1.0. Question Develop a model that can be used to schedule production for The Mill, and at the same time, determine how many yards of each fabric must be purchased from another mill. Include a discussion and analysis of the following items in your answer: 1. The final production schedule and loom assignments for each fabric 2. The projected total contribution to profit 3. A discussion of the value of additional loom time (The Mill is considering purchasing a ninth dobbie loom. What is your estimate of the monthly profit contribution of this additional loom?) 4. A discussion of the objective coefficient ranges 5. A discussion of how the objective of minimizing total costs would provide a different model than the objective of maximizing total profit contribution: How would the interpretation of the objective coefficients ranges differ for these two models? Table 1.0 Monthly Demand, Selling Price, Variable Cost, and Purchase Price Data for The Mill Demand Selling Price Variable Cost Purchase Price Fabric (yards) ($/yard) ($/yard) ($/yard) 1 16,500 0.99 0.66 0.80 2 22,000 0.86 0.55 0.70 3 62,000 1.10 0.49 0.60 4 7,500 1.24 0.51 0.70 5 62,000 0.70 0.50 0.70 Table 1.1 Loom Production Rates for The Mill Loom Rate (yards/hour) Fabric Dobbie Regular 1 4.63 * 2 4.63 * 3 5.23 5.23 4 5.23 5.23 5 4.17 4.17 * Fabrics 1 and 2 can be manufactured only on the dobbie loom. Hello, I require the answer ASAP. There will be a generous tip depending on how long it takes you to answer this question: if it's answered within 3 days there will be a $US50 tip. If you have ANY questions... Thank you!  
 
 
 
 


Subject:
Re: Linear Programing / Decision Making Question
Answered By: answerguruga on 20 Oct 2003 10:05 PDT Rated: 
Hi thanksmatega, I have completed the analysis and discussion of the question you have posed. The analysis was done in Excel, but I have brought everything into text format including a description of all the variables used etc. Preliminary Definitions of Variables and Constraints: x1d – length of fabric 1 produced on dobbie looms x2d – length of fabric 2 produced on dobbie looms x3d – length of fabric 3 produced on dobbie looms x4d – length of fabric 4 produced on dobbie looms x5d – length of fabric 5 produced on dobbie looms x3r – length of fabric 3 produced on regular looms x4r – length of fabric 4 produced on regular looms x5r – length of fabric 5 produced on regular looms Fabric 1 Fabric 2 Fabric 3 Fabric 4 Fabric 5 Demand, D 16500 22000 62000 7500 62000 Selling price, s 0.99 0.86 1.1 1.24 0.7 Variable cost, v 0.66 0.55 0.49 0.51 0.5 Purchase price, p 0.8 0.7 0.6 0.7 0.7 x1m = quantity manufactured of fabric 1 = x1d x2m = quantity manufactured of fabric 2 = x1d x3m = quantity manufactured of fabric 3 = x3d + x3r x4m = quantity manufactured of fabric 4 = x4d + x4r x5m = quantity manufactured of fabric 5 = x5d + x5r x1p = quantity purchased of fabric 1 = D1 – x1m x2p = quantity purchased of fabric 2 = D2 – x2m x3p = quantity purchased of fabric 3 = D3 – x3m x4p = quantity purchased of fabric 4 = D4 – x4m x5p = quantity purchased of fabric 5 = D5 – x5m P1 = profit from fabric 1 = D1*s1 – xm1*v1 – xp1*p1 P2 to P5 are defined similarly P = P1 + P2 + P3 + P4 + P5 Tda = time available on 8 dobbie looms in a month = 8*24*30 = 5760 hours Tra = time available on 30 regular looms in a month = 30*24*30 = 21600 hours m1d = production rate of fabric 1 on dobbie looms = 4.63 yards/hour m2d = 4.63 yards/hour m3d = 5.23 yards/hour m4d = 5.23 yards/hour m5d = 4.17 yards/hour m3r = 5.23 yards/hour m4r = 5.23 yards/hour m5r = 4.17 yards/hour T1d = time required to produce x1d yards of fabric 1 on dobbie looms = x1d/m1d T2d to T5d and T3r to T5r are defined similarly T d = T1d + …… + T5d Tr = T3r + T4r + T5r Time constraints Td ≤ Tda Tr ≤ Tra Total manufactured quantity cannot be greater than demand xm1 ≤ D1 xm2 ≤ D2 xm3 ≤ D3 xm4 ≤ D4 xm5 ≤ D5 Quantity manufactured on each machine cannot be negative xd1 ≥ 0 Similarly for others The assignment is summarised in the table below Fabric 1 Fabric 2 Fabric 3 Fabric 4 Fabric 5 D 16500 22000 62000 7500 62000 xd 4668.8 22000 0 0 0 no of looms 1.4 6.6 0.0 0.0 0.0 xr NA NA 27707.81 7500 62000 no of looms 7.4 2.0 20.6 xm 4668.8 22000 27707.81 7500 62000 xp 11831.2 0 34292.19 0 0 P 3788.632 6820 34047.86 5475 12400 Total P for all fabrics = 62531.49 1. Assignment of looms 1 dobbie loom to make fabric 1 throughout the month 6 dobbie looms to make fabric 2 throughout the month 8th dobbie loom to make fabric 1 for 40% of the time and fabric 2 for 60% of the time 7 regular looms to make fabric 3 throughout the month 2 regular looms to make fabric 4 throughout the month 20 regular looms to make fabric 5 throughout the month 30th regular loom to make fabric 3 for 40% of the time and fabric 5 for 60% of the time 2. Contribution to profit The projected total contribution to profit is $ 62,531.49 per month 3. Additional loom time Fabric 1 and fabric 3 are being purchased at a price greater than the variable cost of manufacture. Hence it makes sense to install another loom. Since the difference between purchase prise and variable cost is greater for fabric 1 it would be prudent to allocate the new loom to fabric 1. The linear programme is run again using 9 dobbie looms. Tda would now becomes 6480 hours as a result of this change The additional dobbie loom will be used to make fabric 1 for all 30 days. A total of 8002.4 yards of fabric 1 will be produced, the additional quantity being 3333.6 yards. The additional contribution to profit will be $ 466.70 per month. 4. Objective coefficient ranges By expanding xm and xp the objective function can be rewritten as ∑ Di*(si – pi) + ∑xid*(pi – vi) + ∑xjr*(pj – vj) (where i varies from 1 to 5 and j from 3 to 5) ∑ Di*(si – pi) is a constant (pi – vi) & (pj – vj) are the objective coefficients. The range of the objective coefficient refers to the range in which the coefficient can be varied without changing the optimal solution. Consider the objective coefficient associated with x3d which is 0.11. If this is reduced to 0, then the optimal solution remains the same. If it is increased to 0.13, then the optimal solution changes. Hence the range of this objective coefficient is 0 to 0.12. A similar analysis can be done for other objective coefficients. 5. Minimising total costs The objective function consists of three parts. • Sales of the fabric • Cost of manufacture • Cost of purchase The sales quantity is constant for each fabric. The total demand is met either from purchase or from manufacture. As a result sales revenue remains constant. Hence maximising profit is same as minimising total cost (purchase plus manufacture). This is shown in Sheet 4 of lp.xls, where the same optimal solution is reached for minimising cost. Converting the objective function in Question 4 to a cost function ∑ Di*pi + ∑xid*(vi – pi) + ∑xjr*(vj – pj) (where i varies from 1 to 5 and j from 3 to 5) The optimal solution is actually the same in value but with a negative sign. The objective coefficient range has also changed to reflect the negative of the maximisation problem range. The range for this problem is 0 to –0.12. Hopefully the discussion and calculations are clear  if you have any trouble understanding the information above please do post a clarification. This was certainly an interesting problem to work on! Cheers! answerguruga  
 
 

thanksmatega
rated this answer:
and gave an additional tip of:
$60.00
Very well answered! Thank you! 

Subject:
Re: Linear Programing / Decision Making Question
From: answerguruga on 20 Oct 2003 12:34 PDT 
Thanks you very much for you kind words, rating, and generous tip! If you would me to answer this question specifically please do mention my name in your question (answerguruga). 
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