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!

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Request for Question Clarification by answerguru-ga on 18 Oct 2003 19:26 PDT

Hi there,

I was just wondering if a written description of the model would be
sufficient for you - since this forum doesn't currently have the
capabilities to transfer files etc. it may be difficult to achieve a
more robust representation of this model.

Please let me know what you have in mind :)

answerguru-ga

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Clarification of Question by thanksmate-ga on 19 Oct 2003 03:03 PDT

Hello,

A written description as in the linear progam model whereby the
decision variables, constraints and the objective function is defined
is fine.

It would also be great if you could copy and paste the output
(results) from whatever decision analysis software or spread sheet you
use to answer my question. And please discuss all the items listed in
the question.

Thanks for taking my question!

---

Clarification of Question by thanksmate-ga on 20 Oct 2003 03:41 PDT

Hi Answerguru,
Can you please give me an estimate as to when you'll have an answer?
Thanks.

---

Request for Question Clarification by answerguru-ga on 20 Oct 2003 06:59 PDT

I'm aiming to have it for you sometime today - thank you for your patience :)

answerguru-ga

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Clarification of Question by thanksmate-ga on 20 Oct 2003 09:59 PDT

That would be perfect!!

Will check back later :)

Hi thanksmate-ga,

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!

answerguru-ga

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Clarification of Answer by answerguru-ga on 20 Oct 2003 10:13 PDT

Hi again,

I just noticed that upon posting, some of the symbols didn't come out
right so here is a repost of those parts:

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 

<and further down in the answer...>

4. Objective coefficient ranges 
 
By expanding xm and xp the objective function can be rewritten as  
 
(Summation) Di*(si – pi) + (Summation)xid*(pi – vi) +
(Summation)xjr*(pj – vj)
 
(where i varies from 1 to 5 and j from 3 to 5) 
 
(Summation) Di*(si – pi) is a constant  
 
(pi – vi) & (pj – vj) are the objective coefficients. 

<then for question #5...>

Converting the objective function in Question 4 to a cost function 
 
(Summation) Di*pi + (Summation)xid*(vi – pi) + (Summation)xjr*(vj –
pj)
(where i varies from 1 to 5 and j from 3 to 5) 

Sorry about that :)

answerguru-ga

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Request for Answer Clarification by thanksmate-ga on 20 Oct 2003 11:56 PDT

Very well done!
I'm impressed!
100% for you (and an extra tip) and I would be delighted if you would
answer my next question which I will post tomorrow for a similar
price. I hope you have time. The topic of my next question will also
be "Linear Programing / Decision Making Question".
Thanks!

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Clarification of Answer by answerguru-ga on 20 Oct 2003 12:35 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 next question (answerguru-ga).

Very well answered! Thank you!

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 (answerguru-ga).