Quantitative methods for Busimess(Linear Programming)

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July 24th 2007, 03:22 PM
lillian

Quantitative methods for Busimess(Linear Programming)

:(
The NEC Company makes two cheese spreads by blending mild cheddar cheese with extra sharp cheddar cheese. The cheese spreads are packaged in 12-ounce containers, which are then sold to distributors. The regular blend contains 75% mild cheddar and 25% extra sharp, and the Zesty blend contains 55% mild cheddar and 45% extra sharp. A local dairy company offers to provide up to 8000 pounds of mild cheddar cheese for $1.30 per pound and up to 3500 pounds of extra sharp cheddar for $1.50 per pound. The cost to blend and package the cheese spreads, excluding the cost of cheese, is $0.30 per container. If each container of Regular is sold at $2.15 and each container of Zesty is sold for $2.35

(a) how many containers of Regular and Zesty should NEC produce?
(b) What is the optimal profit?
August 4th 2007, 01:32 AM
CaptainBlack

Quote:

Originally Posted by lillian

:(
The NEC Company makes two cheese spreads by blending mild cheddar cheese with extra sharp cheddar cheese. The cheese spreads are packaged in 12-ounce containers, which are then sold to distributors. The regular blend contains 75% mild cheddar and 25% extra sharp, and the Zesty blend contains 55% mild cheddar and 45% extra sharp. A local dairy company offers to provide up to 8000 pounds of mild cheddar cheese for $1.30 per pound and up to 3500 pounds of extra sharp cheddar for $1.50 per pound. The cost to blend and package the cheese spreads, excluding the cost of cheese, is $0.30 per container. If each container of Regular is sold at $2.15 and each container of Zesty is sold for $2.35

(a) how many containers of Regular and Zesty should NEC produce?
(b) What is the optimal profit?

First we write down the cost of producing $r$ units of regular and $z$ units of zesty.

$ C=(r+z) \times 0.30 + \frac{12}{16} [(0.75\times 1.30+0.25 \times 1.5) \times r + (0.55\times 1.30+0.45 \times 1.5) \times z] $

or:

$ C=1.3425 z + 1.3125 r $

Now the revenue for the sale of $r$ units of mild and $z$ units of zesty is:

$ R=2.35 z+ 2.15 r $

So the profit is:

$ P=R-C=1.0075 z + 0.8375 r $

which is what we wish to maximise subject to the constraints:

$z, r \ge 0$

the amount of mild cheese used is less that or equal 8000 pounds:

$r\times \frac{12}{16}\times 0.75 + z\times \frac{12}{16}\times 0.55 \le 8000$

and the amount of extra sharp cheese used is less that or equal 3500 pounds:

$r\times \frac{12}{16}\times 0.25 + z\times \frac{12}{16}\times 0.45 \le 3500$

RonL
August 4th 2007, 03:35 AM
ticbol

Quote:

Originally Posted by lillian

:(
The NEC Company makes two cheese spreads by blending mild cheddar cheese with extra sharp cheddar cheese. The cheese spreads are packaged in 12-ounce containers, which are then sold to distributors. The regular blend contains 75% mild cheddar and 25% extra sharp, and the Zesty blend contains 55% mild cheddar and 45% extra sharp. A local dairy company offers to provide up to 8000 pounds of mild cheddar cheese for $1.30 per pound and up to 3500 pounds of extra sharp cheddar for $1.50 per pound. The cost to blend and package the cheese spreads, excluding the cost of cheese, is $0.30 per container. If each container of Regular is sold at $2.15 and each container of Zesty is sold for $2.35

(a) how many containers of Regular and Zesty should NEC produce?
(b) What is the optimal profit?

Let me assume you know simple (not simplex) linear programming.

Regular spread, R = 75% Mild(M) and 25% Sharp(S)
Zesty spread, Z = 55% M and 45% S
So,
M = 0.75R +0.55Z
S = 0.25R +0.45Z

M <= 8000 lbs
S <= 3500 ibs
So,
0.75R +0.55Z <= 8000 -----------constraint (1)
0.25R +0.45Z <= 3500 -----------constraint (2)

Or, multiplying (1) and (2) by 20,
15R +11Z = 160,000 --------------(1a)
5R +9Z = 70,000 -----------------(2a)

R > 0, and Z > 0 -------------non-zero (because there should be R or Z), non-negative constraints.

Plot those on the same (R,Z) rectangular axes, in lbs.
Intersection of (1a) and (2a) is point (8375,3125)

The feasible region is the quadrilateral bounded by (1a), (2a) and the R and Z axes. The optimum point can only be (8375,3125).

--------------------------
a) how many containers of Regular and Zesty should NEC produce?

1 container = 12 oz.
1 pound = 16 oz.

Regular spread = (8375*16)/12 = 11,166.67 containers .............answer.
Zesty spread = (3125*16)/12 = 4,166.67 containers ................answer.

--------------------------
b) What is the optimal profit?

(b.1) Total Cost:
Mild cheddar ========> [0.75(8375) +0.55(3125)](1.30) = $10,400.
Sharp cheddar =======> [0.25(8375) +0.45(3125)](1.50) = $5,250.
Blending and packaging ==> [11,166.67 +4,166.67](0.30) = $4,600.
So, Total Cost = 10,400 +5250 +4600 = $20,250

(b.2) Total Revenue:
Regular spread ==> (11,166.67)(2.15) = $24,008.34
Zesty spread =====> (4,166.67)(2.35) = $9,791.67
So, Total Revenue = 24,008.34 +9,791.67 = $33,800.

Therefore, optimal profit = 33,800 -20,250 = $13,550 .......answer.
August 4th 2007, 05:08 AM
CaptainBlack

Quote:

Originally Posted by ticbol

Let me assume you know simple (not simplex) linear programming.

Regular spread, R = 75% Mild(M) and 25% Sharp(S)
Zesty spread, Z = 55% M and 45% S
So,
M = 0.75R +0.55Z
S = 0.25R +0.45Z

M <= 8000 lbs
S <= 3500 ibs
So,
0.75R +0.55Z <= 8000 -----------constraint (1)
0.25R +0.45Z <= 3500 -----------constraint (2)

Or, multiplying (1) and (2) by 20,
15R +11Z = 160,000 --------------(1a)
5R +9Z = 70,000 -----------------(2a)

R > 0, and Z > 0 -------------non-zero (because there should be R or Z), non-negative constraints.
.

We have been through this before, the way a linear programming
programming problem is formulated there should be no strict inequality
constraints. As I have explained this is because when a general method of
solution is presented or set up in some programming environment you cannot
guarantee the existance of a solution with strict inequality.

That is what would otherwise be the solution may not ne in the feasible
region.

It may make no difference here, but that is not relevant, the poster is
learing about linear programming, so we teach them techniques that
wont at a latter date turn and bite them. In fact we won't have to wait
for that later date, if the poster hands in an answer based on your method
they will lose marks even though the final answer ir right, because the method
is flawed.

If the objactive were changed so that its slope were greater than ~ -0.556, or
less than ~ -1.364 then given the way you have formulated the problem there
would be no optimum as the best points would have been excluded from the
feasible region. There would of course be integer points close by that did
better that your optimal point.

RonL
August 4th 2007, 05:37 AM
CaptainBlack

Quote:

Originally Posted by ticbol

Let me assume you know simple (not simplex) linear programming.

Regular spread, R = 75% Mild(M) and 25% Sharp(S)
Zesty spread, Z = 55% M and 45% S
So,
M = 0.75R +0.55Z
S = 0.25R +0.45Z

M <= 8000 lbs
S <= 3500 ibs
So,
0.75R +0.55Z <= 8000 -----------constraint (1)
0.25R +0.45Z <= 3500 -----------constraint (2)

Or, multiplying (1) and (2) by 20,
15R +11Z = 160,000 --------------(1a)
5R +9Z = 70,000 -----------------(2a)

R > 0, and Z > 0 -------------non-zero (because there should be R or Z), non-negative constraints.

Plot those on the same (R,Z) rectangular axes, in lbs.
Intersection of (1a) and (2a) is point (8375,3125)

The feasible region is the quadrilateral bounded by (1a), (2a) and the R and Z axes. The optimum point can only be (8375,3125).

Only because you have excluded either R or Z being zero. If they are allowed
to be zero then (10666.7,0) and (0,7777.8) are candidates (OK round down
to nearest integer), as they are corners of the feasible region.

(I wont worry about (0,0) as there is no profit there)

RonL
August 4th 2007, 10:58 PM
CaptainBlack

Quote:

Originally Posted by ticbol

Let me assume you know simple (not simplex) linear programming.

Regular spread, R = 75% Mild(M) and 25% Sharp(S)
Zesty spread, Z = 55% M and 45% S
So,
M = 0.75R +0.55Z
S = 0.25R +0.45Z

M <= 8000 lbs
S <= 3500 ibs
So,
0.75R +0.55Z <= 8000 -----------constraint (1)
0.25R +0.45Z <= 3500 -----------constraint (2)

Or, multiplying (1) and (2) by 20,
15R +11Z = 160,000 --------------(1a)
5R +9Z = 70,000 -----------------(2a)

R > 0, and Z > 0 -------------non-zero (because there should be R or Z), non-negative constraints.

Plot those on the same (R,Z) rectangular axes, in lbs.
Intersection of (1a) and (2a) is point (8375,3125)

The feasible region is the quadrilateral bounded by (1a), (2a) and the R and Z axes. The optimum point can only be (8375,3125).

--------------------------
a) how many containers of Regular and Zesty should NEC produce?

1 container = 12 oz.
1 pound = 16 oz.

Regular spread = (8375*16)/12 = 11,166.67 containers .............answer.
Zesty spread = (3125*16)/12 = 4,166.67 containers ................answer.

--------------------------
b) What is the optimal profit?

(b.1) Total Cost:
Mild cheddar ========> [0.75(8375) +0.55(3125)](1.30) = $10,400.
Sharp cheddar =======> [0.25(8375) +0.45(3125)](1.50) = $5,250.
Blending and packaging ==> [11,166.67 +4,166.67](0.30) = $4,600.
So, Total Cost = 10,400 +5250 +4600 = $20,250

(b.2) Total Revenue:
Regular spread ==> (11,166.67)(2.15) = $24,008.34
Zesty spread =====> (4,166.67)(2.35) = $9,791.67
So, Total Revenue = 24,008.34 +9,791.67 = $33,800.

Therefore, optimal profit = 33,800 -20,250 = $13,550 .......answer.

In fact since the number of containers must be an integer this profit cannot
be achived, the maximum profit occurs at a production 4167 containers of
zesty and 11166 containers of mild, and the corresponding profit is $13549.80

RonL