Solution of 9-12

Set Up of 9-12

Viewing the answers for 9-12 we can offer some suggest= ions for improvement.

In general, the key difficulty in this problem was the identification of the attribute table with Beef Pork and Lamb as the attrib= utes

Attribut= e Table

Ingredient	Beef	Pork	Lamb
Rib (R)	1	0	0
Shank (S)=	1	0	0
Tongue (T= )	1	0	0
Pork (P)<= /p>	0	1	0
Lamb (L)<= /p>	0	0	1

Final products are: German (G) and Italian (I) sausage= s.

1) Decision Variables. You need the three Ns.

In general students identified the Notations correctly= . However many provided an invalid Name. In LP, A Name must be d= efined quantitatively in a precise manner. The name “amount of rib in German sausage in unacceptable. The name “the amount in pounds of rib in German sausage” is proper.

The unit of measurement has been sometimes missing. E.= g., “U.O.M =3D One= pound of rib in a German s= ausage”.

2) The objective was identified typically correctly bu= t one must define the name of the objective and its UOM properly.

“ The total cost in dollars of producing German and Italian sausages&#= 8221; and

“U.O.M=3D = $1 of total cost dollar= s of producing German and It= alian sausages”

3) As to the constraints, see below, let us not miss t= he implicit constraints that restrict the ingredients.

These are easy group specialized resource constraints.= It should be noted that most respo= nding students have correctly reported the more difficult mixture constraints.

I should emphasize that when coming to complex constra= ints it is preferred to process them in detail. This way, even if one arrives at= the wrong Std. format, it is still possible to receive partial credit, perhaps = most of the credit, for correct translation of the constraint from its verbal representation to algebraic form, and for correct algebraic processing. It is not possible to receive the = credit if the steps are missing and the expression in the standard format is wrong= . While I do not necessary advocate = this lengthy approach, for every constraint, under time scarcity, I do recommend it in w= hen complex constraints are present and you are not sure of your processing.

Here is an example.

Step 1: Verbal expression of the constraint,

At least 60% of the German sausage is beef.

Step 2: A translation into a logically equivalent verb= al statement where the direction of the constraint is indicated in the middle = of the sentence,

The weight of beef in the German sausage is at least 60% of the= total weight of the German sausage. To save time you may want to express the idea more succinctly,

Beef in G at least 60%of G

Step 3: Translation of the constraint into algebraic expression,

Based on= the attribute table since only R, S and T are beef,

XRG + XSG + XTG > .6(XRG + XSG + XTG + XPG + XLG)

Step 4: Algebraic Processing of the constraint.

(The purpose of this step is collecting like terms of the variables on the left hand side of the inequality, with the free coefficient place on the other side of the inequality.)

.4XRG + = .4XSG + .4XTG - .6XPG - .6XLG > 0

Step 5: Expressing the cons= traint in a standardized format.

.4XRG +.4XSG +.4XTG -.6XPG -.6XLG +0XR=
I +0XSI +0XTI +0XPI +0XLI > 0

This constraint appears as constraint (3) in the std.

format<= /span> below.

&nbs= p; &= nbsp;

4) Comme= nts regarding the Std. Format.

One must terminate the solution with the regular standardized format.

Please notice, in multivariate LP (Ch, 9), the non negativity constraints should not be expressed in the standardized format, = as the software is written only for non negative decision variables only.

The Set = Up in Standard Format

1. Decision Variables:

Xij amount in pounds of Ingredient i in final product j;  i=3D G,I; j=3D R,S,T,P,L.

Hence, the D.V. are:

XRG, XSG, XTG, XPG, XLG, XRI, XSI, XTI, XPI, XLI.=

The U.O.M. for Xij is=
, One pound of ingredient i final product j. i=3D G=
,I; j=3D R,S,T,P,L.

2. Objective

Minimize C =3D 1 XRG +1.5 XSG +1 XTG +.9 XPG +1.2=
 XLG +1 XRI +1.5 XSI +1 XTI +.9XPI + 1.2 XLI.

<=
span
style=3D'font-size:9.5pt'>

C is total do=
llar cost of producing the G and I sausages. U.O.M. is, $1 of total costs of producing =
G and I sausages.

3. Subject to Constraints

(1) 1XRG +1XSG +1XTG +1XPG +1XLG +0XRI +0XSI +0XT=
I +0XPI +0XLI > 500

(2) 0XRG +0XSG +0XTG +0XPG +0XLG +1XRI +1XSI +1XT=
I +1XPI +1XLI > 300

(3) .4XRG +.4XSG +.4XTG -.6XPG -.6XLG +0XRI +0XSI=
 +0XTI +0XPI +0XLI > 0

(4) 0XRG +0XSG +0XTG +0XPG+0XLG +.2XRI +.2XSI +.2=
XTI -.8XPI -.8XLI > 0

(5) -.1XRG -.1XSG -.1XTG -.1XPG +.9XLG +0XRI +0XS=
I +0XTI +0XPI +0XLI < 0

(6) 0XRG +0XSG +0XTG +0XPG +0XLG -.2XRI -.2XSI -.=
2XTI +.8XPI -.2XLI < 0

(7) 1XRG +0XSG +0XTG +0XPG +0XLG +1XRI +0XSI +0XT=
I +0XPI +0XLI < 200

(8) 0XRG +1XSG +0XTG +0XPG +0XLG +0XRI +1XSI +0XT=
I +0XPI +0XLI < 500

(9) 0XRG +0XSG +1XTG +0XPG +0XLG +0XRI +0XSI +1XT=
I +0XPI +0XLI < 700

(10) 0XRG +0XSG +0XTG +1XPG +0XLG +0XRI +0XSI +0X=
TI +1XPI +0XLI < 1000

(11) 0XRG +0XSG +0XTG +0XPG +1XLG +0XRI +0XSI +0X=
TI +0XPI +1XLI < 800

XRG, XSG, XTG, XPG, XLG, XRI, XSI, XTI=
, XPI, XLI > 0

Please see also in this regards, “Handling Mixtu= re Constraints” on my Home Page Site