Modeling III: blending, pooling and quality specifications

Week 4 · 2105623 Optimization of Chemical Processes

Soorathep Kheawhom

31 August 2026

Three lectures

Outline

Linear blending and quality specifications — why a ratio constraint is an LP in disguise, and the three conditions under which that is exactly true
The pooling problem — one shared tank, bilinear terms, and the end of every guarantee linear programming gave you
Local versus global in practice — multistart, relaxation bounds, and when a plant engineer should distrust a converged answer

Session plan, 180 minutes

Today

15 min Opening: HW1 and the Week 3 pack released, the ipopt clinic, today’s objectives
30 min Lecture 1: blending as a linear model, quality specifications, the multiply-through rule
20 min Activity 1 — specification negotiation
30 min Lecture 2: the pooling problem, bilinear terms, p- and q-formulations, nonconvexity
12 min Break
25 min Activity 2 — blend it, in OpenSolver
23 min Lecture 3: local versus global, multistart, when to distrust a converged answer
20 min Activity 3 — the pooling trap
5 min Wrap-up, and what is due when

The designed crossover session: OpenSolver before the break, Python and ipopt after it. Notebook w04-modeling3-blending-pooling.ipynb; workbook w04-blending.xlsx.

By the end you should be able to

Learning objectives

Formulate a blending problem with quality specifications as a linear program, and explain why the specification rows must be written in unnormalized form.
Apply the multiply-through rule, state the three conditions under which it is exact, and recognise where each one fails.
Explain why a pool introduces bilinear terms, and write both the p-formulation and the q-formulation.
Demonstrate that a local NLP solver returns different answers from different starting points, and quantify the penalty of accepting a local solution.
Compare the McCormick relaxations of the two formulations and report the bound each one delivers.

CLO 1 (formulation) and CLO 4 (implementation and solution in Pyomo). Reference: Rao Ch. 1 and Ch. 10; Edgar, Himmelblau and Lasdon Ch. 7; Haverly (1978); Tawarmalani and Sahinidis Ch. 9.

Blending: the most common linear program in the process industries

Where the money actually is

Choose the least costly mixture of purchased component streams that still meets every quality specification of every finished product.

Gasoline and diesel blending the classic case, and still the largest single LP application in refining.
Feed blending animal nutrition, cement raw meal.
Electrolyte formulation salt concentration, water content, conductivity, viscosity.
Cathode precursor blending metal ratios, impurity ceilings.

The one modeling assumption that makes it linear. Properties blend linearly on a mass basis. Good for sulfur and density, acceptable for RVP if a blending index is used, and only rough for octane, where real blending is measurably nonlinear.

The second, usually unstated, assumption

Every component reaches the blender as a separate stream, perfectly segregated up to the blend header.

Lecture 2 removes exactly this assumption, and the model stops being linear.

The instance used all session

Five component streams, two finished grades

Component	RON	RVP kPa	S ppm	cost USD/t	avail t
Reformate	98.0	3.0	110	760	3200
FCC naphtha	92.0	6.5	380	730	4000
Alkylate	95.0	4.5	15	805	2000
Butane	93.0	62.0	5	510	900
Straight-run naphtha	68.0	9.0	520	695	2600

	Regular	Premium
RON minimum	91.0	95.0
RVP maximum, kPa	9.0	8.5
Sulfur maximum, ppm	290	190
Price, USD/t	742	806
Minimum volume, t	4000	2500
Maximum volume, t	7000	5000

The same instance appears in the notebook (Section 1) and in the solution workbook. The two routes must agree to the cent.

The linear blending model

Nine constraint families, and one of them is doing something unusual

Decision variables

x_cg ≥ 0 tonnes of component c sent to grade g · y_g ≥ 0 tonnes of grade g produced

The model

max Σ_g price_g y_g − Σ_c Σ_g cost_c x_cg

s.t. Σ_c x_cg = y_g

Σ_g x_cg ≤ avail_c

d_g^min ≤ y_g ≤ d_g^max

Σ_c ron_c x_cg ≥ ron_g^min y_g

Σ_c rvp_c x_cg ≤ rvp_g^max y_g

Σ_c sul_c x_cg ≤ sul_g^max y_g

objective revenue minus component purchase cost, in USD
grade material balance what goes in comes out
component availability what is on site
contract volume window the commercial envelope
octane floor, RVP ceiling, sulfur ceiling the three specifications, per grade

Look at the last three rows. Every specification row has the decision variable y_g on its right-hand side. That is not an accident, and the next slide explains it.

Quality specifications are ratios, and the ratio is removed by hand

The single manipulation that makes blending an LP

The physical requirement

( Σ_c ron_c x_cg ) ⁄ ( Σ_c x_cg ) ≥ ron_g^min

is a ratio of decision variables, and therefore nonlinear.

The rule

Multiply through by the denominator, which is y_g and is non-negative:

Σ_c ron_c x_cg − ron_g^min y_g ≥ 0

or equivalently Σ_c ( ron_c − ron_g^min ) x_cg ≥ 0

The linear form is not a linearization. It is the same feasible set, intersected with y_g > 0.

Solving the ratio form with ipopt returns the same optimum to within solver tolerance — while the constraint reports a polynomial degree of None instead of 1.

On a spreadsheet. The specification RHS cell contains a changing cell: =C11*C26. That is what “multiply the ratio through by its denominator” looks like on a sheet, and the plain RHS-column convention does not anticipate it.

Three conditions, and each of them fails somewhere

When the rule is exact, and when it quietly is not

Let a, b be data vectors and x the decision vector. Then (aᵀx) ⁄ (bᵀx) ≤ β is equivalent to aᵀx − β bᵀx ≤ 0 provided that bᵀx > 0 at every point of interest.

Sign. Multiplying an inequality by a negative number reverses it. The rule needs bᵀx > 0, not merely bᵀx ≠ 0. In blending the denominator is a total mass and the contract window forces it strictly positive.
Zero denominator. If bᵀx = 0 is attainable the ratio is undefined, while the multiplied-through form reads 0 ≤ 0 and is satisfied trivially. The linear model is then a strict relaxation — it permits a blend of size zero whose quality is meaningless. A grade that may be shut off honestly needs a disjunction, and a binary variable: Week 5.
Same balance. The denominator must be a linear function of the same variables that appear in the numerator. When the denominator belongs to a different balance, as in a pool, the product of two decision variables survives and the model is genuinely nonlinear.

The practical test, one question. Is the denominator the same aggregate that the numerator is normalised by, in the same balance? If yes, the constraint is linear in disguise. If no, expect bilinear terms.

Translation table: worth copying into your notes

Reusable, and the last row is the rest of this session

Engineering requirement	Natural form	Linear equivalent	Condition
average property at most β	Σq_cx_c ⁄ Σx_c ≤ β	Σ(q_c−β)x_c ≤ 0	Σx_c > 0
average property at least β	Σq_cx_c ⁄ Σx_c ≥ β	Σ(q_c−β)x_c ≥ 0	Σx_c > 0
recycle ratio at most r	R ⁄ (R+F) ≤ r	(1−r)R − rF ≤ 0	R+F > 0
conversion at least η	(F_in−F_out) ⁄ F_in ≥ η	F_out ≤ (1−η)F_in	F_in > 0
fixed split fraction θ	S₁ ⁄ (S₁+S₂) = θ	(1−θ)S₁ − θS₂ = 0	S₁+S₂ > 0
purity of a mixed stream at least π	m_key ⁄ m_tot ≥ π	m_key − π m_tot ≥ 0	m_tot > 0
ratio objective	max (cᵀx+c₀) ⁄ (dᵀx+d₀)	Charnes–Cooper LP	dᵀx+d₀ > 0
property of a stream leaving a pool	p = Σq_iF_i ⁄ ΣP_g, and p multiplies P_g	no linear equivalent	denominator is in another balance

A ratio in the objective is a different objective

Charnes–Cooper, and a plant that deliberately makes less

Charnes–Cooper

Maximising (cᵀx+c₀) ⁄ (dᵀx+d₀) over {Ax ≤ b, x ≥ 0} with dᵀx+d₀ > 0 is a linear-fractional program. Substituting

t = 1 ⁄ (dᵀx+d₀) > 0, x′ = t x

gives the equivalent LP max cᵀx′ + c₀t

s.t. Ax′ − bt ≤ 0, dᵀx′ + d₀t = 1, x′, t ≥ 0

and x = x′⁄t recovers the original solution.

On this instance

Total margin (Section 1 LP)	413,439.75 USD
Margin per tonne (CC LP)	45.8708 USD/t
its plan	4,000.00 t Regular
	5,000.00 t Premium
its total margin	412,837.10 USD
margin given up	602.65 USD

Read this carefully. The plant that maximises the ratio deliberately declines the last 80.6 t of Regular, because that increment earns less than the current average. Which objective is correct is a business question, not a mathematical one — but the two must never be confused.

The linear optimum, and why binding specifications are the real output

Quality above specification is quality given away

Optimum: 413,439.75 USD

Component, t	Regular	Premium	slack
Reformate	107.63	3092.37	0
FCC naphtha	2623.63	1376.37	0
Alkylate	833.01	0	1166.99
Butane	206.67	367.83	325.50
Straight-run naphtha	309.67	163.42	2126.91
Volume y_g, t	4080.60	5000.00

This recipe is one of several. The optimum is degenerate: your solver will very likely report different numbers in this table at exactly the same margin. Compare objectives, not recipes.

All six quality rows bind exactly. Regular sits at RON 91.00, RVP 9.00 kPa, 290.00 ppm sulfur; Premium at RON 95.00, RVP 8.50, 190.00 ppm.

That is normal, and it is the point of a blending model: quality above specification is quality given away, so the optimizer gives away nothing.

Marginal values

Availability duals: reformate 41.9412, FCC naphtha 13.3609 USD/t.

Alkylate, butane and straight-run have slack and a dual of zero.

Activity 1 · specification negotiation

18 min · two halves of the room · paper

The refinery blends the five component streams of slide 6 into Regular and Premium. Component costs, availabilities and properties as tabulated. Regular sells at 742, Premium at 806 USD/t. Contractual volumes: Regular 4,000–7,000 t, Premium 2,500–5,000 t. The plan must cover one month.

Brief for the Production half

Your objective is minimum cost, that is maximum blending margin. You are accountable for meeting the contracted volumes from the components actually on site, and for not buying more of the expensive streams than you must.

Write down the constraint set you would insist the model contain. Anything you leave out, you are agreeing to give away.

Brief for the Quality Control half

The product must meet three named specifications per grade. Regular: RON ≥ 91.0, RVP ≤ 9.0 kPa, sulfur ≤ 290 ppm. Premium: RON ≥ 95.0, RVP ≤ 8.5, sulfur ≤ 190 ppm.

The octane floor is a customer contract term: a delivery below it is a rejected cargo. The RVP and sulfur ceilings are internal laboratory targets, set inside the legal limits with a deliberate margin.

Two halves, working in pairs within each half. 10 min separately, then compare and agree one constraint set, explicitly labelling each constraint hard or soft. Hand back the agreed list, one sheet per side, with H or S against every row.

Activity 1 · [ANSWER] · hard and soft is a modeling decision

Nothing in the physics distinguishes the octane floor from the sulfur ceiling

Constraint	H/S
grade material balance	H (physics)
component availability	H (physics)
contract volume window	H (contract)
octane floor	H (contract)
vapour pressure ceiling	S (internal target)
sulfur ceiling	S (internal target)
non-negativity	H

A soft constraint belongs in the objective

Σ_c sul_cx_cg − sul_g^maxy_g ≤ v_g, v_g ≥ 0, objective − π_gv_g

What distinguishes them is who wrote them down, and what happens when they are missed.

What the numbers say once all three are hard

Margin 413,439.75 USD, all six quality rows binding. The value of relaxing each row by one unit — the LP dual of the multiplied-through row, identical for both grades — is

row	USD
octane	0.69276
vapour pressure	5.12983
sulfur	0.13507

So the internal RVP target is by far the most expensive of the three to hold — and it is precisely the one nobody in the room called a contract term.

Real plants do not have that many tanks

One physical fact destroys linearity

Segregated: linear

Quality delivered to X is Σ_i q_ix_iX, a linear function. Multiply the ratio through by Σ_i x_iX and the model is an LP.

Pooled: bilinear

Everything drawn from the tank has the same composition p, whatever it is fed by. p is a decision, and it multiplies flows.

Streams are collected in a common pool. The pool quality satisfies p Σ_gP_g = Σ_iq_iF_i, and the quality delivered to product g is p P_g. Both expressions contain the product of two decision variables.

Why multiplying through does not rescue it

Condition 3 of slide 9, failing

The pool quality is a ratio

p = Σ_i q_iF_i ⁄ Σ_g P_g

Its denominator is the total pool throughput. But the quantity that has to be constrained is the pool’s contribution to the balance of a different stream, namely product g:

p P_g + q_CC_g ≤ spec_g ( P_g + C_g )

Multiplying by Σ_gP_g clears the ratio and leaves p P_g untouched.

The feasible set is no longer convex.

A nonconvex NLP with bilinear terms, and every guarantee that came with linear programming is gone:

no vertex theorem the optimum need not be at a corner
no duality certificate nothing proves the point is best
no reason a converged point is the best one

Where this bites in practice. Refinery gasoline pools, crude tank farms, LPG caverns, hydrogen headers, and any electrolyte make-up tank feeding more than one cell chemistry.

Haverly’s instance, 1978

Seven variables, three bilinear rows, and it has defeated people for fifty years

Stream	route	S %	USD
Feed A	into the pool	3.0	6
Feed B	into the pool	1.0	16
Feed C	bypasses the pool	2.0	10
Product X	≤ 2.5 % S		9
Product Y	≤ 1.5 % S		15

Feeds are costs, products are prices. Demand limits X ≤ 100, Y ≤ 200. Seven variables: A, B into the pool; P_X, P_Y out; C_X, C_Y direct; and the pool sulfur p. The bounds 1 ≤ p ≤ 3 are physical.

The p-formulation

max 9(P_X+C_X) + 15(P_Y+C_Y) − 6A − 16B − 10(C_X+C_Y)

s.t. A + B = P_X + P_Y

3A + B = p (P_X+P_Y)

p P_X + 2C_X ≤ 2.5(P_X+C_X)

p P_Y + 2C_Y ≤ 1.5(P_Y+C_Y)

P_X+C_X ≤ 100, P_Y+C_Y ≤ 200

1 ≤ p ≤ 3, all flows ≥ 0

Three of these rows are bilinear, and each contains a product of two decision variables.

The q-formulation: carry proportions instead of quality

Same feasible set, same global optimum, different relaxation

Change of variables

Let q_i be the fraction of the pool inlet contributed by feed i, so Σ_iq_i = 1 and 0 ≤ q_i ≤ 1. Then F_i = q_i Σ_gP_g, and the amount of feed i reaching product g is q_iP_g.

max 9(P_X+C_X) + 15(P_Y+C_Y) − 6q_A(P_X+P_Y) − 16q_B(P_X+P_Y) − 10(C_X+C_Y)

s.t. q_A + q_B = 1

(3q_A+q_B)P_X + 2C_X ≤ 2.5(P_X+C_X)

(3q_A+q_B)P_Y + 2C_Y ≤ 1.5(P_Y+C_Y)

Every bilinear term is now a product of a bounded proportion and a flow.
The pool mass balance disappears it is implied by the proportions summing to one.
Same feasible set, same global optimum since the pool quality is recovered as the proportion-weighted sulfur.

But they are not the same relaxation. And that, not the algebra, is the whole reason for preferring one over the other. Reparametrizing a nonconvex problem does not make it convex.

Nonconvexity: what a solver can and cannot promise you

“Optimal” is a statement about a point, not about the problem

A termination status of optimal from a local NLP solver is a statement about the KKT conditions at the point it stopped.

It is not, and never was, a statement about global optimality.

Two runs of the same solver on the same model, differing only in the starting values, can converge to two different objective values, both reported optimal.
There is no gap, no bound and no warning in the output to tell them apart.
A spreadsheet is worse, not better Solver silently abandons Simplex LP for GRG Nonlinear and reports “Solver found a solution. All constraints and optimality conditions are satisfied.” That sentence is true and misleading.

What a global solver adds. A spatial branch-and-bound code (couenne, BARON, ANTIGONE) maintains a convex relaxation of the nonconvex problem, so it can report an incumbent and a bound, and therefore a gap. That is the only kind of guarantee available here.

The discipline this forces

Never report a single local solve as the optimum.
Always report the starting point with the answer.
Run a multistart.
Report the relaxation bound alongside the incumbent, so the reader knows how much room is left.

McCormick envelopes: how a bilinear term is relaxed

Turn the nonconvex problem into an LP whose optimum is a valid bound

The construction

Replace a product w = uv with u ∈ [u^L, u^U], v ∈ [v^L, v^U] by a new variable w and four linear inequalities:

w ≥ u^Lv + v^Lu − u^Lv^L

w ≥ u^Uv + v^Uu − u^Uv^U

w ≤ u^Uv + v^Lu − u^Uv^L

w ≤ u^Lv + v^Uu − u^Lv^U

This is exactly the convex hull of the graph of uv over the box, so it is the tightest possible relaxation of one product in isolation.

Reformulation–linearization (RLT)

In the q-formulation the bilinear variables v_ig = q_iP_g inherit the identity Σ_iq_i = 1. Multiplying it by P_g and by the pool throughput limit gives

Σ_i v_ig = P_g ∀g, Σ_g v_ig ≤ Cap · q_i ∀i

These rows are redundant for the nonlinear model but not for its relaxation.

Why this is the practical lever. The bound is what a spatial branch-and-bound tree spends its time closing, so choosing the formulation with the tighter relaxation is worth more than any solver setting.

Break

12 minutes.

Next: open w04-blending.xlsx with OpenSolver loaded, and check that ipopt is on your path. Activity 2 starts on the spreadsheet. Activity 3, after Lecture 3, is the reason this course leaves the spreadsheet behind.

Activity 2 · blend it, in OpenSolver

25 min · computer · individual

Part 1 — reproduce the optimum and read off what binds

Open w04-blending.xlsx, sheet Model. Complete the yellow cells: the availability VALUE column E19:E23, the constraint VALUE column E33:E44 — every quality row in multiplied-through form, so its RHS is 0 — and the objective C29.

OpenSolver with CBC: objective $C$29 set to Max, changing cells $C$19:$D$24, constraints as listed on the Readme sheet, and tick Get sensitivity report.

The objective cell must read 413,439.75 USD.

Then write down, from the check block in rows 46–51, which specifications bind.

Part 2 — predict before you solve

Change one specification by a small amount — for example tighten the Regular sulfur ceiling from 290 to 250 ppm.

Before re-solving, write on paper: will the margin rise or fall, and roughly by how much? Use the sensitivity report from Part 1 to justify the number. Then re-solve and compare.

Your recipe will probably not match slide 12, and that is not an error. The optimum is degenerate: several recipes achieve the same margin. Check the objective and the binding set, not the recipe.

Individual. Hand back nothing — the instructor walks the room. Anyone whose objective differs has a formula error.

Activity 2 · [ANSWER] · the binding set is the interesting output

And the linear prediction is not a bound in the direction you expect

Part 1

Objective C29 = 413,439.75 USD, matching the Pyomo model to the cent (413,439.749277).

Binding: all six quality rows; the Premium volume ceiling at 5,000 t; and the availability rows for reformate (3,200 t) and FCC naphtha (4,000 t).

Not binding: the Regular volume window — it settles at 4,080.60 t, above its floor and below its ceiling — and the availability of alkylate, butane and straight-run naphtha.

The optimum is degenerate: several different recipes achieve the same margin, and a different solver reports a different one. The binding set and the objective are stable; the recipe is not.

Part 2 — the prediction rule, and its direction

Predicted change = (dual of the row) × (change in the row). Tightening a ceiling that binds can only reduce the margin, so the sign is known before any arithmetic.

Regular S ceiling	actual	linear prediction
290 → 250 ppm	−27,445	−22,047
290 → 200 ppm	−63,231	−49,606
290 → 150 ppm	−110,056	−77,164

The linear prediction under-states the loss, and always will. The LP value function is concave in the right-hand side for a maximisation, so the tangent lies above it. Push the specification far enough and a different set of constraints binds, the dual rises, and the real cost runs away from the prediction.

What a nonconvex landscape looks like

And why a gradient method cannot see across it

Schematic. Every marked point satisfies the KKT conditions, and a local solver reports optimal at each.

The mechanism

A gradient-based method climbs the hill it starts on. It has no information about the landscape beyond the neighbourhood it has visited, and no mechanism for crossing a valley of infeasible or worse points.

The starting point selects the answer.

On a nonconvex problem the initial values are not a numerical detail — they are part of the specification of what you computed. A result reported without them is not reproducible.

Multistart: the poor person’s global search

Evidence, and a good incumbent. Not a proof.

The procedure

Sample N starting points over the physically meaningful ranges of the variables that drive the nonconvexity.
Solve from each, and keep every terminal point, not just the best.
Report the histogram of terminal objective values, the best, the worst, and the fraction of starts reaching the best.

It gives no bound, so it cannot prove optimality. A multistart that always returns the same value has found either a convex problem or a badly chosen sampling range.

Practical advice

Prefer a physical starting point Initialising a pool quality from the actual current plant operation is worth a hundred random draws.
Sample the quality and split variables Sampling flows alone often lands in the same basin every time.
Fix the random seed and record it Otherwise the run is not reproducible.
If the problem is small use a global solver and stop guessing.

When a plant engineer should distrust a converged answer

Six warning signs

A product of two decision variables appears anywhere in the model, including in a spreadsheet cell.
A quality, composition, temperature or split fraction is itself a variable.
A recycle or a shared tank connects two balances.
The solver engine changed from linear to nonlinear without you asking.
The answer moves when you change only the initial values.
The solution is much better than anything operations has ever achieved.

A blend recipe from a local optimum is not infeasible. It meets every specification.

It is wrong in the sense of leaving margin on the table, silently, every day, with no signal that it is doing so.

And for electrolyte blending: the same structure appears whenever one make-up tank feeds more than one cell chemistry. The tank composition is a variable, and it multiplies every draw from the tank.

Activity 3 · the pooling trap

15 min · computer · groups of 3 to 4

What to do

Open w04-modeling3-blending-pooling.ipynb and run everything up to and including the Haverly section. The model builder is build_haverly(p0=...), where p0 is the starting value of the pool sulfur p, and the solver is ipopt.

Solve the same model twice: once from p₀ = 1.0, once from p₀ = 3.0.

Nothing else changes. Same data, same constraints, same solver, same tolerances.

Record, for each of the two runs

the solver’s termination status,
the objective value,
the final pool quality p,
the six flows A, B, P_X, P_Y, C_X, C_Y.

Do not compare with the group next to you until the poll. When both columns are on the board, the question to the room is simply: which one is right?

Groups of 3 to 4, one machine per group. Hand back nothing written — each group reports its two numbers out loud when polled.

Activity 3 · [ANSWER] 1 of 2 · both are right, and that is the problem

Same model, same solver, same tolerances

	start p₀ = 1.0	start p₀ = 3.0
termination	`optimal`	`optimal`
profit, USD	400.00	100.00
final p, % S	1.000	3.000
A	0	50
B	100	0
P_X	0	50
P_Y	100	0
C_X	0	50
C_Y	100	0

300 USD out of 400 — 75 percent of the achievable margin, on a problem with seven variables. Both points genuinely satisfy the KKT conditions, so both status words are honest.

Fixing p turns the pooling problem back into an ordinary LP, so sweeping p traces the true profile. Two separated hills, and a valley over which no feasible plan makes any money.

Activity 3 · [ANSWER] 2 of 2 · the split was not random

Four of nine starts reach the better solution, and every run reports optimal

starting p₀	profit	final p
1.00	400.00	1.00
1.25	400.00	1.00
1.50	400.00	1.00
1.75	400.00	1.00
2.00	100.00	3.00
2.25	100.00	3.00
2.50	100.00	3.00
2.75	100.00	3.00
3.00	100.00	3.00

On the 81-point sweep the profit is exactly zero for p from 1.525 to 2.400.

That flat valley is the boundary between the two basins of attraction, and it is why no gradient step crosses it.

The left hill is reached by feeding the pool with the low-sulfur stream B and selling the high-value product Y.
The right hill is reached by feeding the pool with the cheap high-sulfur stream A and selling only the low-value product X.

The closing point of the session. A spreadsheet solver would have returned one of these two numbers and never mentioned the other. No message, no bound, no gap. That is why the course moves to Python from here — and the reason is not size. It is that a spreadsheet cannot tell you its answer is local.

Closing the loop: what a bound would have told the room

“Profit 400” is an opinion. Here is what a result looks like.

Instance	form	global	bound	gap
Haverly, 2 feeds, no pool limit	p	400	500.00	25.0 %
Haverly, 2 feeds, no pool limit	q	400	500.00	25.0 %
extended, 4 feeds, pool ≤ 150	p	400	555.56	38.9 %
extended, 4 feeds, pool ≤ 150	q	400	466.67	16.7 %

On Haverly the two bounds coincide a single pool fed by two streams and limited only by product demand leaves the RLT constraints nothing to add.
On the extended instance the q-formulation cuts the gap from 38.9 to 16.7 percent, for the cost of extra rows. That is why global solvers build the pq-formulation internally.

The honest report.

“Profit 400” is an opinion.

“Profit 400, from p₀ = 1.0, with a valid upper bound of 500 from the McCormick relaxation, and a multistart over nine starts in which four reached 400 and five reached 100” is a result.

Forward

Week 10 supplies the KKT theory behind the word optimal. Week 11 supplies the branch-and-bound machinery that closes a gap. Week 13 returns to what to do when the model is not known at all.

Summary

What today established

A quality specification is naturally a ratio Multiplying by the blend volume, which belongs to the same balance, makes it linear. That single manipulation is why blending is an LP.
The rule is exact only when the denominator is strictly positive and lives in the same balance as the numerator.
A ratio objective is a different objective Charnes-Cooper turns it into an LP, and its plan is generally not the plan that maximises the total.
A pool forces every product drawn from it to see the same composition, the pool quality multiplies flows belonging to other balances, and the feasible set stops being convex.
Optimal from a local NLP solver asserts the KKT conditions, not global optimality.
The p- and q-formulations share a feasible set and a global optimum but not a relaxation. The bound is what a global solver works to close.
Report the bound with the incumbent An answer to a nonconvex problem without a bound is an opinion.

Numbers established today

Linear blending margin	413,439.75 USD
Regular / Premium volume	4,080.60 / 5,000.00 t
Binding quality rows	all six
Availability duals, reformate / FCC	41.9412 / 13.3609
Spec duals, RON / RVP / S	0.69276 / 5.12983 / 0.13507
Margin per tonne (Charnes-Cooper)	45.8708 USD/t
margin given up	602.65 USD
Haverly from p₀ = 1.0	400.00 USD, p = 1
Haverly from p₀ = 3.0	100.00 USD, p = 3
Multistart, 9 starts	4 reach 400, 5 reach 100
Flat zero valley	p ∈ [1.525, 2.400]
McCormick, Haverly p / q	500.00 / 500.00
McCormick, extended p / q	555.56 / 466.67

Next: Week 5 — logical and discrete decisions, big-M versus convex hull, and formulation strength.

Before next week

Reading, notebook, and HW2

Reading

Edgar, Himmelblau and Lasdon Ch. 7, blending and refinery planning models.
Rao Ch. 1 for problem classification and Ch. 10 for the nonlinear background.
Haverly (1978) the original two-page note that named the pooling problem.
Tawarmalani and Sahinidis Convexification and Global Optimization, Ch. 9, for the p- and q-formulations.

Notebook to finish

w04-modeling3-blending-pooling.ipynb, all sections, plus Exercises 2, 4 and 6. Exercise 4 is the proper 100-point multistart on Haverly and is the one that settles the argument started today.

HW1 is released today, due Monday 7 September — the start of Week 5.

The Week 3 self-study pack is released today too, due Friday 11 September, midnight — the session I was away for. The files have been on the course page without a deadline; the deadline starts now.

HW2 is not released today. It comes in Week 5, so you meet these one at a time.