Download Marginal analysis in decision

DECISION MAKING WITH MARGINAL ANALYSIS
When we have a large number of decision alternatives and states of nature, we have recourse to
marginal analysis to obtain the best decision without using payoff tables. Marginal analysis is a decisionmaking approach that helps select the optimal inventory level. With a manageable number of decision
alternatives and states of nature where the probability of each state of nature is known, we use marginal
analysis with discrete distributions. For cases in which there are a very large number of possible decision
alternatives and states of nature and the probability distribution can be described with a normal
distribution, we make use of marginal analysis with the normal distribution.
Marginal analysis with discrete distributions
Given any inventory level, it is natural to add an additional unit if it is found that its expected
marginal profit equals, or exceeds, its expected marginal loss.
If we let p be the probability that the demand will be greater than or equal to a given supply, the
expected marginal profit (MP) may be easily found by multiplying p by the marginal profit obtained for
every unit sold. Similarly, the expected marginal loss is determined by multiplying the marginal loss
(ML) by (1 – p).
The optimal decision rule is p.MP ≥ (1 − p ).ML . This inequality may be simplified to
p≥
ML
MP + ML
The above inequality implies that, as long as the value of p exceeds the ratio on the right-hand
side, any additional unit will be kept in stock.
Example
Joseph has just opened a new bakery called Morning Fresh. In performing an economic analysis, Joseph
has determined that the marginal loss for each dozen of breadrolls that is sold, is R4.00. The marginal
profit is estimated to be R2.75 per dozen. Joseph is considering stocking 10, 15, 20, 25 or 30 dozen
breadrolls. The chance of selling 10 dozen rolls per day is 10%. The chance of selling 15 dozen rolls per
day is 20%. There is a 30% chance that Morning Fresh will sell either 20 or 25 dozen rolls per day.
Finally, there is a 10% of selling 30 dozen breadrolls per day, which Joseph considers to be the most that
Morning Fresh would be able to accommodate. What would your recommendation be?
Solution
Marginal loss per dozen = 4.00
Marginal profit per dozen = 2.75
The probability that the demand will be greater or equal to a given supply, p, is given by
4
p≥
⇒ p ≥ 0.5926
4 + 2.75
The probability distribution for Morning Fresh is given in the table below.
Daily sales
(dozens of breadrolls)
30
25
20
15
10
Probability that sales will
be at this level
0.10
0.30
0.30
0.20
0.10
Probability that sales will
at this level or greater
0.10
0.40
0.70
0.90
1.00
Morning Fresh should therefore keep an optimal quantity of 20 dozens of breadrolls per day in stock since
this is the first value that exceeds 0.5926. (The probability of selling more than 25 dozens is 0.70, which
exceeds 0.5926.)
Marginal analysis with the normal distribution
In most business environment, the product demand or sales follow a normal distribution so that
marginal analysis with a normal distribution can be applied. We need to know a priori the
1.
Mean sales for the product ( µ )
2.
Standard deviation of sales ( σ )
3.
Marginal profit for the product (MP)
4.
Marginal loss for the product (ML)
When the above values are known, the procedure for finding the best stocking policy is similar to
marginal analysis with discrete distributions. First, the value of p is determined from the equation
p=
ML
MP + ML
The area equal to p is then shaded, starting from the upper tail of the normal distribution. The
corresponding z-value is then read from the standard normal table. By using the standardisation formula
z=
we then solve for x, the optimal stocking level.
x−µ
,
σ
Example
Paula produces a weekly stock exchange report, which she sells to investors. She normally sells
3000 per week, and, 70% of the time, her sales range between 2990 and 3010 reports per week. The
report costs Paula R15 per copy to produce and she can sell it for R350 per copy. Any reports not sold at
the end of the week have no value. How many reports should Paula produce per week?
Solution
Marginal profit = R350 – R15 = R335
Marginal loss = R15
Mean number of reports sold = 3000
Given that P[2990 < X < 3010] = 0.70 (X = number of reports), we can find the standard deviation of X
since it is easy to deduce that P[3000 < X < 3010] = P[2990 < X < 3000] = 0.35, the interval [2990, 3010]
being centred.
At 3010, z =
3010 − 3000
= 1.037 ⇒ σ = 9.6432 .
σ
Using marginal analysis with the normal distribution, the value of p is
15
p=
= 0.0429
335 + 15
From the standard normal distribution, the corresponding z-value = 1.718.
Therefore, if P[X > b] = 0.0429, then
b − 3000
= 1.718 ⇒ b = 3011.35 . The optimal stock is 3011.
9.6432